Question

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

Answer 1

`(63)/(65)`

Explanation:

The angle theta terminates in quadrant 4, so we know
`(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` :

`(-(16)/(65))^2+\cos^2\theta=1`.

Solving for
`\cos\theta`, we have

`\cos^2\theta=1-(256)/(4225)\\`,

`\cos^2\theta=(3969)/(4225)`.

Taking the square root of both sides gives

`\cos\theta=\pm(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=(63)/(65)`.