Algebra/Trig Review - Pauls Online Math Notes - Lamar University

Algebra/Trig ReviewIntroductionThis review was originally written for my Calculus I class but it should be accessible toanyone needing a review in some basic algebra and trig topics. The review contains theoccasional comment about how a topic will/can be used in a calculus class. If you aren’tin a calculus class you can ignore these comments. I don’t cover all the topics that youwould see in a typical Algebra or Trig class, I’ve mostly covered those that I feel wouldbe most useful for a student in a Calculus class although I have included a couple that arenot really required for a Calculus class. These extra topics were included simply becausethe do come up on occasion and I felt like including them. There are also, in alllikelihood, a few Algebra/Trig topics that do arise occasionally in a Calculus class that Ididn’t include.Because this review was originally written for my Calculus students to use as a test oftheir algebra and/or trig skills it is generally in the form of a problem set. The solution tothe first problem in a set contains detailed information on how to solve that particulartype of problem. The remaining solutions are also fairly detailed and may contain furtherrequired information that wasn’t given in the first problem, but they probably won’tcontain explicit instructions or reasons for performing a certain step in the solutionprocess. It was my intention in writing the solutions to make them detailed enough thatsomeone needing to learn a particular topic should be able to pick the topic up from thesolutions to the problems. I hope that I’ve accomplished this.So, why did I even bother to write this?The ability to do basic algebra is absolutely vital to successfully passing a calculus class.As you progress through a calculus class you will see that almost every calculus probleminvolves a fair amount of algebra. In fact, in many calculus problems, 90% or more ofthe problem is algebra.So, while you may understand the basic calculus concepts, if you can’t do the algebra youwon’t be able to do the problems. If you can’t do the problems you will find it verydifficult to pass the course.Likewise you will find that many topics in a calculus class require you to be able to basictrigonometry. In quite a few problems you will be asked to work with trig functions,evaluate trig functions and solve trig equations. Without the ability to do basic trig youwill have a hard time doing these problems.Good algebra and trig skills will also be required in Calculus II or Calculus III. So, if youdon’t have good algebra or trig skills you will find it very difficult to complete thissequence of courses.© 2006 Paul Dawkins 1http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewMost of the following set of problems illustrates the kinds of algebra and trig skills thatyou will need in order to successfully complete any calculus course here at LamarUniversity. The algebra and trig in these problems fall into three categories :• Easier than the typical calculus problem,• similar to a typical calculus problem, and• harder than a typical calculus problem.Which category each problem falls into will depend on the instructor you have. In mycalculus course you will find that most of these problems falling into the first twocategories.Depending on your instructor, the last few sections (Inverse Trig Functions throughSolving Logarithm Equations) may be covered to one degree or another in your class.However, even if your instructor does cover this material you will find it useful to havegone over these sections. In my course I spend the first couple of days covering thebasics of exponential and logarithm functions since I tend to use them on a regular basis.This problem set is not designed to discourage you, but instead to make sure you have thebackground that is required in order to pass this course. If you have trouble with thematerial on this worksheet (especially the Exponents - Solving Trig Equations sections)you will find that you will also have a great deal of trouble passing a calculus course.Please be aware that this problem set is NOT designed to be a substitute for an algebra ortrig course. As I have already mentioned I do not cover all the topics that are typicallycovered in an Algebra or Trig course. The most of the topics covered here are those thatI feel are important topics that you MUST have in order to successfully complete acalculus course (in particular my Calculus course). You may find that there are otheralgebra or trig skills that are also required for you to be successful in this course that arenot covered in this review. You may also find that your instructor will not require all theskills that are listed here on this review.Here is a brief listing and quick explanation of each topic covered in this review.AlgebraExponents – A brief review of the basic exponent properties.Absolute Value – A couple of quick problems to remind you of how absolutevalue works.Radicals – A review of radicals and some of their properties.Rationalizing – A review of a topic that doesn’t always get covered all that wellin an algebra class, but is required occasionally in a Calculus class.Functions – Function notation and function evaluation.Multiplying Polynomials – A couple of polynomial multiplication problemsillustrating common mistakes in a Calculus class.Factoring – Some basic factoring.© 2006 Paul Dawkins 2http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSimplifying Rational Expressions – The ability to simplify rational expressionscan be vital in some Calculus problems.Graphing and Common Graphs – Here are some common functions and how tograph them. The functions include parabolas, circles, ellipses, and hyperbolas.Solving Equations, Part I – Solving single variable equations, including thequadratic formula.Solving Equations, Part II – Solving multiple variable equations.Solving Systems of Equations – Solving systems of equations and someinterpretations of the solution.Solving Inequalities – Solving polynomial and rational inequalities.Absolute Value Equations and Inequalities – Solving equations and inequalitiesthat involve absolute value.TrigonometryTrig Function Evaluation – How to use the unit circle to find the value of trigfunctions at some basic angles.Graphs of Trig Functions – The graphs of the trig functions and some niceproperties that can be seen from the graphs.Trig Formulas – Some important trig formulas that you will find useful in aCalculus course.Solving Trig Equations – Techniques for solving equations involving trigfunctions.Inverse Trig Functions – The basics of inverse trig functions.Exponentials / LogarithmsBasic Exponential Functions – Exponential functions, evaluation of exponentialfunctions and some basic properties.Basic Logarithm Functions – Logarithm functions, evaluation of logarithms.Logarithm Properties – These are important enough to merit their own section.Simplifying Logarithms – The basics for simplifying logarithms.Solving Exponential Equations – Techniques for solving equations containingexponential functions.Solving Logarithm Equations – Techniques for solving equations containinglogarithm functions.AlgebraExponentsSimplify each of the following as much as possible.1.1 34 −3 −19 3 42x y x+ yy −© 2006 Paul Dawkins 3http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSolutionAll of these problems make use of one or more of the following properties.nn m n+ m p n−m1p p = p = p =mm−np pn( )mnm0p p p p( pq)p−n= = 1, provided ≠ 0n n n⎛ p⎞p= p q⎜ ⎟ =⎝ q ⎠ q1 1 n= = pn−npp−n⎛ p⎞ ⎛ q ⎞ q⎜ ⎟ = ⎜ ⎟ =⎝ q ⎠ ⎝ p⎠pThis particular problem only uses the first property.nnn1 3 1 35−4 3 19 3 4 4 19 3 3 −−− − − − 4 −15 −3122x y x + y y = 2x y + y = 2x y + yRemember that the y’s in the last two terms can’t be combined! You can onlycombine terms that are products or quotients. Also, while this would be an acceptableand often preferable answer in a calculus class an algebra class would probably wantyou to get rid of the negative exponents as well. In this case your answer would be.1 3 5−−4 −3 −19 3 4 −15 −32 1122xy x + yy = 2x y + y = +15 3 5x y12yThe 2 will stay in the numerator of the first term because it doesn’t have a negativeexponent.nnn2.3 15 2 2xxx −Solution3 1 3 1 6 20 5 21− + 2− + −5 2 2 5 2 10 10 10 10xxx = x = x = xNot much to this solution other than just adding the exponents.3.1−3xx2x5Solution1 2 2 13−−3 3 3 −53xx x x x x 1= = = = x5 52x2x2 2 2Note that you could also have done the following (probably is easier….).13−3© 2006 Paul Dawkins 4http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review1 1 1 13− − − −3 3 3 −43xx x x x x 1= = = = x5 42x2x2 2 2In the second case I first canceled an x before doing any simplification.In both cases the 2 stays in the denominator. Had I wanted the 2 to come up to thenumerator with the x I would have used ( ) 52x in the denominator. So watchparenthesis!13−34.⎛⎜2x x y⎜ x+y⎝4−2 5 6⎞⎟⎟⎠−3SolutionThere are a couple of ways to proceed with this problem. I’m going to first simplifythe inside of the parenthesis a little. At the same time I’m going to use the lastproperty above to get rid of the minus sign on the whole thing.4−3 3⎛−2 5 6⎞ ⎛ ⎞⎜2x x y ⎟ x+y= ⎜ ⎟6x y−⎜ + ⎟ ⎜ 5 62xy⎟⎝ ⎠ ⎝ ⎠Now bring the exponent in. Remember that every term (including the 2) needs to getthe exponent.4−3⎛−2 6⎞3 35⎜2x x y ⎟ ( x+ y)( x+y)= =63 18x y−⎜ + ⎟−3⎛ ⎞63 5 185⎝ ⎠ 2 ⎜x⎟ ( y )8xy⎝ ⎠Recall that ( ) 3 3 3x+ y ≠ x + y so you can’t go any further with this.5.⎛⎜ xxx − xx x⎜ x + 1⎝4 9 101−7 2 3 2 −920⎞⎟⎟⎠SolutionDon’t make this one harder than it has to be. Note that the whole thing is raised to thezero power so there is only one property that needs to be used here.4 9 1010⎛ −7 2 3 2 −9⎞2⎜ xxx − xx x ⎟ = 1x + 1⎜⎟⎝⎠© 2006 Paul Dawkins 5http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewAbsolute Value1. Evaluate 5 and − 123SolutionTo do these evaluations we need to remember the definition of absolute value.⎧ p if p≥0p = ⎨⎩ − p if p < 0With this definition the evaluations are easy.5 = 5 because 5 ≥ 0( )− 123 =− − 123 = 123 because -123 < 0Remember that absolute value takes any nonzero number and makes sure that it’spositive.2. Eliminate the absolute value bars from 3−8xSolutionThis one is a little different from the first example. We first need to address a verycommon mistake with these.3−8x≠ 3+8xAbsolute value doesn’t just change all minus signs to plus signs. Remember thatabsolute value takes a number and makes sure that it’s positive or zero. To convinceyourself of this try plugging in a number, say x = − 1083 = 83 = 3+ 80 = 3−8( −10) ≠ 3 + 8( − 10)= 3 − 80 =− 77There are two things wrong with this. First, is the fact that the two numbers aren’teven close to being the same so clearly it can’t be correct. Also note that if absolutevalue is supposed to make nonzero numbers positive how can it be that we got a -77of out of it? Either one of these should show you that this isn’t correct, but togetherthey show real problems with doing this, so don’t do it!That doesn’t mean that we can’t eliminate the absolute value bars however. We justneed to figure out what values of x will give positive numbers and what values of xwill give negative numbers. Once we know this we can eliminate the absolute valuebars. First notice the following (you do remember how to Solve Inequalities right?)33−8x≥0 ⇒ 3≥8x ⇒ x≤833− 8x< 0 ⇒ 3< 8x ⇒ x>833So, if x ≤ then 3−8x≥ 0 and if x > then 3− 8x< 0. With this information we88can now eliminate the absolute value bars.© 2006 Paul Dawkins 6http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review⎧33 −8xif x≤⎪83− 8x=⎨⎪− 3( 3 − 8x)if x >⎪⎩8Or,⎧33 −8xif x≤⎪83− 8x=⎨⎪− 33 + 8xif x >⎪⎩8So, we can still eliminate the absolute value bars but we end up with two differentformulas and the formula that we will use will depend upon what value of x thatwe’ve got.On occasion you will be asked to do this kind of thing in a calculus class so it’simportant that you can do this when the time comes around.3. List as many of the properties of absolute value as you can.SolutionHere are a couple of basic properties of absolute value.p ≥0− p = pThese should make some sense. The first is simply restating the results of thedefinition of absolute value. In other words, absolute value makes sure the result ispositive or zero (if p = 0). The second is also a result of the definition. Since takingabsolute value results in a positive quantity or zero it won’t matter if there is a minussign in there or not.We can use absolute value with products and quotients as followsa aab = a b=b bNotice that I didn’t include sums (or differences) here. That is because in generala+ b ≠ a + bTo convince yourself of this consider the following example7 =− 7 = 2 − 9 = 2 + ( −9)≠ 2 +− 9 = 2 + 9 = 11Clearly the two aren’t equal. This does lead to something that is often called thetriangle inequality. The triangle inequality isa+ b ≤ a + bThe triangle inequality isn’t used all that often in a Calculus course, but it’s a niceproperty of absolute value so I thought I’d include it.© 2006 Paul Dawkins 7http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewRadicalsEvaluate the following.1. 3125SolutionIn order to evaluate radicals all that you need to remember isnny = x is equivalent to x=yIn other words, when evaluating n x we are looking for the value, y, that we raise tothe n to get x. So, for this problem we’ve got the following.3 3125 = 5 because 5 = 1252. 6643.5Solution664 = 2 because62 = 64− 243Solution5− 243 =−3 because − 3 =− 2434. 2100( ) 5Solution2100 = 100 = 102Remember that x = x45. − 16Solution4− 16 = n/aTechnically, the answer to this problem is a complex number, but in most calculusclasses, including mine, complex numbers are not dealt with. There is also the factthat it’s beyond the scope of this review to go into the details of getting a complexanswer to this problem.Convert each of the following to exponential form.6. 7xSolution© 2006 Paul Dawkins 8http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewTo convert radicals to exponential form you need to remember the following formula1np p n =For this problem we’ve got.27 = 7 = 7( ) 12x x xThere are a couple of things to note with this one. Remember p =2p and noticethe parenthesis. These are required since both the 7 and the x was under the radical soboth must also be raised to the power. The biggest mistake made here is to convertthis as127xhowever this is incorrect because127x= 7 xAgain, be careful with the parenthesis7.5 2xSolution1 25 2 2 5 5( )x = x = xNote that I combined exponents here. You will always want to do this.8. 34x + 8Solution3( x ) 134x+ 8 = 4 + 81 13 3You CANNOT simplify further to ( 4 ) ( 8)( ) n n na+ b ≠ a + b !!!!Simplify each of the following.6 139. 