Question

1.)The angle θ lies in Quadrant IV. sinθ=−2/3 What is cosθ?

a. −2/3
b. 2/3
c. −5/√3
d.5/√3


2.)The angle θ lies in Quadrant I. cosθ=3/5 What is tanθ?

a.4/3
b.−4/5
c. −4/3
d. 4/5

3.)Given sin(−θ)=−1/6 and tanθ=−35/√35. What is the value of cosθ?

a.−35/√210
b.−35/√6
c.35/√210
d.35/√6


4.)cos(−θ)=3/√4, sinθ<0 What is the value of sinθ?

a.13/√4
b. 13/√16
c. −13/√16
d.−13/√4


5.)Marty is proving that the following trigonometric identity is true: tan^2 θ⋅cos^2 θ=1−cos^2 θ

Which step would be the first line of his proof?

a. tan^2 θ⋅cos^2 θ=tan^2 θ
b. tan^2 θ⋅cos^2 θ=sin^2 θ
c.tan^2 θ⋅cos^2 θ=1−sin^2 θ
d.tan^2 θ=1−cos^2 θ⋅cos^2 θ

Answer 1

1. From sin²θ +cos²θ =1 and sinθ=-2/3, we see that cosθ=√(1-sin²θ) or cosθ=√5/3, where the sign of cosine is positive as it is in Quadrant IV. x lies in 4th quadrant , cos x is +ve. , cos x = √5/3. Answer.

answer : cos x = √5/3

2. 4/3

3. sin (- theta) = - sin (x) so sin x = 1/6

tan = sin / cos = 1/6 / cos = - sqrt35/35 solve for cos

cos = 1/6 * (-35/sqrt35)

= -35 sqrt35 /210

answer : −35/√210

4. The cosine function is an even function, so cos(θ) = cos(-θ).

The relationship between sin(θ) and cos(θ) is sin(θ) = ±√(1 -cos(θ)^2)

For sin(θ) < 0 and cos(θ) = (√3)/4, sin(θ) = -√(1 -3/16) = -√(13/16)

sin(θ) = -(√13)/4 For sin(θ) < 0 and cos(0) = √(3/4), ...

sin(θ) = -√(1 -3/4) = -√(1/4) sin(θ) = -1/2

answer : -13/√4

5. answer : tan^2 θ ⋅ cos^2 θ = 1 − cos^2 θ would be the first step