Answer:
[tex]\frac{63}{65}[/tex]
Step-by-step explanation:
The angle theta terminates in quadrant 4, so we know [tex]\frac{3\pi}{2} < \theta < 2\pi[/tex] and that the sin is negative and the cos is positive.
Using the Pythagorean identity [tex]\sin^2\theta+\cos^2\theta=1[/tex], we substitute [tex]\sin\theta[/tex] to find [tex]\cos\theta[/tex] :
[tex](-\frac{16}{65})^2+\cos^2\theta=1[/tex].
Solving for [tex]\cos\theta[/tex], we have
[tex]\cos^2\theta=1-\frac{256}{4225}\\[/tex],
[tex]\cos^2\theta=\frac{3969}{4225}[/tex].
Taking the square root of both sides gives
[tex]\cos\theta=\pm\frac{63}{65}[/tex].
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
[tex]\cos\theta=\frac{63}{65}[/tex].