316x y Assume that x ≥ 0 and y ≥ 0 for this problem.x + so don’t do that!!!! Remember thatSolutionThe property to use here isA similar property for quotients isn nxy = xnynx=ynnxy© 2006 Paul Dawkins 9http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewBoth of these properties require that at least one of the following is true x ≥ 0 and/ory ≥ 0 . To see why this is the case consider the following example2( )( ) ( i)( i) i4 = 16 = −4 −4 ≠ −4 − 4 = 2 2 = 4 =− 4If we try to use the property when both are negative numbers we get an incorrectanswer. If you don’t know or recall complex numbers you can ignore this example.The property will hold if one is negative and the other is positive, but you can’t haveboth negative.I’ll also need the following property for this problem.n nx = x provided n is oddIn the next example I’ll deal with n even.Now, on to the solution to this example. I’ll first rewrite the stuff under the radical alittle then use both of the properties that I’ve given here.16xy = 8xx yyyy 2y3 6 13 3 3 3 3 3 3 3===8 x x y y y y 2y3 3 3 3 3 3 3 3 3 3 3 3 3 3xxyyyy232xy2 42 32ySo, all that I did was break up everything into terms that are perfect cubes and termsthat weren’t perfect cubes. I then used the property that allowed me to break up aproduct under the radical. Once this was done I simplified each perfect cube and dida little combining.y10.4 8 1516x ySolutionI did not include the restriction that x ≥ 0 and y ≥ 0 in this problem so we’re going tohave to be a little more careful here. To do this problem we will need the followingproperty.n nx = x provided n is evenTo see why the absolute values are required consider 4 . When evaluating this weare really asking what number did we square to get four? The problem is there are infact two answer to this : 2 and -2! When evaluating square roots (or any even rootfor that matter) we want a predicable answer. We don’t want to have to sit down eachand every time and decide whether we want the positive or negative number.Therefore, by putting the absolute value bars on the x we will guarantee that theanswer is always positive and hence predictable.© 2006 Paul Dawkins 10http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSo, what do we do if we know that we want the negative number? Simple. We add aminus sign in front of the square root as follows− 4 =− ( 2)=− 2This gives the negative number that we wanted and doesn’t violate the rule thatsquare root always return the positive number!Okay, let’s finally do this problem. For the most part it works the same as theprevious one did, we just have to be careful with the absolute value bars.4 8 15 4 4 4 4 4 4 316xy= 16xx yyy y= 2x x y y y4y3= 22 34 3x y y2 34 3= 2x y yNote that I could drop the absolute value on the2x term because the power of 2 will2give a positive answer for x regardless of the sign of x. They do need to stay on they term however because of the power.RationalizingRationalize each of the following.1.3xyx +ySolutionThis is the typical rationalization problem that you will see in an algebra class. Inthese kinds of problems you want to eliminate the square roots from the denominator.To do this we will use2 2( a+ b)( a− b) = a − b .So, to rationalize the denominator (in this case, as opposed to the next problem) wewill multiply the numerator and denominator by x − y. Remember, that torationalize we simply multiply numerator and denominator by the term containing theroots with the sign between them changed. So, in this case, we had x + y and sowe needed to change the “+” to a “-”.Now, back to the problem. Here’s the multiplication.3xy( x − y)3xy x − y=x + y x − y x−y( )( )( )© 2006 Paul Dawkins 11http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewNote that the results will often be “messier” than the original expression. However,as you will see in your calculus class there are certain problems that can only beeasily worked if the problem has first been rationalized.Unfortunately, sometimes you have to make the problem more complicated in orderto work with it.t + 2−22.2t − 4SolutionIn this problem we’re going to rationalize the numerator. Do NOT get too locked intoalways rationalizing the denominator. You will need to be able to rationalize thenumerator occasionally in a calculus class. It works in pretty much the same wayhowever.( t+ 2− 2)( t+ 2+ 2)t + 2−4=2 2t − 4 t+ 2+ 2 t − 4 t+ 2+2( )( ) ( )( )==t − 2( t− 2)( t+ 2)( t+ 2+2)1( t+ 2)( t+ 2+2)Notice that, in this case there was some simplification we could do after therationalization. This will happen occasionally.Functions21. Given f ( x) =− x + 6x− 11 and g( x) 4x3(a) f ( 2)(b) g ( 2)(c) f ( − 3)(d) g ( 10)(e) f ( t ) (f) f ( t− 3)(g) f ( x− 3)(h) f ( 4x−1)= − find each of the following.SolutionAll throughout a calculus sequence you will be asked to deal with functions so makesure that you are familiar and comfortable with the notation and can evaluatefunctions.First recall that the f ( x ) in a function is nothing more than a fancy way of writingthe y in an equation so© 2006 Paul Dawkins 12http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review( )2f x =− x + 6x−11is equivalent to writing2y =− x + 6x−11except the function notation form, while messier to write, is much more convenientfor the types of problem you’ll be working in a Calculus class.In this problem we’re asked to evaluate some functions. So, in the first case f ( 2)isasking us to determine the value of2y x x=− + 6 − 11 when x = 2 .The key to remembering how to evaluate functions is to remember that you whateveris in the parenthesis on the left is substituted in for all the x’s on the right side.So, here are the function evaluations.(a) f ( 2) =− ( 2) 2+ 6(2) − 11 =− 3(b) g ( 2)= 4(2) − 3 = 5(c) f ( − 3) =−( − 3) 2+ 6( −3) − 11 =− 38(d) g ( 10)= 4(10) − 3 = 37f t =− t + 6t−11Remember that we substitute for the x’s WHATEVER is in the parenthesis on theleft. Often this will be something other than a number. So, in this case we put t’sin for all the x’s on the left.(e) ( )2This is the same as we did for (a) – (d) except we are now substituting insomething other than a number. Evaluation works the same regardless of whetherwe are substituting a number or something more complicated.2 2(f) ( ) ( ) ( )f t− 3 =− t− 3 + 6 t−3 − 11 =− t + 12t−38Often instead of evaluating functions at numbers or single letters we will havesome fairly complex evaluations so make sure that you can do these kinds ofevaluations.2 2(g) ( ) ( ) ( )f x− 3 =− x− 3 + 6 x−3 − 11 =− x + 12x−38The only difference between this one and the previous one is that I changed the tto an x. Other than that there is absolutely no difference between the two!© 2006 Paul Dawkins 13http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewDo not let the fact that there are x’s in the parenthesis on the left get you workedup! Simply replace all the x’s in the formula on the right side with x − 3 . Thisone works exactly the same as(f).(h) ( ) ( ) ( )2 2f 4x− 1 =− 4x− 1 + 6 4x−1 − 11 =− 16x + 32x−18Do not get excited by problems like (e) – (h). This type of problem works the same as(a) – (d) we just aren’t using numbers! Instead of substituting numbers you aresubstituting letters and/or other functions. So, if you can do (a) – (d) you can dothese more complex function evaluations as well!2. Given f ( x ) = 10 find each of the following.(a) f ( 7)(b) f ( 0)(c) f ( − 14)SolutionThis is one of the simplest functions in the world to evaluate, but for some reasonseems to cause no end of difficulty for students. Recall from the previous problemhow function evaluation works. We replace every x on the right side with what everis in the parenthesis on the left. However, in this case since there are no x’s on theright side (this is probably what causes the problems) we simply get 10 out of each ofthe function evaluations. This kind of function is called a constant function. Just tobe clear here are the function evaluations.( ) ( ) ( )f 7 = 10 f 0 = 10 f − 14 = 1023. Given f ( x) = 3x − x+ 10 and g( x) 1 20⎛ f ⎞⎜ ⎟⎝ g ⎠f g x(a) ( f − g)( x)(b) ( x)= − x find each of the following.(c) ( fg )( x )(d) ( f g)( 5)(e) ( )( ) (f) ( gf )( x)SolutionThis problem makes sure you are familiar with notation commonly used withfunctions. The appropriate formulas are included in the answer to each part.(a)( f − g)( x) = f ( x) −g( x)2= 3x − x+ 10 −( 1−20x)2= 3x − x+ 10 − 1+20x2= 3x+ 19x+9(b)© 2006 Paul Dawkins 14http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review⎛ f ⎞⎜ ⎟⎝ g ⎠( x)==f( x)( )g xx − x+1−20x23 10(c)( fg )( x) = f ( x) g ( x)= ( 3x 2 − x+ 10)( 1−20x)=− + − +3 260x 23x 201x10(d) For this problems (and the next two) remember that the little circle , , in thisproblem signifies that we are doing composition NOT multiplication!The basic formula for composition is( f g)( x) = f ( g( x))In other words, you plug the second function listed into the first function listedthen evaluate as appropriate.In this case we’ve got a number instead of an x but it works in exactly the sameway.f g 5 = f g 5( )( ) ( ( ))= f ( −99)= 29512(e) Compare the results of this problem to (c)! Composition is NOT the same asmultiplication so be careful to not confuse the two!f g x = f g x( )( ) ( ( ))= f ( 1−20x)2( x) ( x)= 3 1−20 − 1− 20 + 102( )= 3 1− 40x+ 400x − 1+ 20x+1021200 100 12= x − x+(f) Compare the results of this to (e)! The order in which composition is written isimportant! Make sure you pay attention to the order.gf x = g f x( )( ) ( ( ))2( 3 10)2( x x )= g x − x+= 1−20 3 − + 10=− x + x−260 20 199© 2006 Paul Dawkins 15http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewMultiplying PolynomialsMultiply each of the following.1. ( 7x− 4)( 7x+4)SolutionMost people remember learning the FOIL method of multiplying polynomials froman Algebra class. I’m not very fond of the FOIL method for the simple reason that itonly works when you are multiplying two polynomials each of which has exactly twoterms (i.e. you’re multiplying two binomials). If you have more than twopolynomials or either of them has more or less that two terms in it the FOIL methodfails.The FOIL method has its purpose, but you’ve got to remember that it doesn’t alwayswork. The correct way to think about multiplying polynomials is to remember therule that every term in the second polynomial gets multiplied by every term in thefirst polynomial.So, in this case we’ve got.2 2( 7x− 4)( 7x+ 4) = 49x + 28x−28x− 16 = 49x− 16Always remember to simplify the results if possible and combine like terms.This problem was to remind you of the formulaa+ b a− b = a − b2. ( 2x − 5) 2Solution( )( )2 22Remember that 3 = ( 3)( 3)and so( ) ( )( )2 22x− 5 = 2x−5 2x− 5 = 4x − 20x+25This problem is to remind you thatnn( a−b) ≠a − b and ( a+ b)≠ a + bso do not make that mistake!( ) 2 22x−5 ≠4x− 25There are actually a couple of formulas here.2 2 2a + b = a + 2ab + bn n n n( )( )2 2 2a− b = a − 2ab+b© 2006 Paul Dawkins 16http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewYou can memorize these if you’d like, but if you don’t remember them you canalways just FOIL out the two polynomials and be done with it…3. 2( x + 3) 2SolutionBe careful in dealing with the 2 out in front of everything. Remember that order ofoperations tells us that we first need to square things out before multiplying the 2through.2 2 22( x+ 3) = 2( x + 6x+ 9)= 2x + 12x+18Do, do not do the following2 2 22( x+ 3) ≠ ( 2x+ 6)= 4x + 24x+36It is clear that if you multiply the 2 through before squaring the term out you will getvery different answers!There is a simple rule to remember here. You can only distribute a number through aset of parenthesis if there isn’t any exponent on the term in the parenthesis.⎞2x − x ⎜ x + ⎟⎝ x ⎠4. ( )3 ⎛ 2SolutionWhile the second term is not a polynomial you do the multiplication in exactly sameway. The only thing that you’ve got to do is first convert everything to exponentsthen multiply.13 ⎛ 2 ⎞ 3⎛2 −1⎞( 2x − x) ⎜ x + ⎟= ( 2x − x)⎜x + 2x⎟⎝ x ⎠ ⎝ ⎠5. ( 3x+ 2)( x 2 − 9x+12)7 32 2 2= 4x + 2x −x−2SolutionRemember that the FOIL method will not work on this problem. Just multiply everyterm in the second polynomial by every term in the first polynomial and you’ll bedone.3x+ 2 x 2 − 9x+ 12 = x 2 3x+ 2 − 9x 3x+ 2 + 12 3x+2( )( ) ( ) ( ) ( )= − + +3 23x 25x 18x24© 2006 Paul Dawkins 17http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewFactoringFactor each of the following as much as possible.1.2.2100x − 81SolutionWe have a difference of squares and remember not to make the following mistake.2100x−81 ≠( 10x−9) 2This just simply isn’t correct. To convince yourself of this go back to Problems 1 and2 in the Multiplying Polynomials section. Here is the correct answer.2100x − 81 = 10x− 9 10x+92100x + 81( )( )SolutionThis is a sum of squares and a sum of squares can’t be factored, except in rare cases,so this is as factored as it will get. As noted there are some rare cases in which a sumof squares can be factored but you will, in all likelihood never run into one of them.3.23x+ 13x−10SolutionFactoring this kind of polynomial is often called trial and error. It will factor as( ax + b)( cx + d )where ac = 3 and bd = − 10 . So, you find all factors of 3 and all factors of -10 andtry them in different combinations until you get one that works. Once you do enoughof these you’ll get to the point that you can usually get them correct on the first orsecond guess. The only way to get good at these is to just do lot’s of problems.Here’s the answer for this one.23x + 13x− 10 = 3x− 2 x+5( )( )4.225x+ 10x+1SolutionThere’s not a lot to this problem.225x + 10x+ 1= ( 5x+ 1)( 5x+ 1) = ( 5x+1) 2When you run across something that turns out to be a perfect square it’s usually bestwrite it as such.5.4x −8x − 32x5 4 3© 2006 Paul Dawkins 18http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review6.SolutionIn this case don’t forget to always factor out any common factors first before goingany further.5 4 3 3 2 34x −8x − 32x = 4x x −2x− 8 = 4x x− 4 x+23125x − 8( ) ( )( )SolutionRemember the basic formulas for factoring a sum or difference of cubes.3 3 2 2a − b = a− b a + ab+bIn this case we’ve got( )( )( )( )a + b = a + b a − ab + b3 3 2 2( )( )3 2125 8 5 2 25 10 4x − = x− x + x+Simplifying Rational ExpressionsSimplify each of the following rational expressions.1.x22x−82− 4x+4SolutionThere isn’t a lot to these problems. Just factor the numerator and denominator asmuch as possible then cancel all like terms. Recall that you can only cancel termsthat multiply the whole numerator and whole denominator. In other words, you can’t2just cancel the x and you can’t cancel a 4 from the 8 in the numerator and the 4 inthe denominator. Things just don’t work this way.If you need convincing of this consider the following number example.17 8 + 9 4 + 98.5 = = ≠ = 132 2 1Here is the answer to this problem.222x−82( x − 4) 2( x− 2)( x+ 2)2( x+2)= = =2 22x − 4x+ 4 x − 4x+ 4 x − 2 x−2( )2.2x −5x−626x−xSolution© 2006 Paul Dawkins 19http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewIn this one you’ve got to be a little careful. First factor the numerator anddenominator.2x −5x− 6 ( x− 6)( x+1)2 =6x−x x 6−x( )At first glance it doesn’t look like anything will cancel. However remember thata− b=− b−a( )Using this on the term in the denominator gives the following.2x −5x− 6 ( x− 6)( x+ 1)( x− 6)( x+1)x+ 1 x+1= = = = −26x−x x 6−x −x x−6−x x( )Also recall thata −a a= = −−b b bso it doesn’t matter where you put the minus sign.