396. Find the values of $\sin x, \cos x, \operatorname{ctg} x$, if it is known that $\operatorname{tg} x=-8 / 15,3 \pi / 2<x<2 \pi$.
Solution. First, find $\operatorname{ctg} x = -15 / 8$. Since the angle $x$ ends in the IV. quadrant, we conclude that $\sin x < 0$. According to identity V, we have $1 + \operatorname{tg}^{2} x = \frac{1}{\cos ^{2} x}$, from which $\cos ^{2} x = \frac{1}{1 + \operatorname{tg}^{2} x}$. Therefore, $\cos ^{2} x = \frac{1}{1 + \frac{64}{225}} = \frac{1}{\frac{289}{225}} = \frac{225}{289}$, i.e., $\cos x = \frac{15}{17}$. Using the formula $\sin ^{2} x = 1 - \cos ^{2} x$, we get $\sin ^{2} x = 1 - \frac{225}{289} = \frac{64}{289}$, i.e., $\sin x = -\frac{8}{17}$.

$397-400$. Find the values of all trigonometric functions if it is known that: