## Calculus 8th Edition

$sinx=-\frac{2\sqrt 2}{3}$ $tan x=2\sqrt 2$ $cscx=-\frac{3}{2\sqrt 2}=-\frac{3\sqrt 2}{4}$ $sec x=-3$ $cot x=\frac{1}{2\sqrt 2}=\frac{\sqrt 2}{4}$
Since $cosx=-\frac{1}{3}$, we can label the adjacent side as having length 1 and the hypotenuse as having length 5. Then Pythagorean Theorem gives opposite side $=\sqrt {3^{2}-1^{2}}=2\sqrt 2$ As $\pi0$ The other five trigonometric functions are given as follows: $sinx=-\frac{2\sqrt 2}{3}$ $tan x=2\sqrt 2$ $cscx=-\frac{3}{2\sqrt 2}=-\frac{3\sqrt 2}{4}$ $sec x=-3$ $cot x=\frac{1}{2\sqrt 2}=\frac{\sqrt 2}{4}$