( )Graphing and Common GraphsSketch the graph of each of the following.1.2y =− x+35SolutionThis is a line in the slope intercept formy = mx + bIn this case the line has a y intercept of (0,b) and a slope of m. Recall that slope canbe thought of asrisem =runIf the slope is negative we tend to think of the rise as a fall.The slope allows us to get a second point on the line. Once we have any point on theline and the slope we move right by run and up/down by rise depending on the sign.This will be a second point on the line.2In this case we know (0,3) is a point on the line and the slope is − . So starting at5(0,3) we’ll move 5 to the right (i.e. 0→ 5) and down 2 (i.e. 3→ 1) to get (5,1) as asecond point on the line. Once we’ve got two points on a line all we need to do isplot the two points and connect them with a line.Here’s the sketch for this line.© 2006 Paul Dawkins 20http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review2. ( ) 2y = x+ 3 − 1SolutionThis is a parabola in the formy = ( x− h) 2+ kParabolas in this form will have the vertex at (h, k). So, the vertex for our parabola is(-3,-1). We can also notice that this parabola will open up since the coefficient of thesquared term is positive (and of course it would open down if it was negative).In graphing parabolas it is also convenient to have the x-intercepts, if they exist. Inthis case they will exist since the vertex is below the x-axis and the parabola opensup. To find them all we need to do is solve.2x + 3 − 1=0( )( x )2+ 3 = 1x + 3=±1x=− 4 and x=−2With this information we can plot the main points and sketch in the parabola. Here isthe sketch.© 2006 Paul Dawkins 21http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review3.2y x x=− + 2 + 3SolutionThis is also the graph of a parabola only it is in the more general form.2y = ax + bx + cbIn this form, the x-coordinate of the vertex is x = − and we get the y-coordinate by2aplugging this value back into the equation. So, for our parabola the coordinates of thevertex will be.2x =− = 12 −1( )2( ) ( )y =− 1 + 21+ 3=4So, the vertex for this parabola is (1,4).We can also determine which direction the parabola opens from the sign of a. If a ispositive the parabola opens up and if a is negative the parabola opens down. In ourcase the parabola opens down.This also means that we’ll have x-intercepts on this graph. So, we’ll solve thefollowing.2− x + 2x+ 3=02x −2x− 3=0( x− 3)( x+ 1)= 0So, we will have x-intercepts at x = − 1 and x = 3. Notice that to make my life easier2in the solution process I multiplied everything by -1 to get the coefficient of the xpositive. This made the factoring easier.Here’s a sketch of this parabola.© 2006 Paul Dawkins 22http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewWe could also use completing the square to convert this into the form used in theprevious problem. To do this I’ll first factor out a -1 since it’s always easier to2complete the square if the coefficient of the x is a positive 1.2y =− x + 2x+32( x 2x3)=− − −Now take half the coefficient of the x and square that. Then add and subtract this tothe quantity inside the parenthesis. This will make the first three terms a perfectsquare which can be factored.2y =− x −2x−3( )2( x 2x1 1 3)2( x 2x1 4)=− − + − −=− − + −2(( x 1)4)=− − −The final step is to multiply the minus sign back through.y =− x− 1 + 4f x =− x + 2x+34. ( )2( ) 2SolutionThe only difference between this problem and the previous problem is that we’ve gotf x is just a fancyan f ( x ) on the left instead of a y. However, remember that ( )2way of writing the y and so graphing f ( x) x 2x32y x x=− + 2 + 3.=− + + is identical to graphingSo, see the previous problem for the work. For completeness sake here is the graph.© 2006 Paul Dawkins 23http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review5.2x y y= − 6 + 5SolutionMost people come out of an Algebra class capable of dealing with functions in theform y = f( x). However, many functions that you will have to deal with in aCalculus class are in the form x= f( y)and can only be easily worked with in thatform. So, you need to get used to working with functions in this form.The nice thing about these kinds of function is that if you can deal with functions inthe form y = f( x)then you can deal with functions in the form x= f( y).So,2y = x − 6x+5is a parabola that opens up and has a vertex of (3,-4).Well our function is in the form2x = ay + by + cand this is a parabola that opens left or right depending on the sign of a (right if a isbpositive and left if a is negative). The y-coordinate of the vertex is given by y = −2aand we find the x-coordinate by plugging this into the equation.Our function is a parabola that opens to the right (a is positive) and has a vertex at(-4,3). To graph this we’ll need y-intercepts. We find these just like we found x-intercepts in the previous couple of problems.2y − 6y+ 5=0( y−5)( y− 1)= 0The parabola will have y-intercepts at y = 1 and y = 5 . Here’s a sketch of the graph.© 2006 Paul Dawkins 24http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review26. ( ) 2x+ y+ 5 = 4SolutionThis is a circle in it s standard form( ) ( )2 2 2x− h + y− k = rWhen circles are in this form we can easily identify the center : (h, k) and radius : r.Once we have these we can graph the circle simply by starting at the center andmoving right, left, up and down by r to get the rightmost, leftmost, top most andbottom most points respectively.Our circle has a center at (0, -5) and a radius of 2. Here’s a sketch of this circle.7.2 2x x y y+ 2 + − 8 + 8=0Solution© 2006 Paul Dawkins 25http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewThis is also a circle, but it isn’t in standard form. To graph this we’ll need to do it putit into standard form. That’s easy enough to do however. All we need to do iscomplete the square on the x’s and on the y’s.2 2x + 2x+ y − 8y+ 8=02 2x + 2x+ 1− 1+ y − 8y+ 16 − 16 + 8 = 0( x ) ( y )2 2+ 1 + − 4 = 9So, it looks like the center is (-1, 4) and the radius is 3. Here’s a sketch of the circle.x − 2928. ( ) ( y )+ 4 + 2 = 12SolutionThis is an ellipse. The standard form of the ellipse is( x−h) 2 ( y−k)2+ = 12 2a bThis is an ellipse with center (h, k) and the right most and left most points are adistance of a away from the center and the top most and bottom most points are adistance of b away from the center.The ellipse for this problem has center (2, -2) and has a = 3 andsketch of the ellipse.1b = . Here’s a2© 2006 Paul Dawkins 26http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review2 29. ( x+ 1 ) ( y−2)− = 19 4SolutionThis is a hyperbola. There are actually two standard forms for a hyperbola.2 2− = 12 2Form( x−h) ( y−k)ab( y−h) ( x−k)b2 2− = 12 2aCenter (h, k) (h, k)Opens Opens right and left Opens up and downVerticesSlope of Asymptotesa units right and leftof center.b±ab units up and downfrom center.b±aSo, what does all this mean? First, notice that one of the terms is positive and theother is negative. This will determine which direction the two parts of the hyperbolaopen. If the x term is positive the hyperbola opens left and right. Likewise, if the yterm is positive the hyperbola opens up and down.Both have the same “center”. Hyperbolas don’t really have a center in the sense thatcircles and ellipses have centers. The center is the starting point in graphing ahyperbola. It tells up how to get to the vertices and how to get the asymptotes set up.The asymptotes of a hyperbola are two lines that intersect at the center and have theslopes listed above. As you move farther out from the center the graph will get closerand closer to they asymptotes.© 2006 Paul Dawkins 27http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewFor the equation listed here the hyperbola will open left and right. Its center is(-1, 2). The two vertices are (-4, 2) and (2, 2). The asymptotes will have slopesHere is a sketch of this hyperbola.2± .310. y = xSolutionThere isn’t much to this problem. This is just the square root and it’s a graph that youneed to be able to sketch on occasion so here it is. Just remember that you can’t plugany negative x’s into the square root. Here is the graph.11.y = x3SolutionAnother simple graph that doesn’t really need any discussion, it is here simply tomake sure you can sketch it.© 2006 Paul Dawkins 28http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review12. y = xSolutionThis last sketch is not that difficult if you remember how to evaluate Absolute Valuefunctions. Here is the sketch for this one.Solving Equations, Part ISolve each of the following equations.1.3 2 2x 3x x 21x− = +SolutionTo solve this equation we’ll just get everything on side of the equation, factor thenuse the fact that if ab = 0 then either a = 0 or b = 0 .3 2 2x − 3x = x + 21x−4 − 21 = 03 2x x x2( x )x x−4 − 21 = 0( )( x )x x− 7 + 3 = 0So, the solutions are x = 0 , x = 7 , and x = − 3.© 2006 Paul Dawkins 29http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewRemember that you are being asked to solve this not simplify it! Therefore, makesure that you don’t just cancel an x out of both sides! If you cancel an x out as thiswill cause you to miss x = 0 as one of the solutions! This is one of the more commonmistakes that people make in solving equations.2.23x− 16x+ 1 = 0SolutionIn this case the equation won’t factor so we’ll need to resort to the quadratic formula.Recall that if we have a quadratic in standard form,2ax + bx + c = 0the solution is,2− b ± b −4acx =2aSo, the solution to this equation is2( 16) ( 16) 4( 3)( 1)23 ( )− − ± − −x =16 ± 244=616 ± 2 61=68 ± 61=3Do not forget about the quadratic formula! Many of the problems that you’ll beasked to work in a Calculus class don’t require it to make the work go a little easier,but you will run across it often enough that you’ll need to make sure that you can useit when you need to. In my class I make sure that the occasional problem requiresthis to make sure you don’t get too locked into “nice” answers.3.x2− 8x+ 21 = 0SolutionAgain, we’ll need to use the quadratic formula for this one.8 ± 64 −4( 1)( 21)x =28 ± −20=28±2 5i=2= 4±5i© 2006 Paul Dawkins 30http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewComplex numbers are a reality when solving equations, although we won’t often seethem in a Calculus class, if we see them at all.Solving Equations, Part IISolve each of the following equations for y.1.2y− 5x =6 − 7ySolutionHere all we need to do is get all the y’s on one side, factor a y out and then divide bythe coefficient of the y.2y− 5x =6 − 7yx( 6− 7y)= 2y−56x − 7xy = 2y−56x+ 5= ( 7x+2)y6x+ 5y =7x+ 2Solving equations for one of the variables in it is something that you’ll be doing onoccasion in a Calculus class so make sure that you can do it.3x 3− 5y + sin x = 3xy+ 822. ( )SolutionThis one solves the same way as the previous problem.23x 3− 5y + sin x = 3xy+ 8( )x − x y + x = xy +2 29 15 sin 3 8( )9 + sin − 8 = 3 + 152 2x x x x yy =x + x−3x+15x29 sin 823.2 22x+ 2y= 5SolutionSame thing, just be careful with the last step.© 2006 Paul Dawkins 31http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewDon’t forget the “ ± ” in the solution!2 22 2 5x+ y =2y= 5−2xy2 21= 5 − 22( x )2 25y =± −x22Solving Systems of Equations1. Solve the following system of equations. Interpret the solution.2x− y− 2z=−3x+ 3y+ z =−15x− 4y+ 3z= 10SolutionThere are many possible ways to proceed in the solution process to this problem. Allwill give the same solution and all involve eliminating one of the variables andgetting down to a system of two equations in two unknowns which you can thensolve.Method 1The first solution method involves solving one of the three original equations for oneof the variables. Substitute this into the other two equations. This will yield twoequations in two unknowns that can be solve fairly quickly.For this problem we’ll solve the second equation for x to get.x=−z−3y−1Plugging this into the first and third equation gives the following system of twoequations.2( −z−3y−1)− y− 2z=−35( −z−3y−1)− 4y+ 3z= 10Or, upon simplification−7y− 4z=−1−19y− 2z= 15Multiply the second equation by -2 and add.−7y− 4z=−138y+ 4z=−3031y= −31© 2006 Paul Dawkins 32http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewFrom this we see that y = − 1. Plugging this into either of the above two equationsyields z = 2 . Finally, plugging both of these answers into x=−z−3y− 1 yieldsx = 0 .Method 2In the second method we add multiples of two equations together in such a way toeliminate one of the variables. We’ll do it using two different sets of equationseliminating the same variable in both. This will give a system of two equations intwo unknowns which we can solve.So we’ll start by noticing that if we multiply the second equation by -2 and add it tothe first equation we get.2x− y− 2z=−3−2x−6y− 2z= 2−7y− 4z=−1Next multiply the second equation by -5 and add it to the third equation. This gives−5x−15y− 5z= 55x− 4y+ 3z= 10−19y− 2z= 15This gives the following system of two equations.−7y− 4z=−1−19y− 2z= 15We can now solve this by multiplying the second by -2 and adding−7y− 4z=−138y+ 4z=−3031y= −31From this we get that y = − 1, the same as the first solution method. Plug this intoeither of the two equations involving only y and z and we’ll get that z = 2 . Finallyplug these into any of the original three equations and we’ll get x = 0 .You can use either of the two solution methods. In this case both methods involvedthe same basic level of work. In other cases on may be significantly easier than theother. You’ll need to evaluate each system as you get it to determine which methodwill work the best.InterpretationRecall that one interpretation of the solution to a system of equations is that thesolution(s) are the location(s) where the curves or surfaces (as in this case) intersect.So, the three equations in this system are the equations of planes in 3D space (you’lllearn this in Calculus II if you don’t already know this). So, from our solution weknow that the three planes will intersect at the point (0,-1,2). Below is a graph of thethree planes.