## Trigonometry 7th Edition

$\sin\theta$ = $\frac{\sqrt 7}{4}$ $\cos\theta$ = -$\frac{3}{4}$ $\tan\theta$ = - $\frac{\sqrt 7}{3}$ $\cot\theta$ = - $\frac{3}{\sqrt 7}$ $\sec\theta$ = -$\frac{4}{3}$ $\csc\theta$ =$\frac{4}{\sqrt 7}$
Given, point $(-3, \sqrt 7)$ is on the terminal side of $\theta$, we may apply Definition I to find all six trigonometric functions- We got $x = -3, y = \sqrt 7$ Therefore r= $\sqrt (x^{2} + y^{2})$ = $\sqrt ((-3)^{2} + (\sqrt 7)^{2})$ = $\sqrt (9 + 7)$ = $\sqrt (16)$ = 4 i.e. $x = -3, y = \sqrt 7,$ and $r= 4$ Applying Definition I- $\sin\theta$ =$\frac{y}{r}$ = $\frac{\sqrt 7}{4}$ $\cos\theta$ =$\frac{x}{r}$ =$\frac{-3}{4}$ = -$\frac{3}{4}$ $\tan\theta$ =$\frac{y}{x}$ =$\frac{\sqrt 7}{-3}$ = - $\frac{\sqrt 7}{3}$ $\cot\theta$ =$\frac{x}{y}$ =$\frac{-3}{\sqrt 7}$ = - $\frac{3}{\sqrt 7}$ $\sec\theta$ =$\frac{r}{x}$ =$\frac{4}{-3}$ = -$\frac{4}{3}$ $\csc\theta$ =$\frac{r}{y}$ =$\frac{4}{\sqrt 7}$