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

Answer:

$\cos(\theta)= \frac{\sqrt{377}}{29}$

Step-by-step explanation:

Given

$\sin(\theta) = -\frac{4}{\sqrt{29}}$ -- 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: $\sin(\theta) = -\frac{4}{\sqrt{29}}$

$\cos^2(\theta)= 1 - (-\frac{4}{\sqrt{29}})^2$

Evaluate all squares

$\cos^2(\theta)= 1 - (\frac{16}{29})$

Take LCM

$\cos^2(\theta)= \frac{29 - 16}{29}$

$\cos^2(\theta)= \frac{13}{29}$

Take square roots of both sides

$\cos(\theta)= \±\sqrt{\frac{13}{29}}$

cosine is positive in the 4th quadrant;

So:

$\cos(\theta)= \sqrt{\frac{13}{29}}$

Split

$\cos(\theta)= \frac{\sqrt{13}}{\sqrt{29}}$

Rationalize

$\cos(\theta)= \frac{\sqrt{13}}{\sqrt{29}} * \frac{\sqrt{29}}{\sqrt{29}}$

$\cos(\theta)= \frac{\sqrt{13*29}}{29}$

$\cos(\theta)= \frac{\sqrt{377}}{29}$