© 2006 Paul Dawkins 33http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewIn this graph the red plane is the graph of 2x− y− 2z=− 3, the green plane is thegraph of x+ 3y+ z = 1 and the blue plane is the graph of 5x− 4y+ 3z= 10 . You cansee from this figure that the three planes do appear to intersect at a single point. It is asomewhat hard to see what the exact coordinates of this point. However, if we couldzoom in and get a better graph we would see that the coordinates of this point are(0, − 1, 2) .2. Determine where the following two curves intersect.2 2x + y = 132y = x −1SolutionIn this case we’re looking for where the circle and parabola intersect. Here’s a quickgraph to convince ourselves that they will in fact. This is not a bad idea to do withthis kind of system. It is completely possible that the two curves don’t cross and wewould spend time trying to find a solution that doesn’t exist! If intersection points doexist the graph will also tell us how many we can expect to get.© 2006 Paul Dawkins 34http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSo, now that we know they cross let’s proceed with finding the TWO intersectionpoints. There are several ways to proceed at this point. One way would be tosubstitute the second equation into the first as follows then solve for x.2 2x + ( x − 1) 2= 13While, this may be the way that first comes to mind it’s probably more work than isrequired. Instead of doing it this way, let’s rewrite the second equation as follows2x = y+1Now, substitute this into the first equation.2y+ 1+ y = 132y + y− 12 = 0( y+ 4)( y− 3)= 0This yields two values of y: y = 3 and y = − 4 . From the graph it’s clear (I hope….)that no intersection points will occur for y = − 4 . However, let’s suppose that wedidn’t have the graph and proceed without this knowledge. So, we will need to2substitute each y value into one of the equations (I’ll use x = y+ 1) and solve for x.First, y = 3 .2x = 3+ 1=4x = ± 2So, the two intersection points are (-2,3) and (2,3). Don’t get used to these being“nice” answers, most of the time the solutions will be fractions and/or decimals.Now, y = − 4x2=− 4+ 1=−3x= ±3 iSo, in this case we get complex solutions. This means that while ( 3, i − 4)and( 3, i 4)− − are solutions to the system they do not correspond to intersection points.© 2006 Paul Dawkins 35http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewOn occasion you will want the complex solutions and on occasion you won’t want thecomplex solutions. You can usually tell from the problem statement or the type ofproblem that you are working if you need to include the complex solutions or not. Inthis case we were after where the two curves intersected which implies that we areafter only the real solutions.3. Graph the following two curves and determine where they intersect.2x= y −4y−8x= 5y+28SolutionBelow is the graph of the two functions. If you don’t remember how to graphx= f y go back to the Graphing and Common Graphs sectionfunction in the form ( )for quick refresher.There are two intersection points for us to find. In this case, since both equations areof the form x= f( y)we’ll just set the two equations equal and solve for y.2y −4y− 8 = 5y+282y −9y− 36 = 0( y+ 3)( y− 12)= 0The y coordinates of the two intersection points are then y = − 3 and y = 12 . Now,plug both of these into either of the original equations and solve for x. I’ll use the line(second equation) since it seems a little easier.First, y = − 3 .Now, y = 12.x = 5( − 3)+ 28 = 13x = 5( 12)+ 28 = 88© 2006 Paul Dawkins 36http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSo, the two intersection are (13,-3) and (88,12).Solving InequalitiesSolve each of the following inequalities.1.x2− 10 > 3xSolutionTo solve a polynomial inequality we get a zero on one side of the inequality, factorand then determine where the other side is zero.2 2x − 10 > 3x ⇒ x −3x− 10 > 0 ⇒ ( x− 5)( x+ 2)> 0So, once we move everything over to the left side and factor we can see that the leftside will be zero at x = 5 and x = − 2. These numbers are NOT solutions (since weonly looking for values that will make the equation positive) but are useful to findingthe actual solution.To find the solution to this inequality we need to recall that polynomials are nicesmooth functions that have no breaks in them. This means that as we are movingacross the number line (in any direction) if the value of the polynomial changes sign(say from positive to negative) then it MUST go through zero!So, that means that these two numbers ( x = 5 and x = − 2) are the ONLY placeswhere the polynomial can change sign. The number line is then divided into threeregions. In each region if the inequality is satisfied by one point from that region thenit is satisfied for ALL points in that region. If this wasn’t true (i.e it was positive atone point in region and negative at another) then it must also be zero somewhere inthat region, but that can’t happen as we’ve already determined all the places wherethe polynomial can be zero! Likewise, if the inequality isn’t satisfied for some pointin that region that it isn’t satisfied for ANY point in that region.This means that all we need to do is pick a test point from each region (that are easyto work with, i.e. small integers if possible) and plug it into the inequality. If the testpoint satisfies the inequality then every point in that region does and if the test pointdoesn’t satisfy the inequality then no point in that region does.One final note here about this. I’ve got three versions of the inequality above. Youcan plug the test point into any of them, but it’s usually easiest to plug the test pointsinto the factored form of the inequality. So, if you trust your factoring capabilitiesthat’s the one to use. However, if you HAVE made a mistake in factoring, then youmay end up with the incorrect solution if you use the factored form for testing. It’s atrade-off. The factored form is, in many cases, easier to work with, but if you’vemade a mistake in factoring you may get the incorrect solution.So, here’s the number line and tests that I used for this problem.© 2006 Paul Dawkins 37http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewFrom this we see that the solution to this inequality is −∞ < x < − 2 and 5 < x

Algebra/Trig Review3.23x−2x− 11 > 0SolutionThis one is a little different, but not really more difficult. The quadratic doesn’tfactor and so we’ll need to use the quadratic formula to solve for where it is zero.Doing this gives1±34x =3Reducing to decimals this is x = 2.27698 and x = − 1.61032. From this point on it’sidentical to the previous two problems. In the number line below the dashed lines areat the approximate values of the two numbers above and the inequalities show thevalue of the quadratic evaluated at the test points shown.From the number line above we see that the solution is⎛1+34 ⎞⎜ , ∞3 ⎟.⎝ ⎠⎛ 1−34 ⎞⎜−∞,3 ⎟⎝ ⎠and4.x − 3 ≥ 0x + 2SolutionThe process for solving inequalities that involve rational functions is nearly identicalto solving inequalities that involve polynomials. Just like polynomial inequalities,rational inequalities can change sign where the rational expression is zero. However,they can also change sign at any point that produces a division by zero error in therational expression. A good example of this is the rational expression 1 x . Clearly,there is division by zero at x = 0 and to the right of x = 0 the expression is positiveand to the left of x = 0 the expression is negative.It’s also important to note that a rational expression will only be zero for values of xthat make the numerator zero.© 2006 Paul Dawkins 39http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSo, what we need to do is first get a zero on one side of the inequality so we can usethe above information. For this problem that has already been done. Now, determinewhere the numerator is zero (since the whole expression will be zero there) and wherethe denominator is zero (since we will get division by zero there).At this point the process is identical to polynomial inequalities with one exceptionwhen we go to write down the answer. The points found above will divide thenumber line into regions in which the inequality will either always be true or alwaysbe false. So, pick test points from each region, test them in the inequality and get thesolution from the results.For this problem the numerator will be zero at x = 3 and the denominator will be zeroat x = − 2. The number line, along with the tests is shown below.So, from this number line it looks like the two outer regions will satisfy theinequality. We need to be careful with the endpoints however. We will include x = 3because this will make the rational expression zero and so will be part of the solution.On the other hand, x = − 2 will give division by zero and so MUST be excluded fromthe solution since division by zero is never allowed.The solution to this inequality is −∞ < x < − 2 and 3 ≤ x

Algebra/Trig ReviewIn this case we won’t include any endpoints in the solution since they either givedivision by zero or make the expression zero and we want strictly less than zero forthis problem.The solution is then ( −∞, − 2) and (1, 5) .6.2x≥ 3x + 1SolutionWe need to be a little careful with this one. In this case we need to get zero on oneside of the inequality. This is easy enough to do. All we need to do is subtract 3 fromboth sides. This gives2x−3≥0x + 12x− 3( x+1)≥ 0x + 1−x−3≥ 0x + 1Notice that I also combined everything into a single rational expression. You willalways want to do this. If you don’t do this it can be difficult to determine where therational expression is zero. So once we’ve gotten it into a single expression it’s easyto see that the numerator will be zero at x = − 3 and the denominator will be zero atx = − 1. The number line for this problem is below.© 2006 Paul Dawkins 41http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewThe solution in this case is [ −3, − 1) . We don’t include the -1 because this is wherethe solution is zero, but we do include the -3 because this makes the expression zero.Absolute Value Equations and InequalitiesSolve each of the following.1. 3x + 8 = 2SolutionThis uses the following factp = d ≥0⇒ p=±dThe requirement that d be greater than or equal to zero is simply an acknowledgementthat absolute value only returns number that are greater than or equal to zero. SeeProblem 3 below to see what happens when d is negative.So the solution to this equation is3x+ 8= 2 3x+ 8=−23x=− 6 OR 3x=−10x = −2 10x = −3So there were two solutions to this. That will almost always be the case. Also, do notget excited about the fact that these solutions are negative. This is not a problem. Wecan plug negative numbers into an absolute value equation (which is what we’redoing with these answers), we just can’t get negative numbers out of an absolutevalue (which we don’t, we get 2 out of the absolute value in this case).2. 2x − 4 = 10© 2006 Paul Dawkins 42http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSolutionThis one works identically to the previous problem.2x− 4 = 10 2x− 4 =−102x= 14 OR 2x= −6x= 7 x= −3Do not make the following very common mistake in solve absolute value equationsand inequalities.2x−4 ≠ 2x+ 4 = 102x= 6x = 3Did you catch the mistake? In dropping the absolute value bars I just changed every“-” into a “+” and we know that doesn’t work that way! By doing this we get a singleanswer and it’s incorrect as well. Simply plug it into the original equation toconvince yourself that it’s incorrect.2( 3)− 4 = 6 − 4 = 2 = 2 ≠ 10When first learning to solve absolute value equations and inequalities people tend tojust convert all minus signs to plus signs and solve. This is simply incorrect and willalmost never get the correct answer. The way to solve absolute value equations is theway that I’ve shown here.3. x + 1 =− 15SolutionThis question is designed to make sure you understand absolute values. In this casewe are after the values of x such that when we plug them into x + 1 we will get -15.This is a problem however. Recall that absolute value ALWAYS returns a positivenumber! In other words, there is no way that we can get -15 out of this absolutevalue. Therefore, there are no solutions to this equation.4. 7x −10 ≤ 4SolutionTo solve absolute value inequalities with < or ≤ in them we usep < d ⇒ − d < p

Algebra/Trig Review−4 ≤7x−10 ≤46 ≤7x≤146≤ x ≤27In solving these make sure that you remember to add the 10 to BOTH sides of theinequality and divide BOTH sides by the 7. One of the more common mistakes hereis to just add or divide one side.5. 1− 2x< 7SolutionThis one is identical to the previous problem with one small difference.− 7< 1− 2x< 7− 8 x >−3Don’t forget that when multiplying or dividing an inequality by a negative number (-2in this case) you’ve got to flip the direction of the inequality.6. x −9 ≤− 1SolutionThis problem is designed to show you how to deal with negative numbers on theother side of the inequality. So, we are looking for x’s which will give us a number(after taking the absolute value of course) that will be less than -1, but as withProblem 3 this just isn’t possible since absolute value will always return a positivenumber or zero neither of which will ever be less than a negative number. So, thereare no solutions to this inequality.7. 4x + 5 > 3SolutionAbsolute value inequalities involving > and ≥ are solved as follows.p > d ⇒ pdp ≥d ⇒ p≤−d or p≥dNote that you get two separate inequalities in the solution. That is the way that itmust be. You can NOT put these together into a single inequality. Once I get thesolution to this problem I’ll show you why that is.Here is the solution© 2006 Paul Dawkins 44http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review4x+ 534x−2x −2So the solution to this inequality will be x’s that are less than -2 or greater than1− .2Now, as I mentioned earlier you CAN NOT write the solution as the following doubleinequality.1− 2 > x >−2When you write a double inequality (as we have here) you are saying that x will be anumber that will simultaneously satisfy both parts of the inequality. In other words,in writing this I’m saying that x is some number that is less than -2 and AT THE1SAME TIME is greater than − . I know of no number for which this is true. So,2this is simply incorrect. Don’t do it. This is however, a VERY common mistake thatstudents make when solving this kinds of inequality.8. 4 −11x≥ 9SolutionNot much to this solution. Just be careful when you divide by the -11.4 −11x≤−9 4 −11x≥9−11x≤−13 or −11x≥513 5x≥x≤−11 119. 