Question

consider this equation sin(theta) = -4 square root of 29/29 if theta is an angle in quadrant IV what is the value of cos (theta)

Answer 1

`\cos(\theta)= (√(377))/(29)`

Explanation:

Given

`\sin(\theta) = -(4)/(√(29))` -- the correct expression

Required

`\cos(\theta)`

We know that:

`\sin^2(\theta) + \cos^2(\theta)= 1`

Make
`\cos^2(\theta)` the subject

`\cos^2(\theta)= 1 - \sin^2(\theta)`

Substitute:
`\sin(\theta) = -(4)/(√(29))`

`\cos^2(\theta)= 1 - (-(4)/(√(29)))^2`

Evaluate all squares

`\cos^2(\theta)= 1 - ((16)/(29))`

Take LCM

`\cos^2(\theta)= (29 - 16)/(29)`

`\cos^2(\theta)= (13)/(29)`

Take square roots of both sides

`\cos(\theta)= \±\sqrt{(13)/(29)}`

cosine is positive in the 4th quadrant;

So:

`\cos(\theta)= \sqrt{(13)/(29)}`

Split

`\cos(\theta)= (√(13))/(√(29))`

Rationalize

`\cos(\theta)= (√(13))/(√(29)) * (√(29))/(√(29))`

`\cos(\theta)= (√(13*29))/(29)`

`\cos(\theta)= (√(377))/(29)`