College Trigonometry, 2011a

978 Applications of Trigonometry
2. From 16x 2 +24xy +9y 2 +15x − 20y =0,wegetA = 16, B =24andC =9sothat
cot(2θ) = 7
24
. Since this isn’t one of the values of the common angles, we will need to use
inverse functions. Ultimately, we need to find cos(θ) and sin(θ), which means we have two
options. If we use the arccotangent function immediately, after the usual calculations we
get θ = 1 2 arccot ( 7
24)
. To get cos(θ) andsin(θ) from this, we would need to use half angle
identities. Alternatively, we can start with cot(2θ) = 7
24
, use a double angle identity, and
then go after cos(θ) and sin(θ). We adopt the second approach. From cot(2θ) = 7
24 ,wehave
tan(2θ) = 24
7 . Using the double angle identity for tangent, we have 2 tan(θ)
1−tan 2 (θ) = 24
7 ,which
gives 24 tan 2 (θ)+14tan(θ)−24 = 0. Factoring, we get 2(3 tan(θ)+4)(4 tan(θ)−3) = 0 which
gives tan(θ) =− 4 3 or tan(θ) = 3 4
. While either of these values of tan(θ) satisfies the equation
cot(2θ) = 7
24 , we choose tan(θ) = 3 4 , since this produces an acute angle,4 θ =arctan ( 3
4)
.To
find the rotation equations, we need cos(θ) =cos ( arctan ( ( (
3
4))
and sin(θ) =sin arctan
3
4))
.
Using the techniques developed in Section 10.6 we get cos(θ) = 4 5 and sin(θ) = 3 5
. Our rotation
equations are x = x ′ cos(θ) − y ′ sin(θ) = 4x′
5 − 3y′
5
and y = x ′ sin(θ) +y ′ cos(θ) = 3x′
5 + 4y′
5 .
As usual, we now substitute these quantities into 16x 2 +24xy +9y 2 +15x − 20y = 0 and
simplify. As a first step, the reader can verify
x 2 = 16(x′ ) 2
− 24x′ y ′ ) 2
25 25 +9(y′ 25 , xy = 12(x′ ) 2
+ 7x′ y ′ ) 2
25 25 −12(y′ , y 2 = 9(x′ ) 2 y ′ ) 2
25
25 +24x′ 25 +16(y′ 25
Once the dust settles, we get 25(x ′ ) 2 − 25y ′ =0,ory ′ =(x ′ ) 2 , whose graph is a parabola
opening along the positive y ′ -axis with vertex (0, 0). We graph this equation below.
y
y
x ′
y ′ x ′
θ = arctan ( ) 3
4
y ′ θ = π 4
x
x
5x 2 +26xy +5y 2 − 16x √ 2+16y √ 2 − 104 = 0
16x 2 +24xy +9y 2 +15x − 20y =0
4 As usual, there are infinitely many solutions to tan(θ) = 3 . We choose the acute angle θ = arctan ( )
3
4 4 . The
reader is encouraged to think about why there is always at least one acute answer to cot(2θ) = A−C and what this
B
means geometrically in terms of what we are trying to accomplish by rotating the axes. The reader is also encouraged
to keep a sharp lookout for the angles which satisfy tan(θ) =− 4 in our final graph. (Hint: ( )( )
3
3 4 −
4
3 = −1.)