10x + 1 >− 4SolutionThis is another problem along the lines of Problems 3 and 6. However, the answerthis time is VERY different. In this case we are looking for x’s that when plugged inthe absolute value we will get back an answer that is greater than -4, but sinceabsolute value only return positive numbers or zero the result will ALWAYS begreater than any negative number. So, we can plug any x we would like into thisabsolute value and get a number greater than -4. So, the solution to this inequality isall real numbers.Trigonometry© 2006 Paul Dawkins 45http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewTrig Function EvaluationOne of the problems with most trig classes is that they tend to concentrate on righttriangle trig and do everything in terms of degrees. Then you get to a calculus coursewhere almost everything is done in radians and the unit circle is a very useful tool.So first off let’s look at the following table to relate degrees and radians.Degree 0 30 45 60 90 180 270 360π π π π 3πRadians 0π 2π6 4 3 2 2Know this table! There are, of course, many other angles in radians that we’ll see duringthis class, but most will relate back to these few angles. So, if you can deal with theseangles you will be able to deal with most of the others.Be forewarned, everything in most calculus classes will be done in radians!Now, let’s look at the unit circle. Below is the unit circle with just the first quadrantfilled in. The way the unit circle works is to draw a line from the center of the circleoutwards corresponding to a given angle. Then look at the coordinates of the point wherethe line and the circle intersect. The first coordinate is the cosine of that angle and thesecond coordinate is the sine of that angle. There are a couple of basic angles that areπ π π π 3πcommonly used. These are 0, , , , , π, , and 2π and are shown below along6 4 3 2 2with the coordinates of the intersections. So, from the unit circle below we can see that⎛π⎞ 3 ⎛π⎞ 1cos⎜⎟ = and sin ⎜ ⎟ = .⎝ 6 ⎠ 2 ⎝ 6 ⎠ 2© 2006 Paul Dawkins 46http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewRemember how the signs of angles work. If you rotate in a counter clockwise directionthe angle is positive and if you rotate in a clockwise direction the angle is negative.Recall as well that one complete revolution is 2π , so the positive x-axis can correspondto either an angle of 0 or 2π (or 4π , or 6π , or − 2π , or − 4π , etc. depending on thedirection of rotation). Likewise, the angle 6π (to pick an angle completely at random)can also be any of the following angles:π 13ππ+ 2π= (start at then rotate once around counter clockwise)6 6 6π 25ππ+ 4π= (start at then rotate around twice counter clockwise)6 6 6π 11ππ− 2π=− (start at then rotate once around clockwise)6 6 6π 23ππ− 4π=− (start at then rotate around twice clockwise)6 6 6etc.© 2006 Paul Dawkins 47http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Reviewπ πIn fact can be any of the following angles + 2 π n , n= 0, ± 1, ± 2, ± 3, In this case6 6n is the number of complete revolutions you make around the unit circle starting at 6π .Positive values of n correspond to counter clockwise rotations and negative values of ncorrespond to clockwise rotations.So, why did I only put in the first quadrant? The answer is simple. If you know the firstquadrant then you can get all the other quadrants from the first. You’ll see this in thefollowing examples.Find the exact value of each of the following. In other words, don’t use a calculator.10.⎛2π⎞sin ⎜ ⎟⎝ 3 ⎠ and ⎛ 2π⎞sin ⎜−⎟⎝ 3 ⎠SolutionThe first evaluation here uses the angle 2 π 2ππ. Notice that = π − . So 2 π is33 3 3π 2πfound by rotating up from the negative x-axis. This means that the line for 3 3will be a mirror image of the line for 3π only in the second quadrant. The coordinatesfor 2 π π will be the coordinates for except the x coordinate will be negative.332π2ππLikewise for − we can notice that − =− π + , so this angle can be found by33 3π 2πrotating down from the negative x-axis. This means that the line for − will be3 3a mirror image of the line for 3π only in the third quadrant and the coordinates will bethe same as the coordinates for 3π except both will be negative.Both of these angles along with their coordinates are shown on the following unitcircle.© 2006 Paul Dawkins 48http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewFrom this unit circle we can see that⎛2π⎞ 3sin ⎜ ⎟ = and⎝ 3 ⎠ 2⎛ 2π⎞ 3sin ⎜− ⎟=−.⎝ 3 ⎠ 2This leads to a nice fact about the sine function. The sine function is called an oddfunction and so for ANY angle we havesin − θ =− sin θ( ) ( )11.⎛7π⎞cos⎜⎟⎝ 6 ⎠ and ⎛ 7π⎞cos⎜−⎟⎝ 6 ⎠SolutionFor this example notice that 7 π ππ= π + so this means we would rotate down6 667ππfrom the negative x-axis to get to this angle. Also − =−π− so this means we6 6would rotate up 6π from the negative x-axis to get to this angle. These are bothshown on the following unit circle along with appropriate coordinates for theintersection points.© 2006 Paul Dawkins 49http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review⎛7π⎞ 3 ⎛ 7π⎞ 3From this unit circle we can see that cos⎜⎟ = − and cos⎜− ⎟=−. In⎝ 6 ⎠ 2 ⎝ 6 ⎠ 2this case the cosine function is called an even function and so for ANY angle we havecos − θ = cos θ .( ) ( )⎛ π ⎞12. tan ⎜−⎟⎝ 4 ⎠ and ⎛7π⎞tan ⎜ ⎟⎝ 4 ⎠SolutionHere we should note that 7 π π= 2π− so 7 π π and − are in fact the same angle!4 4 4 4The unit circle for this angle is© 2006 Paul Dawkins 50http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewNow, if we remember that tan ( x)values the tangent function. So,( x)( x)sin= we can use the unit circle to find thecos( −π)( −π)⎛7π⎞ ⎛ π ⎞ sin 4 − 22tan ⎜ ⎟= tan ⎜− ⎟= = =−1.⎝ 4 ⎠ ⎝ 4 ⎠ cos 4 22⎛π⎞On a side note, notice that tan ⎜ ⎟ = 1 and we se can see that the tangent function is⎝ 4 ⎠also called an odd function and so for ANY angle we will havetan ( − θ) =− tan ( θ).⎛9π⎞13. sin ⎜ ⎟⎝ 4 ⎠SolutionFor this problem let’s notice that 9 π π= 2π+ . Now, recall that one complete4 4revolution is 2π . So, this means that 9 π π and are at the same point on the unit4 4circle. Therefore,⎛9π ⎞ ⎛ π ⎞ ⎛π⎞ 2sin ⎜ ⎟= sin ⎜2π+ ⎟= sin ⎜ ⎟=⎝ 4 ⎠ ⎝ 4⎠ ⎝ 4⎠2This leads us to a very nice fact about the sine function. The sine function is anexample of a periodic function. Periodic function are functions that will repeat© 2006 Paul Dawkins 51http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Reviewthemselves over and over. The “distance” that you need to move to the right or leftbefore the function starts repeating itself is called the period of the function.In the case of sine the period is 2π . This means the sine function will repeat itselfevery 2π . This leads to a nice formula for the sine function.sin ( x+ 2πn) = sin ( x)n= 0, ± 1, ± 2, Notice as well that since1csc( x)=sin ( x)we can say the same thing about cosecant.csc( x+ 2πn) = csc( x)n= 0, ± 1, ± 2, Well, actually we should be careful here. We can say this provided x≠ nπ since sinewill be zero at these points and so cosecant won’t exist there!14.⎛25π⎞sec⎜⎟⎝ 6 ⎠SolutionHere we need to notice that 25 π ππ= 4π+ . In other words, we’ve started at and6 66rotated around twice to end back up at the same point on the unit circle. This meansthat⎛25π ⎞ ⎛ π ⎞ ⎛π⎞sec⎜ ⎟= sec⎜4π+ ⎟=sec⎜ ⎟⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠Now, let’s also not get excited about the secant here. Just recall that1sec( x)=cos xand so all we need to do here is evaluate a cosine! Therefore,⎛25π⎞ ⎛π⎞ 1 1 2sec⎜ ⎟= sec⎜ ⎟= = =⎝ 6 ⎠ ⎝ 6 ⎠ ⎛π⎞cos3 3⎜ ⎟⎝ 6 ⎠2We should also note that cosine and secant are periodic functions with a period of2π . So,cos( x+ 2πn) = cos( x)n = 0, ± 1, ± 2, sec x+ 2πn = sec x( ) ( )( )15.⎛4π⎞tan ⎜ ⎟⎝ 3 ⎠Solution© 2006 Paul Dawkins 52http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewTo do this problem it will help to know that tangent (and hence cotangent) is also aperiodic function, but unlike sine and cosine it has a period of π .tan ( x+ π n) = tan ( x)n = 0, ± 1, ± 2, cot x+ π n = cot x( ) ( )So, to do this problem let’s note that 4 π π= π + . Therefore,3 3⎛4π ⎞ ⎛ π ⎞ ⎛π⎞tan ⎜ ⎟= tan ⎜π+ ⎟= tan ⎜ ⎟=3⎝ 3 ⎠ ⎝ 3⎠ ⎝ 3⎠Trig Evaluation Final ThoughtsAs we saw in the previous examples if you know the first quadrant of the unit circle youcan find the value of ANY trig function (not just sine and cosine) for ANY angle that canbe related back to one of those shown in the first quadrant. This is a nice idea toremember as it means that you only need to memorize the first quadrant and how to getthe angles in the remaining three quadrants!In these problems I used only “basic” angles, but many of the ideas here can also beapplied to angles other than these “basic” angles as we’ll see in Solving Trig Equations.Graphs of Trig FunctionsThere is not a whole lot to this section. It is here just to remind you of the graphs of thesix trig functions as well as a couple of nice properties about trig functions.Before jumping into the problems remember we saw in the Trig Function Evaluationsection that trig functions are examples of periodic functions. This means that all wereally need to do is graph the function for one periods length of values then repeat thegraph.Graph the following function.1. y = cos( x)SolutionThere really isn’t a whole lot to this one other than plotting a few points between 0and 2π , then repeat. Remember cosine has a period of 2π (see Problem 6 in TrigFunction Evaluation).Here’s the graph for −4π≤ x ≤ 4π.© 2006 Paul Dawkins 53http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewNotice that graph does repeat itself 4 times in this range of x’s as it should.Let’s also note here that we can put all values of x into cosine (which won’t be thecase for most of the trig functions) and let’s also note that−1 ≤cos( x)≤ 1It is important to notice that cosine will never be larger than 1 or smaller than -1.This will be useful on occasion in a calculus class.2. y = cos( 2x)SolutionWe need to be a little careful with this graph. ( )not dealing with cos( x ) here. We are dealing with ( )cos x has a period of 2π , but we’recos 2x . In this case notice thatif we plug in x = π we will getcos( 2( π)) = cos( 2π) = cos( 0)= 1In this case the function starts to repeat itself after π instead of 2π ! So, this functionhas a period of π . So, we can expect the graph to repeat itself 8 times in the range−4π≤ x ≤ 4π. Here is that graph.Sure enough, there are twice as many cycles in this graph.In general we can get the period of cos( x)ω using the following.2πPeriod =ω© 2006 Paul Dawkins 54http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewIf ω > 1we can expect a period smaller than 2π and so the graph will oscillate faster.Likewise, if ω < 1 we can expect a period larger than 2π and so the graph willoscillate slower.Note that the period does not affect how large cosine will get. We still have−1 ≤cos 2x≤ 13. y = 5cos( 2x)( )SolutionIn this case I added a 5 in front of the cosine. All that this will do is increase how bigcosine will get. The number in front of the cosine or sine is called the amplitude.Here’s the graph of this function.Note the scale on the y-axis for this problem and do not confuse it with the previousgraph. The y-axis scales are different!In general,−R≤ Rcos( ω x)≤ R4. y = sin ( x)SolutionAs with the first problem in this section there really isn’t a lot to do other than graphit. Here is the graph on the range −4π≤ x ≤ 4π.© 2006 Paul Dawkins 55http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewFrom this graph we can see that sine has the same range that cosine does. In general−R≤ R ω x ≤ Rsin ( )As with cosine, sine itself will never be larger than 1 and never smaller than -1.⎛ x ⎞5. y = sin ⎜ ⎟⎝ 3 ⎠SolutionSo, in this case we don’t have just an x inside the parenthesis. Just as in the case ofsin ω x by usingcosine we can get the period of ( )2π2πPeriod = = = 6πω 1 3In this case the curve will repeat every 6π . So, for this graph I’ll change the range to−6π≤ x ≤ 6πso we can get at least two traces of the curve showing. Here is thegraph.6. y = tan ( x)Solution© 2006 Paul Dawkins 56http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewIn the case of tangent we have to be careful when plugging x’s in since tangentsin xdoesn’t exist wherever cosine is zero (remember that tan x = ). Tangent will notcos xexist at5π 3π π π 3π 5πx = , − , − , − , , , , 2 2 2 2 2 2and the graph will have asymptotes at these points. Here is the graph of tangent on5π5πthe range − < x < .2 2Finally, a couple of quick properties about Rtan( x)−∞ < Rtan ( ω x)< ∞For the period remember that ( )ω .πPeriod =ωtan x has a period of π unlike sine and cosine andthat accounts for the absence of the 2 in the numerator that was there for sine andcosine.7. y = sec( x)SolutionAs with tangent we will have to avoid x’s for which cosine is zero (remember that1sec x = cos x). Secant will not exist at5π 3π π π 3π 5πx = , − , − , − , , , , 2 2 2 2 2 2and the graph will have asymptotes at these points. Here is the graph of secant on the5π5πrange − < x < .2 2© 2006 Paul Dawkins 57http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewNotice that the graph is always greater than 1 and less than -1. This should not be−1 ≤cos x ≤ 1. So, 1 divided by something less thanterribly surprising. Recall that ( )1 will be greater than 1. Also, 1 1±1 = ± and so we get the following ranges out ofsecant.Rsec ωx ≥R and Rsecωx ≤− R8. y = csc( x)( ) ( )SolutionFor this graph we will have to avoid x’s where sine is zerograph of cosecant will not exist forx = , −2 π, −π,0, π,2 π,Here is the graph of cosecant.⎛ 1 ⎞⎜cscx = ⎟. So, the⎝ sin x ⎠Cosecant will have the same range as secant.Rcsc ωx ≥R and Rcscωx ≤− R9. y = cot ( x)( ) ( )© 2006 Paul Dawkins 58http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSolutionCotangent must avoidx = , −2 π, −π,0, π,2 π,since we will have division by zero at these points. Here is the graph.Cotangent has the following range.−∞

