# How do you evaluate  sin ( arccos (15/17))?

$\sin x = \pm \frac{8}{17.}$
Call x the arc whose $\cos x = \frac{15}{17.}$
${\sin}^{2} x = 1 - {\cos}^{2} x = 1 - \frac{225}{289} = \frac{64}{289.}$
$\sin x = \pm \frac{8}{17.}$