Answer:
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