Algebra/Trig ReviewIf you know the formula from Problem 1 in this section you can get this one for free.( θ) + ( θ)=( θ)( θ)+ =( θ)( θ) ( θ)2 2tan ( θ) + 1 = sec ( θ)2 2sin cos 12 2sin cos 1cos cos cos2 2 222Can you come up with a similar formula relating cot ( θ ) and ( )3. sin ( 2t ) =Solutionsin 2t = 2sin t cos t( ) ( ) ( )csc θ ?This formula is often used in reverse so that a product of a sine and cosine (with thesame argument of course) can be written as a single sine. For example,( )( x ) ( x ) = ( x ) ( x )sin 3 cos 3 sin 3 cos 33 2 3 2 2 22( ( x ))⎛1 ⎞= ⎜ sin 2 3 ⎟⎝2⎠1 sin3 2= ( 6 x )8You will find that using this formula in reverse can significantly reduce thecomplexity of some of the problems that you’ll face in a Calculus class.334. cos( 2x ) =(Three possible formulas)SolutionAs noted there are three possible formulas to use here.2 2cos 2x = cos x − sin x( ) ( ) ( )2cos( 2x) = 2cos ( x)−12cos( 2x) = 1−2sin ( x)You can get the second formula by substituting sin 2 ( x) 1 cos2( x)from this section) into the first. Likewise, you can substitute cos 2 ( x) = 1−sin2( x)into the first formula to get the third formula.= − (see Problem 15.2cos ( x ) =(In terms of cosine to the first power)Solution© 2006 Paul Dawkins 60http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review2x = 1 ( + ( x))cos ( ) 1 cos 22This is really the second formula from Problem 4 in this section rearranged and isVERY useful for eliminating even powers of cosines. For example,2 ⎛1⎞5cos ( 3x) = 5⎜( 1+cos( 2( 3x)))⎟⎝2⎠5= ( 1 + cos ( 6 x))2Note that you probably saw this formula written as⎛ x ⎞ 1cos⎜⎟ =± 1 + cos⎝2⎠2in a trig class and called a half-angle formula.( ( x))6.2sin ( x ) =(In terms of cosine to the first power)Solution2x = 1 ( − ( x))sin ( ) 1 cos 22As with the previous problem this is really the third formula from Problem 4 in thissection rearranged and is very useful for eliminating even powers of sine. Forexample,( )( t) = ( t)4 24sin 2 4 sin 2⎛1⎞= 4⎜( 1−cos( 4t))⎟⎝2⎠⎛1⎞2= 4⎜⎟( 1− 2cos( 4t) + cos ( 4t))⎝4⎠1= 1− 2cos( 4t) + ( 1+cos( 8t))23 1= − 2cos( 4t) + cos( 8t)2 2As shown in this example you may have to use both formulas and more than once ifthe power is larger than 2 and the answer will often have multiple cosines withdifferent arguments.Again, in a trig class this was probably called half-angle formula and written as,⎛ x ⎞ 1sin ⎜ ⎟ =± ( 1 − cos( x))⎝2⎠222© 2006 Paul Dawkins 61http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSolving Trig EquationsSolve the following trig equations. For those without intervals listed find ALL possiblesolutions. For those with intervals listed find only the solutions that fall in thoseintervals.1. 2cos( t ) = 3SolutionThere’s not much to do with this one. Just divide both sides by 2 and then go to theunit circle.2cos t = 3cos( )( t)So, we are looking for all the values of t for which cosine will have the value of=32So, let’s take a look at the following unit circle.32 .πFrom quick inspection we can see that t = is a solution. However, as I have shown6on the unit circle there is another angle which will also be a solution. We need to© 2006 Paul Dawkins 62http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Reviewdetermine what this angle is. When we look for these angles we typically wantpositive angles that lie between 0 and 2π . This angle will not be the only possibilityof course, but by convention we typically look for angles that meet these conditions.To find this angle for this problem all we need to do is use a little geometry. Theangle in the first quadrant makes an angle of 6π with the positive x-axis, then so mustπthe angle in the fourth quadrant. So we could use − , but again, it’s more common6π 11πto use positive angles so, we’ll use t = 2π− = .6 6We aren’t done with this problem. As the discussion about finding the second anglehas shown there are many ways to write any given angle on the unit circle.πSometimes it will be − that we want for the solution and sometimes we will want6both (or neither) of the listed angles. Therefore, since there isn’t anything in thisproblem (contrast this with the next problem) to tell us which is the correct solutionwe will need to list ALL possible solutions.This is very easy to do. Go back to my introduction in the Trig Function Evaluationsection and you’ll see there that I usedπ+ 2 π n, n= 0, ± 1, ± 2, ± 3, 6to represent all the possible angles that can end at the same location on the unit circle,i.e. angles that end at 6π . Remember that all this says is that we start at 6π then rotatearound in the counter-clockwise direction (n is positive) or clockwise direction (n isnegative) for n complete rotations. The same thing can be done for the secondsolution.So, all together the complete solution to this problem isπ+ 2 π n, n= 0, ± 1, ± 2, ± 3, 611π+ 2 π n, n= 0, ± 1, ± 2, ± 3, 6πAs a final thought, notice that we can get − by using n = − 1 in the second solution.62. 2cos( ) 3Solutiont = on [ − 2 π,2 π]© 2006 Paul Dawkins 63http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewThis problem is almost identical to the previous except now I want all the solutionsthat fall in the interval [ − 2 π,2 π]. So we will start out with the list of all possiblesolutions from the previous problem.π+ 2 π n, n= 0, ± 1, ± 2, ± 3, 611π+ 2 π n, n= 0, ± 1, ± 2, ± 3, 6Then start picking values of n until we get all possible solutions in the interval.First notice that since both the angles are positive adding on any multiples of 2π (npositive) will get us bigger than 2π and hence out of the interval. So, all positivevalues of n are immediately out. Let’s take a look at the negatives values of n.n = −1π11π+ 2π( − 1)=− >−2π6 611ππ+ 2π( − 1)=− >−2π6 6These are both greater than −2πand so are solutions, but if we subtract another 2πoff (i.e use n = − 2) we will once again be outside of the interval.So, the solutions are :π 11 11, π , − π , − π .6 6 6 63. 2sin ( 5x ) = − 3SolutionThis one is very similar to Problem 1, although there is a very important difference.We’ll start this problem in exactly the same way as we did in Problem 1.2sin(5 x) = − 3− 3sin(5 x)=23So, we are looking for angles that will give − out of the sine function. Let’s2again go to our trusty unit circle.© 2006 Paul Dawkins 64http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review3Now, there are no angles in the first quadrant for which sine has a value of − .2However, there are two angles in the lower half of the unit circle for which sine will3have a value of − . So, what are these angles? A quick way of doing this is to, for23a second, ignore the “-” in the problem and solve sin ( x ) = in the first quadrant2πonly. Doing this give a solution of x = . Now, again using some geometry, this3tells us that the angle in the third quadrant will be 3π below the negative x-axis orπ 4πππ + = . Likewise, the angle in the fourth quadrant will below the positive x-3 33π 5πaxis or 2π − = .3 3Now we come to the very important difference between this problem and Problem 1.The solution is NOT4πx= + 2 π n, n= 0, ± 1, ± 2, 35πx= + 2 π n, n= 0, ± 1, ± 2, 3© 2006 Paul Dawkins 65http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewThis is not the set of solutions because we are NOT looking for values of x for which33sin ( x ) = − , but instead we are looking for values of x for which sin ( 5x ) = − .22Note the difference in the arguments of the sine function! One is x and the other is5x . This makes all the difference in the world in finding the solution! Therefore, theset of solutions is4π5x= + 2 π n, n= 0, ± 1, ± 2, 35π5x= + 2 π n, n= 0, ± 1, ± 2, 3Well, actually, that’s not quite the solution. We are looking for values of x so divideeverything by 5 to get.4π2πnx= + , n= 0, ± 1, ± 2, 15 5π 2πnx= + , n= 0, ± 1, ± 2, 3 5Notice that I also divided the 2πn by 5 as well! This is important! If you don’t dothat you WILL miss solutions. For instance, take n = 1.4π 2π 10π 2π ⎛ ⎛2π ⎞⎞⎛10π⎞ 3x = + = = ⇒ sin 5 sin15 5 15 3⎜ ⎜ ⎟3⎟= ⎜ ⎟=−⎝ ⎝ ⎠⎠⎝ 3 ⎠ 2π 2π 11π ⎛ ⎛11π ⎞⎞⎛11π⎞ 3x = + = ⇒ sin 5 sin3 5 15⎜ ⎜ ⎟ = =−15⎟ ⎜ ⎟⎝ ⎝ ⎠⎠⎝ 3 ⎠ 2I’ll leave it to you to verify my work showing they are solutions. However it makesthe point. If you didn’t divided the 2πn by 5 you would have missed these solutions!4. 2sin ( 5x + 4)=− 3SolutionThis problem is almost identical to the previous problem except this time we have anargument of 5x + 4 instead of 5x . However, most of the problem is identical. In thiscase the solutions we get will be4π5x+ 4 = + 2 π n, n= 0, ± 1, ± 2, 35π5x+ 4 = + 2 π n, n= 0, ± 1, ± 2, 3Notice the difference in the left hand sides between this solution and thecorresponding solution in the previous problem.Now we need to solve for x. We’ll first subtract 4 from both sides then divide by 5 toget the following solution.© 2006 Paul Dawkins 66http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review4π2πn− 4x= + , n= 0, ± 1, ± 2, 15 55π2πn− 4x= + , n= 0, ± 1, ± 2, 15 5It’s somewhat messy, but it is the solution. Don’t get excited when solutions getmessy. They will on occasion and you need to get used to seeing them.5. 2sin ( 3x ) = 1 on [ − ππ , ]SolutionI’m going to leave most of the explanation that was in the previous three out of thisone to see if you have caught on how to do these.( x)2sin 3 = 11sin ( 3x)=2By examining a unit circle we see thatπ3x= + 2 π n, n= 0, ± 1, ± 2, 65π3x= + 2 π n, n= 0, ± 1, ± 2, 6Or, upon dividing by 3,π 2πnx= + , n= 0, ± 1, ± 2, 18 35π2πnx= + , n= 0, ± 1, ± 2, 18 3− ππ , . So, let’s start trying someNow, we are looking for solutions in the range [ ]values of n.n = 0 :n = 1 :n = 2 :π 2π 13πx = + = < π so a solution18 3 185π 2π 17πx = + = < π so a solution18 3 18π 4π 25πx = + = > π so NOT a solution18 3 185π 4π 29πx = + = > π so NOT a solution18 3 18π5πx= & x=18 18© 2006 Paul Dawkins 67http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewOnce, we’ve hit the limit in one direction there’s no reason to continue on. In, otherwords if using n = 2 gets values larger than π then so will all values of n larger than2. Note as well that it is possible to have one of these be a solution and the other tonot be a solution. It all depends on the interval being used.Let’s not forget the negative values of n.n = − 1 :π 2π 11πx = − =− >−πso a solution18 3 185π 2π 7πx = − =− >−πso a solution18 3 18n = − 2 :π 4π 23πx = − =−

Algebra/Trig ReviewSince we want the solutions on [0, 2 π ] negative values of n aren’t needed for thisproblem. So, plugging in values of n will give the following four solutions7. cos( 3x ) = 2π 5π 9π 13πt = , , ,8 8 8 8SolutionThis is a trick question that is designed to remind you of certain properties about sineand cosine. Recall that −1 ≤cos( θ ) ≤ 1 and −1 ≤sin ( θ ) ≤ 1. Therefore, since cosinewill never be greater that 1 it definitely can’t be 2. So THERE ARE NOSOLUTIONS to this equation!8. sin ( 2x) = − cos( 2x)SolutionThis problem is a little different from the previous ones. First, we need to do somerearranging and simplification.sin(2 x) = −cos(2 x)sin(2 x)= −1cos(2 x)tan ( 2x)= −1So, solving sin(2 x) = − cos(2 x)is the same as solving tan(2 x ) = − 1. At some levelwe didn’t need to do this for this problem as all we’re looking for is angles in whichsine and cosine have the same value, but opposite signs. However, for otherproblems this won’t be the case and we’ll want to convert to tangent.Looking at our trusty unit circle it appears that the solutions will be,3π2x= + 2 π n, n= 0, ± 1, ± 2, 47π2x= + 2 π n, n= 0, ± 1, ± 2, 4Or, upon dividing by the 2 we get the solutions3πx= + π n, n= 0, ± 1, ± 2, 87πx= + π n, n= 0, ± 1, ± 2, 8No interval was given so we’ll stop here.9. 2sin ( θ) cos( θ ) = 1© 2006 Paul Dawkins 69http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSolutionAgain, we need to do a little work to get this equation into a form we can handle. Theeasiest way to do this one is to recall one of the trig formulas from the Trig Formulassection (in particular Problem 3).2sin θ cos θ = 1( ) ( )( θ )sin 2 = 1At this point, proceed as we did in the previous problems..π2θ= + 2 πn, n= 0, ± 1, ± 2, 2Or, by dividing by 2,πθ = + πn, n= 0, ± 1, ± 2, 4Again, there is no interval so we stop here.10. ( w) ( w) ( w)sin cos + cos = 0SolutionThis problem is very different from the previous problems.DO NOT DIVIDE BOTH SIDES BY A COSINE!!!!!!If you divide both sides by a cosine you WILL lose solutions! The best way to dealwith this one is to “factor” the equations as follows.sin ( w) cos( w) + cos( w)= 0cos( w) ( sin ( w)+ 1)= 0So, solutions will be values of w for whichcos( w ) = 0or,sin w + 1 = 0 ⇒ sin w =− 1( ) ( )In the first case we will have cos( w ) = 0 at the following values.πw= + 2 π n, n= 0, ± 1, ± 2, 23πw= + 2 π n, n= 0, ± 1, ± 2, 2In the second case we will have sin ( w ) = − 1 at the following values.3πw= + 2 π n, n= 0, ± 1, ± 2, 2Note that in this case we got a repeat answer. Sometimes this will happen andsometimes it won’t so don’t expect this to always happen. So, all together we get thefollowing solutions,© 2006 Paul Dawkins 70http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Reviewπw= + 2 π n, n= 0, ± 1, ± 2, 23πw= + 2 π n, n= 0, ± 1, ± 2, 2As with the previous couple of problems this has no interval so we’ll stop here.Notice as well that if we’d divided out a cosine we would have lost half the solutions.211. ( x) ( x)2cos 3 + 5cos 3 − 3 = 0SolutionThis problem appears very difficult at first glance, but only the first step is differentfor the previous problems. First notice that22t+ 5t− 3=0( 2t− 1)( t+ 3)= 01The solutions to this are t = and t = − 3 . So, why cover this? Well, if you think2about it there is very little difference between this and the problem you are asked todo. First, we factor the equation22cos ( 3x) + 5cos( 3x)− 3 = 0( 2cos( 3x)− 1) ( cos( 3x)+ 3)= 0The solutions to this are1cos( 3x) =2and cos( 3x)= − 3The solutions to the first areπ3x= + 2 π n, 3n= 0, ± 1, ± 2, 5π3x= + 2 π n, 3n= 0, ± 1, ± 2, Or, upon dividing by 3,π 2πnx= + ,9 3n= 0, ± 1, ± 2, 5π2πnx= + ,9 3n= 0, ± 1, ± 2, The second has no solutions because cosine can’t be less that -1. Don’t get used tothis. Often both will yield solutions!Therefore, the solutions to this are (again no interval so we’re done at this point).π 2πnx= + ,9 3n= 0, ± 1, ± 2, 5π2πnx= + ,9 3n= 0, ± 1, ± 2, © 2006 Paul Dawkins 71http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review12. 5sin ( 2x ) = 1SolutionThis problem, in some ways, is VERY different from the previous problems and yetwill work in essentially the same manner. To this point all the problems came downto a few “basic” angles that most people know and/or have used on a regular basis.This problem won’t, but the solution process is pretty much the same. First, get thesine on one side by itself.1sin ( 2x ) =5Now, at this point we know that we don’t have one of the “basic” angles since thoseall pretty much come down to having 0, 1, 1 2 , 22 or 3on the right side of the2equal sign. So, in order to solve this we’ll need to use our calculator. Every1calculator is different, but most will have an inverse sine ( sin − ), inverse cosine1( cos − 1) and inverse tangent ( tan − ) button on them these days. If you aren’t familiarwith inverse trig functions see the next section in this review. Also, make sure thatyour calculator is set to do radians and not degrees for this problem.It is also very important to understand the answer that your calculator will give. First,note that I said answer (i.e. a single answer) because that is all your calculator willever give and we know from our work above that there are infinitely many answers.−1Next, when using your calculator to solve sin ( x)= a, i.e. using sin ( a), we will getthe following ranges for x.