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College Trigonometry, 2011a

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838 Foundations of <strong>Trigonometry</strong><br />

find the angle of inclination, labeled θ below, satisfies tan(θ) = 6<br />

12 = 1 2<br />

.Sinceθ is an acute angle,<br />

we can use the arctangent function and we find θ =arctan ( 1<br />

2)<br />

radians ≈ 26.56 ◦ .<br />

6 feet<br />

θ<br />

12 feet<br />

10.6.4 Solving Equations Using the Inverse Trigonometric Functions.<br />

In Sections 10.2 and 10.3, we learned how to solve equations like sin(θ) = 1 2<br />

for angles θ and<br />

tan(t) =−1 for real numbers t. In each case, we ultimately appealed to the Unit Circle and relied<br />

on the fact that the answers corresponded to a set of ‘common angles’ listed on page 724. If, on<br />

the other hand, we had been asked to find all angles with sin(θ) = 1 3<br />

or solve tan(t) =−2 for<br />

real numbers t, we would have been hard-pressed to do so. With the introduction of the inverse<br />

trigonometric functions, however, we are now in a position to solve these equations. A good parallel<br />

to keep in mind is how the square root function can be used to solve certain quadratic equations.<br />

The equation x 2 = 4 is a lot like sin(θ) = 1 2<br />

in that it has friendly, ‘common value’ answers x = ±2.<br />

The equation x 2 = 7, on the other hand, is a lot like sin(θ) = 1 3 .Weknow8 there are answers, but<br />

we can’t express them using ‘friendly’ numbers. 9 To solve x 2 =7,wemakeuseofthesquareroot<br />

function and write x = ± √ 7. We can certainly approximate these answers using a calculator, but<br />

as far as exact answers go, we leave them as x = ± √ 7. Inthesameway,wewillusethearcsine<br />

function to solve sin(θ) = 1 3<br />

, as seen in the following example.<br />

Example 10.6.7. Solve the following equations.<br />

1. Find all angles θ for which sin(θ) = 1 3 .<br />

2. Find all real numbers t for which tan(t) =−2<br />

3. Solve sec(x) =− 5 3<br />

for x.<br />

Solution.<br />

1. If sin(θ) = 1 3<br />

, then the terminal side of θ, when plotted in standard position, intersects the<br />

Unit Circle at y = 1 3<br />

. Geometrically, we see that this happens at two places: in Quadrant I<br />

and Quadrant II. If we let α denote the acute solution to the equation, then all the solutions<br />

8 How do we know this again?<br />

9 This is all, of course, a matter of opinion. For the record, the authors find ± √ 7 just as ‘nice’ as ±2.

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