Question

Given that sin theta= -16/65 and that angle theta terminates in quadrant 4, then what is the value of cosө

Answer 1

Answer:

$\frac{63}{65}$

Step-by-step explanation:

The angle theta terminates in quadrant 4, so we know $\frac{3\pi}{2} < \theta < 2\pi$ and that the sin is negative and the cos is positive.

Using the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$, we substitute $\sin\theta$ to find $\cos\theta$ :

$(-\frac{16}{65})^2+\cos^2\theta=1$.

Solving for $\cos\theta$, we have

$\cos^2\theta=1-\frac{256}{4225}$,

$\cos^2\theta=\frac{3969}{4225}$.

Taking the square root of both sides gives

$\cos\theta=\pm\frac{63}{65}$.

We found before that since the angle theta terminates in quadrant 4, the cos is positive, so we take the positive square root to get

$\cos\theta=\frac{63}{65}$.