πa≥0 ⇒ 0 ≤ x≤ 1.570796327 =( Quad I)2πa≤0 ⇒ − =−1.570796327 ≤ x≤0 ( Quad IV)2So, when using the inverse sine button on your calculator it will ONLY returnanswers in the first or fourth quadrant depending upon the sign of a.Using our calculator in this problem yields,−1 ⎛1⎞2x= sin ⎜ ⎟=0.2013579⎝5⎠Don’t forget the 2 that is in the argument! We’ll take care of that in a bit.Now, we know from our work above that if there is a solution in the first quadrant tothis equation then there will also be a solution in the second quadrant and that it willbe at an angle of 0.2013579 above the x-axis as shown below.© 2006 Paul Dawkins 72http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewI didn’t put in the x, or cosine value, in the unit circle since it’s not needed for theproblem. I did however note that they will be the same value, except for the negativesign. The angle in the second quadrant will then be,π − 0.2013579 = 2.9402348So, let’s put all this together.2x= 0.2013579 + 2 π n, n= 0, ± 1, ± 2, 2x= 2.9402348 + 2 π n, n= 0, ± 1, ± 2, Note that I added the 2π n onto our angles as well since we know that will be neededin order to get all the solutions. The final step is to then divide both sides by the 2 inorder to get all possible solutions. Doing this gives,x= 0.10067895 + π n, n= 0, ± 1, ± 2, x= 1.4701174 + π n, n= 0, ± 1, ± 2, The answers won’t be as “nice” as the answers in the previous problems but therethey are. Note as well that if we’d been given an interval we could plug in values of nto determine the solutions that actually fall in the interval that we’re interested in.⎛ x ⎞13. 4cos⎜⎟ = − 3⎝5⎠© 2006 Paul Dawkins 73http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSolutionThis problem is again very similar to previous problems and yet has some differences.First get the cosine on one side by itself.⎛ x ⎞ 3cos⎜⎟ = −⎝ 5 ⎠ 4Now, let’s take a quick look at a unit circle so we can see what angles we’re after.I didn’t put the y values in since they aren’t needed for this problem. Note however,that they will be the same except have opposite signs. Now, if this were a probleminvolving a “basic” angle we’d drop the “-” to determine the angle each of the linesabove makes with the x-axis and then use that to find the actual angles. However, inthis case since we’re using a calculator we’ll get the angle in the second quadrant forfree so we may as well jump straight to that one.However, prior to doing that let’s acknowledge how the calculator will work when−1working with inverse cosines. If we’re going to solve cos( x)= a, using cos ( a),then our calculator will give one answer in one of the following ranges, dependingupon the sign of a.πa≥0 ⇒ 0 ≤ x≤ 1.570796 = ( Quad I)2πa≤0 ⇒ π = 3.141593 ≤ x≤ 1.570796 =Quad II2( )© 2006 Paul Dawkins 74http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSo, using our calculator we get the following angle in the second quadrant.x −1 ⎛ 3 ⎞= cos ⎜− ⎟=2.4188585 ⎝ 4⎠Now, we need to get the second angle that lies in the third quadrant. To find thisangle note that the line in the second quadrant and the line in the third quadrant bothmake the same angle with the negative x-axis. Since we know what the angle in thesecond quadrant is we can find the angle that this line makes with the negative x-axisas follows,π − 2.418858 = 0.722735This means that the angle in the third quadrant is,π + 0.722735 = 3.864328Putting all this together gives,x= 2.418858 + 2 π n, 5n= 0, ± 1, ± 2, x= 3.864328 + 2 π n, 5n= 0, ± 1, ± 2, Finally, we just need to multiply both sides by 5 to determine all possible solutions.x= 12.09429 + 10 π n, n= 0, ± 1, ± 2, x= 19.32164 + 10 π n, n= 0, ± 1, ± 2, 14. 10sin ( x − 2)=− 7SolutionWe’ll do this one much quicker than the previous two. First get the sine on one sideby itself.7sin ( x − 2)=−10From a unit circle we can see that the two angles we’ll be looking for are in the thirdand fourth quadrants. Our calculator will give us the angle that is in the fourthquadrant and this angle is,−1 ⎛ 7 ⎞x − 2 = sin ⎜− ⎟=−0.775395⎝ 10 ⎠Note that in all the previous examples we generally wouldn’t have used this answerbecause it is negative. There is nothing wrong with the answer, but as I mentionedseveral times in earlier problems we generally try to use positive angles between 0and 2π . However, in this case since we are doing calculator work we won’t worryabout that fact that it’s negative. If we wanted the positive angle we could always getit as,© 2006 Paul Dawkins 75http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review2π − 0.775395 = 5.5077903Now, the line corresponding to this solution makes an angle with the positive x-axisof 0.775395. The angle in the third quadrant will be 0.775395 radians below thenegative x-axis and so is,x − 2 = π + 0.775395 = 3.916988Putting all this together gives,x− 2 =− 0.775395 + 2 π n, n= 0, ± 1, ± 2, x− 2 = 3.916988 + 2 π n, n= 0, ± 1, ± 2, To get the final solution all we need to do is add 2 to both sides. All possiblesolutions are then,x= 1.224605 + 2 π n, n= 0, ± 1, ± 2, x= 5.916988 + 2 π n, n= 0, ± 1, ± 2, As the last three examples have shown, solving a trig equation that doesn’t give any ofthe “basic” angles is not much different from those that do give “basic” angles. In fact, insome ways there are a little easier to do since our calculator will always give us one forfree and all we need to do is find the second. The main idea here is to always rememberthat we need to be careful with our calculator and understand the results that it gives us.Note as well that even for those problems that have “basic” angles as solutions we couldhave used a calculator as well. The only difference would have been that our answerswould have been decimals instead of the exact answers we got.Inverse Trig FunctionsOne of the more common notations for inverse trig functions can be very confusing.First, regardless of how you are used to dealing with exponentiation we tend to denote aninverse trig function with an “exponent” of “-1”. In other words, the inverse cosine is−1denoted as cos ( x). It is important here to note that in this case the “-1” is NOT anexponent and so,cos( x)≠ .cos1− 1In inverse trig functions the “-1” looks like an exponent but it isn’t, it is simply a notationthat we use to denote the fact that we’re dealing with an inverse trig function. It is anotation that we use in this case to denote inverse trig functions. If I had really wantedexponentiation to denote 1 over cosine I would use the following.( cos( x))( x)− 1 1 =cos( x)© 2006 Paul Dawkins 76http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewThere’s another notation for inverse trig functions that avoids this ambiguity. It is thefollowing.−1cos x = arccos xsin−1( ) ( )( x) = arcsin ( x)( x) = arctan ( x)−1tanSo, be careful with the notation for inverse trig functions!There are, of course, similar inverse functions for the remaining three trig functions, butthese are the main three that you’ll see in a calculus class so I’m going to concentrate onthem.To evaluate inverse trig functions remember that the following statements are equivalent.−1θ = cos ( x) ⇔ x=cos( θ)−1θ = sin ( x) ⇔ x=sin ( θ)−1θ = tan ( x) ⇔ x=tan ( θ)In other words, when we evaluate an inverse trig function we are asking what angle, θ ,did we plug into the trig function (regular, not inverse!) to get x.So, let’s do some problems to see how these work. Evaluate each of the following.1.cos⎛ −1 3 ⎞⎜⎝2⎟⎠SolutionIn Problem 1 of the Solving Trig Equations section we solved the following equation.3cos( t ) =23In other words, we asked what angles, t, do we need to plug into cosine to get2 ?This is essentially what we are asking here when we are asked to compute the inversetrig function.1 3cos⎛ − ⎞⎜2 ⎟⎝ ⎠There is one very large difference however. In Problem 1 we were solving anequation which yielded an infinite number of solutions. These were,π+ 2 π n, n= 0, ± 1, ± 2, ± 3, 611π+ 2 π n, n= 0, ± 1, ± 2, ± 3, 6© 2006 Paul Dawkins 77http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewIn the case of inverse trig functions we are after a single value. We don’t want tohave to guess at which one of the infinite possible answers we want. So, to make surewe get a single value out of the inverse trig cosine function we use the followingrestrictions on inverse cosine.( x)−1θ = cos −1 ≤ x≤1 and 0 ≤θ ≤ πThe restriction on the θ guarantees that we will only get a single value angle andsince we can’t get values of x out of cosine that are larger than 1 or smaller than -1 wealso can’t plug these values into an inverse trig function.So, using these restrictions on the solution to Problem 1 we can see that the answer inthis case iscos⎛ 1 3 ⎞ − = π⎜⎝2 ⎟⎠ 62.cos⎜−⎝⎛ −1 3 ⎞2⎟⎠SolutionIn general we don’t need to actually solve an equation to determine the value of aninverse trig function. All we need to do is look at a unit circle. So in this case we’re3after an angle between 0 and π for which cosine will take on the value − . So,2check out the following unit circle© 2006 Paul Dawkins 78http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewFrom this we can see that1 3 5cos⎛ − π⎞⎜− =2 ⎟⎝ ⎠ 63.− ⎛ ⎞sin ⎜−⎟⎝ 2 ⎠1 1SolutionThe restrictions that we put on θ for the inverse cosine function will not work for theinverse sine function. Just look at the unit circle above and you will see that between0 and π there are in fact two angles for which sine would be 1 2and this is not whatwe want. As with the inverse cosine function we only want a single value.Therefore, for the inverse sine function we use the following restrictions.−1π πθ = sin ( x)−1 ≤ x≤1 and − ≤θ≤2 2By checking out the unit circle© 2006 Paul Dawkins 79http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Reviewwe see−1 ⎛ 1 ⎞ πsin ⎜− ⎟=−⎝ 2⎠6−14. tan ( 1)SolutionThe restriction for inverse tangent is−1π πθ = tan ( x)− < θ

Algebra/Trig Review5.⎛cos cos⎜⎝⎛⎜⎝−1 32⎞⎞⎟⎟⎠⎠SolutionRecalling the answer to Problem 1 in this section the solution to this problem is mucheasier than it look’s like on the surface.⎛ ⎛1 3⎞⎞−⎛π⎞ 3cos cos = cos =⎜ ⎜⎜ ⎟2 ⎟⎟ ⎝ ⎝ ⎠⎠⎝ 6 ⎠ 2This problem leads to a couple of nice facts about inverse cosine−1 −1cos cos x = x AND cos cos θ = θ( ( )) ( ( ))6.1 ⎛πsin sin 4− ⎛ ⎞⎞⎜ ⎜ ⎟⎟⎝ ⎝ ⎠⎠SolutionThis problem is also not too difficult (hopefully…).−1⎛⎛π⎞ ⎞ −1 2 πsin sin sin⎛ ⎞⎜ ⎜ ⎟ = =4⎟⎝ ⎝ ⎠⎠ ⎜ 2 ⎟⎝ ⎠ 4As with inverse cosine we also have the following facts about inverse sine.−1 −1sin sin x = x AND sin sin θ = θ( )−17. tan tan ( 4)− .( ( )) ( ( ))SolutionJust as inverse cosine and inverse sine had a couple of nice facts about them so doesinverse tangent. Here is the fact−1 −1tan tan x = x AND tan tan θ = θ( ( )) ( ( ))−1Using this fact makes this a very easy problem as I couldn’t do tan ( 4)by hand! Acalculator could easily do it but I couldn’t get an exact answer from a unit circle.−1tan tan − 4 =− 4( ( ))Exponentials / LogarithmsBasic Exponential Functions© 2006 Paul Dawkins 81http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewFirst, let’s recall that for b > 0 and b ≠ 1 an exponential function is any function that is inthe formxf ( x) = bWe require b ≠ 1 to avoid the following situation,xf x = =( ) 1 1So, if we allowed b = 1 we would just get the constant function, 1.We require b > 0 to avoid the following situation,x1f ( x) = ( −4) ⇒ f ⎛ ⎞ ( 4) 1 2⎜ ⎟= − = −4⎝2⎠By requiring b > 0 we don’t have to worry about the possibility of square roots ofnegative numbers.1. Evaluate f ( x ) = 4 x , g( x)x⎛1⎞−= ⎜ ⎟ and h( x) = 4 x at x =−2, − 1, 0,1, 2 .⎝4⎠SolutionThe point here is mostly to make sure you can evaluate these kinds of functions. So,here’s a quick table with the answers.x = − 2 x = − 1 x = 0 x = 1 x = 2211f − 1 =16 4f ( 0)= 1 f ( 1)= 4 f ( 2)= 162 16 g − 1 = 4 g ( 0)= 11 1g ( 1)= g ( 2)=4 162 16 h − 1 = 4 h ( 0)= 11 1h ( 1)= h ( 2)=4 16f ( x ) f ( − ) = ( )g( x ) g ( − ) = ( )h( x ) h( − ) = ( )Notice that the last two rows give exactly the same answer. If you think about it thatshould make sense because,x x⎛1⎞1 1 − xg( x) = ⎜ ⎟ = = = 4 = hx x( x)⎝4⎠4 4x⎛1⎞−f x = , g( x)= ⎜ ⎟ and h( x) = 4 x on the same axis system.⎝4⎠2. Sketch the graph of ( ) 4 xSolutionNote that we only really need to graph f ( x ) and ( )previous Problem that g( x) h( x)g x since we showed in the= . Note as well that there really isn’t too much todo here. We found a set of values in Problem 1 so all we need to do is plot the pointsand then sketch the graph. Here is the sketch,© 2006 Paul Dawkins 82http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Reviewx3. List as some basic properties for f ( x) = b .SolutionMost of these properties can be seen in the sketch in the previous Problem.x(a) f ( x) = b > 0 for every x. This is a direct consequence of the requirement thatb > 0 .0 = b = 1.(b) For any b we have ( )0f(c) If b > 1 ( f ( x ) = 4 xxabove, for example) we see that f ( x) b= is an increasingfunction and that,f x →∞as x→∞ and f x →0 as x→−∞( ) ( )x⎛1⎞(d) If 0< b < 1 ( g( x)= ⎜ ⎟⎝4⎠decreasing function and that,f x →0 as x→∞ and f x →∞asx→−∞xabove, for example) we see that f ( x) = b is an( ) ( )Note that the last two properties are very important properties in many Calculustopics and so you should always remember them!x− x1−34. Evaluate f ( x ) = e , g( x) = e and h( x) 5Solutionx= e at 2, 1, 0,1, 2x =− − .© 2006 Paul Dawkins 83http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewAgain, the point of this problem is to make sure you can evaluate these kinds offunctions. Recall that in these problems e is not a variable it is a number! In fact,e = 2.718281828When computing h( x ) make sure that you do the exponentiation BEFOREmultiplying by 5.x = − 2 x = − 1 x = 0 x = 1 x = 2f x 0.135335 0.367879 1 2.718282 7.389056( )g( x ) 7.389056 2.718282 1 0.367879 0.135335h( x ) 5483.166 272.9908 13.59141 0.676676 0.033690x− x5. Sketch the graph of f ( x ) = e and g( x) = e .SolutionAs with the other “sketching” problem there isn’t much to do here other than use thenumbers we found in the previous example to make the sketch. Here it is,Note that from these graphs we can see the following important properties about− xf x = e and g( x) = e .( )xeex−x→∞as x→∞ and e →0 as x→−∞−x→0 as x→∞ and e →∞asx→−∞These properties show up with some regularity in a Calculus course and so should beremembered.x© 2006 Paul Dawkins 84http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewBasic Logarithmic Functions1. Without a calculator give the exact value of each of the following logarithms.(a) log216 (b) log416 (c) log5625127(d) log9(e) log136 (f) log353144168SolutionTo do these without a calculator you need to remember the following.y = log x is equivalent tox=bbWhere, b, is called the base is any number such that b > 0 and b ≠ 1. The first isusually called logarithmic form and the second is usually called exponential form.The logarithmic form is read “y equals log base b of x”.So, if you convert the logarithms to exponential form it’s usually fairly easy tocompute these kinds of logarithms.y2(i)(j)4log216 = 4 because 2 = 162log416 = 2 because 4 = 16Note the difference between (a) and (b)! The base, b, that you use on the logarithm isVERY important! A different base will, in almost every case, yield a differentanswer. You should always pay attention to the base!4(k) log5625 = 4 because 5 = 6251 −61 1(l) log9 =− 6 because 9 = =6531441 9 531441−2⎛1⎞ 2(m) log 36 =− 2 because ⎜ ⎟ = 6 = 36⎝6⎠(n)1627 3 27log 3 because⎛ ⎞= ⎜ ⎟ =8 ⎝2⎠8322. Without a calculator give the exact value of each of the following logarithms.3(a) ln e (b) log1000 (c) log16167(d) log231 (e) log232SolutionThere are a couple of quick notational issues to deal with first.3© 2006 Paul Dawkins 85http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Reviewln x=log x This log is called the natural logarithmelog x=log x This log is called the common logarithm10The e in the natural logarithm is the same e used in Problem 2 above. The commonlogarithm and the natural logarithm are the logarithms are encountered more oftenthan any other logarithm so the get used to the special notation and special names.The work required to evaluate the logarithms in this set is the same as in problem inthe previous problem.13 13 3(a) ln e = becausee = e33(b) log1000 = 3 because 10 = 10001(c) log1616 = 1 because 16 = 160(d) log231 = 0 because 23 = 1155(e) log7232 = because 32 = 32 = ( 2 ) = 271 57 7 7 7Logarithm PropertiesComplete the following formulas.1. log bb =Solution1log b= b1 becauseb = b2. log 1 b=Solutionlog 1 = 0 because0b = 1bx3. log b = b4.Solutionxlog = xb blog b xb =© 2006 Paul Dawkins 86http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSolutionlog b xb = x5. log bxy =Solutionlog xy = log x + log yb b bTHERE IS NO SUCH PROPERTY FOR SUMS OR DIFFERENCES!!!!!⎛ x ⎞6. log b ⎜ ⎟ =⎝ y ⎠( )( )log x+ y ≠ log x+log yb b blog x− y ≠log x−logyb b bSolutionlog ⎛ x⎜⎞ ⎟ = log x − log⎝ y ⎠b b byTHERE IS NO SUCH PROPERTY FOR SUMS OR DIFFERENCES!!!!!r7. log ( )bx =Solutionrlog x = rlogxb( )b( )( )log x+ y ≠ log x+log yb b blog x− y ≠log x−logyb b bNote in this case the exponent needs to be on the WHOLE argument of the logarithm.For instance,However,loglogbb2( x+ y) = 2log ( x+y)2 2( x + y ) ≠ 2logb( x+y)8. Write down the change of base formula for logarithms.Solutionb© 2006 Paul Dawkins 87http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewWrite down the change of base formula for logarithms.logaxlogbx =logabThis is the most general change of base formula and will convert from base b to basea. However, the usual reason for using the change of base formula is so you cancompute the value of a logarithm that is in a base that you can’t easily compute.Using the change of base formula means that you can write the logarithm in terms ofa logarithm that you can compute. The two most common change of base formulasareln xlog xlogbx= and logbx=ln blog bIn fact, often you will see one or the other listed as THE change of base formula!In the problems in the Basic Logarithm Functions section you computed the value ofa few logarithms, but you could do these without the change of base formula becauseall the arguments could be wrote in terms of the base to a power. For instance,2log 49 = 2 because 7 = 497However, this only works because 49 can be written as a power of 7! We would needthe change of base formula to compute log750 .ln 50 3.91202300543log750 = = = 2.0103821378ln 7 1.94591014906ORlog50 1.69897000434log750 = = = 2.0103821378log 7 0.845098040014So, it doesn’t matter which we use, you will get the same answer regardless.Note as well that we could use the change of base formula on log749 if we wanted toas well.ln 49 3.89182029811log749 = = = 2ln 7 1.94591014906This is a lot of work however, and is probably not the best way to deal with this.9. What is the domain of a logarithm?Solution© 2006 Paul Dawkins 88http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewThe domain y = log bx is x > 0 . In other words you can’t plug in zero or a negativenumber into a logarithm. This makes sense if you remember that b > 0 and write thelogarithm in exponential form.yy = logbx ⇒ b = xSince b > 0 there is no way for x to be either zero or negative. Therefore, you can’tplug a negative number or zero into a logarithm!10. Sketch the graph of f ( x) = ln ( x)and g( x) log ( x)= .SolutionNot much to this other than to use a calculator to evaluate these at a few points andthen make the sketch. Here is the sketch.From this graph we can see the following behaviors of each graph.ln x →∞as x→∞ and ln x →−∞as x→ 0 x>0( ) ( ) ( )( x) →∞ x→∞ ( x) →−∞ x→ ( x>)log as and log as 0 0Remember that we require x > 0 in each logarithm.Simplifying LogarithmsSimplify each of the following logarithms.1.3 4 5ln xyzSolution© 2006 Paul Dawkins 89http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewHere simplify means use Property 1 - 7 from the Logarithm Properties section asoften as you can. You will be done when you can’t use any more of these properties.Property 5 can be extended to products of more than two functions so,ln xyz = ln x+ ln y + ln z3 4 5 3 4 5= 3ln x+ 4ln y+5ln z2.⎛49x⎞log3⎜y ⎟⎝ ⎠SolutionIn using property 6 make sure that the logarithm that you subtract is the one thatcontains the denominator as its argument. Also, note that that I’ll be converting theroot to exponents in the first step since we’ll need that done for a later step.41⎛9x⎞4 2log3 = log39x− log3y⎜y ⎟⎝ ⎠14 23 3x3y= log 9 + log −log1= 2 + 4log3 x−log3y2Evaluate logs where possible as I did in the first term.3.⎛log x +y⎜⎝2 2( x−y)3⎞⎟⎠SolutionThe point to this problem is mostly the correct use of property 7.⎛ 2 2x + y ⎞log = log + −log−3⎜( x y)⎟⎝ − ⎠= log + − −2 2( x y ) ( x y)2 2( x y ) 3log ( x y)You can use Property 7 on the second term because the WHOLE term was raised tothe 3, but in the first logarithm, only the individual terms were squared and not theterm as a whole so the 2’s must stay where they are!3Solving Exponential Equations© 2006 Paul Dawkins 90http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewIn each of the equations in this section the problem is there is a variable in the exponent.In order to solve these we will need to get the variable out of the exponent. This meansusing Property 3 and/or 7 from the Logarithm Properties section above. In most cases itwill be easier to use Property 3 if possible. So, pick an appropriate logarithm and take thelog of both sides, then use Property 3 (or Property 7) where appropriate to simplify. Notethat often some simplification will need to be done before taking the logs.Solve each of the following equations.1.4x−22 9e=SolutionxThe first thing to note is that Property 3 is log b = x and NOT blog ( 2 xbb ) = x! Inother words, we’ve got to isolate the exponential on one side by itself with acoefficient of 1 (one) before we take logs of both sides.We’ll also need to pick an appropriate log to use. In this case the natural log would4x−2be best since the exponential in the problem is e . So, first isolate the exponentialon one side.e4x−22 94x−29e =2Now, take the natural log of both sides and use Property 3 to simplify.4x−2 ⎛9⎞ln ( e ) = ln ⎜ ⎟⎝ 2 ⎠⎛9⎞4x− 2 = ln ⎜ ⎟⎝ 2 ⎠Now you all can solve 4x − 2= 9 so you can solve the equation above. All you need⎛9⎞to remember is that ln ⎜ ⎟ is just a number, just as 9 is a number. So add 2 to both⎝ 2 ⎠=sides, then divide by 4 (or multiply by 1 ). 4⎛9⎞4x= 2 + ln ⎜ ⎟⎝ 2 ⎠1⎛⎛9⎞⎞x = 2 ln4 ⎜ + ⎜ ⎟2 ⎟⎝ ⎝ ⎠⎠x = 0.8760193492Note that while the natural logarithm was the easiest (since the left side simplifieddown nicely) we could have used any other log had we wanted to. For instance wecould have used the common log as follows. Remember that in this case we won’t be© 2006 Paul Dawkins 91http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Reviewable to use Property 3 as this requires both the log and the exponential to have thesame base which won’t be the case here. Therefore, we’ll need to use Property 7 todo the simplification.4x−2 ⎛9⎞log ( e ) = log ⎜ ⎟⎝ 2 ⎠⎛9⎞( 4x− 2)log e = log ⎜ ⎟⎝ 2 ⎠As you can see the problem here is that we’ve got a log e left over after usingProperty 7. While this can be dealt with using a calculator, it adds a complexity to theproblem that should be avoided if at all possible. Solving gives us2t t2. 10 − = 1004x− 2=⎛9⎞log⎜⎟⎝ 2 ⎠loge4x= 2+⎛9⎞log⎜⎟⎝ 2 ⎠loge⎛9⎞1 ⎛ log⎜⎟x 2 2⎞⎝ ⎠= +log4 ⎜e⎟⎝ ⎠x = 0.8760193492SolutionNow, in this case it looks like the best logarithm to use is the common logarithmsince left hand side has a base of 10. There’s no initial simplification to do, so justtake the log of both sides and simplify.2tlog10− t= log1002t − t = 2At this point, we’ve just got a quadratic that can be solved2t −t− 2=0( t− 2)( t+ 1)= 0So, it looks like the solutions in this case are t = 2 and t = − 1.As with the last one you could use a different log here, but it would have made thequadratic significantly messier to solve.3.1−3z7 15 10+ e =SolutionThere’s a little more initial simplification to do here, but other than that it’s similar tothe first problem in this section.© 2006 Paul Dawkins 92http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review1−3z7 + 15e= 101−3z15e= 31−3z1e =5Now, take the log and solve. Again, we’ll use the natural logarithm here.1−3z⎛1⎞ln ( e ) = ln ⎜ ⎟⎝ 5 ⎠⎛1⎞1− 3z= ln ⎜ ⎟⎝ 5 ⎠⎛1⎞− 3z=− 1+ ln ⎜ ⎟⎝ 5 ⎠1⎛⎛1⎞⎞z =− 1 ln3 ⎜ − + ⎜ ⎟5 ⎟⎝ ⎝ ⎠⎠z = 0.86981263724.xxe5x+2− =0SolutionThis one is a little different from the previous problems in this section since it’s gotx’s both in the exponent and out of the exponent. The first step is to factor an x out ofboth terms.DO NOT DIVIDE AN x FROM BOTH TERMS!!!!5x+2x− xe= 05x+2x( 1− e ) = 0So, it’s now a little easier to deal with. From this we can see that we get one of twopossibilities.x = 0 OR5x+21− e = 0The first possibility has nothing more to do, except notice that if we had divided bothsides by an x we would have missed this one so be careful. In the second possibilitywe’ve got a little more to do. This is an equation similar to the first few that we didin this section.5x+2e = 15x+ 2 = ln15x+ 2=0Don’t forget that ln1 = 0 !x = −25© 2006 Paul Dawkins 93http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewSo, the two solutions are x = 0 and2x = − .5Solving Logarithm EquationsSolving logarithm equations are similar to exponential equations. First, we isolate thelogarithm on one side by itself with a coefficient of one. Then we use Property 4 fromthe Logarithm Properties section with an appropriate choice of b. In choosing theappropriate b, we need to remember that the b MUST match the base on the logarithm!Solve each of the following equations.1. ( x)4log 1− 5 = 2Solution© 2006 Paul Dawkins 94http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig ReviewThe first step is to divide by the 4, then we’ll convert to an exponential equation.1log ( 1− 5x)=21log( 1−5 x)210 = 101− 5x= 10Note that since we had a common log in the original equation we were forced to use abase of 10 in the exponential equation. Once we’ve used Property 4 to simplify theequation we’ve got an equation that can be solved.121− 5x= 10− 5x=− 1+10121211 ⎛ ⎞2x =− ⎜− 1+10 ⎟5 ⎝ ⎠x = -0.4324555320Now, with exponential equations we were done at this point, but we’ve got a littlemore work to do in this case. Recall the answer to the domain of a logarithm (theanswer to Problem 9 in the Logarithm Properties section). We can’t take thelogarithm of a negative number or zero.This does not mean that x = − 0.4324555320 can’t be a solution just because it’snegative number! The question we’ve got to ask is this : does this solution produce anegative number (or zero) when we plug it into the logarithms in the originalequation. In other words, is 1− 5x negative or zero if we plug x = − 0.4324555320into it? Clearly, (I hope…) 1− 5x will be positive when we plug x = − 0.4324555320in.Therefore the solution to this is x = − 0.4324555320.Note that it is possible for logarithm equations to have no solutions, so if that shouldhappen don’t get to excited!⎛ x ⎞2. 3 + 2ln ⎜ + 3⎟=−4⎝7⎠SolutionThere’s a little more simplification work to do initially this time, but it’s not too bad.© 2006 Paul Dawkins 95http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review⎛ x ⎞2ln ⎜ + 3⎟=−7⎝7⎠⎛ x ⎞ 7ln ⎜ + 3⎟=−⎝7 ⎠ 2e⎛ x ⎞ln 37⎜ + ⎟ −⎝7 ⎠ 2= e7x −2+ 3 = e7Now, solve this.7x −2+ 3 = e77x−2=− 3 + e77⎛ − ⎞2x = 7⎜− 3+e ⎟⎝ ⎠x = -20.78861832xI’ll leave it to you to check that + 3 will be positive upon plugging7x = − 20.78861832 into it and so we’ve got the solution to the equation.3. ( x) ( x)2ln −ln 1− = 2SolutionThis one is a little different from the previous two. There are two logarithms in theproblem. All we need to do is use Properties 5 – 7 from the Logarithm Propertiessection to simplify things into a single logarithm then we can proceed as we did in theprevious two problems.The first step is to get coefficients of one in front of both logs.2ln x −ln 1− x = 2( ) ( )( x) ( x)ln −ln 1− = 2Now, use Property 6 from the Logarithm Properties section to combine into thefollowing log.⎛ x ⎞ln ⎜ ⎟ = 2⎝1−x ⎠Finally, exponentiate both sides and solve.© 2006 Paul Dawkins 96http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Reviewx 2= e1−x2x= e ( 1−x)2 2x= e − e x2 2x( 1+ e ) = e2ex =21 + ex = 0.8807970780Finally, we just need to make sure that the solution, x = 0.8807970780 , doesn’tproduce negative numbers in both of the original logarithms. It doesn’t, so this is infact our solution to this problem.4. x ( x )log + log − 3 = 1SolutionThis one is the same as the last one except we’ll use Property 5 to do thesimplification instead.log x+ log x− 3 = 1( )( x( x ))log − 3 = 12log( x −3x)110 = 102x − 3x=102x −3x− 10 = 0( x− 5)( x+ 2)= 0So, potential solutions are x = 5 and x = − 2. Note, however that if we plug x = − 2into either of the two original logarithms we would get negative numbers so this can’tbe a solution. We can however, use x = 5.Therefore, the solution to this equation is x = 5.It is important to check your potential solutions in the original equation. If you checkthem in the second logarithm above (after we’ve combined the two logs) bothsolutions will appear to work! This is because in combining the two logarithmswe’ve actually changed the problem. In fact, it is this change that introduces the extrasolution that we couldn’t use!So, be careful in finding solutions to equations containing logarithms. Also, do notget locked into the idea that you will get two potential solutions and only one of thesewill work. It is possible to have problems where both are solutions and where neitherare solutions.© 2006 Paul Dawkins 97http://tutorial.math.lamar.edu/terms.aspx

Algebra/Trig Review© 2006 Paul Dawkins 98http://tutorial.math.lamar.edu/terms.aspx