## Trigonometry 7th Edition

$\sin\theta$ = $\frac{-1}{\sqrt {10}}$ $\cos\theta$ =$\frac{-3}{\sqrt {10}}$ $\tan\theta$ = $\frac{1}{3}$ $\cot\theta$ = $3$ $\sec\theta$ = -$\frac{\sqrt {10}}{3}$ $\csc\theta$ = -$\sqrt {10}$
Given, point (-3, -1) is on the terminal side of $\theta$ in standard position, we may apply Definition I to find all six trigonometric functions- We got $x = -3, y = -1$ Therefore r= $\sqrt {x^{2} + y^{2}}$ = $\sqrt {(-3)^{2} + (-1)^{2}}$ = $\sqrt {9 + 1}$ = $\sqrt {10}$ i.e. $x = -3$, $y = -1,$ and $r= \sqrt {10}$ Applying Definition I- $\sin\theta$ =$\frac{y}{r}$ = $\frac{-1}{\sqrt {10}}$ $\cos\theta$ =$\frac{x}{r}$ =$\frac{-3}{\sqrt {10}}$ $\tan\theta$ =$\frac{y}{x}$ =$\frac{-1}{-3}$ = $\frac{1}{3}$ $\cot\theta$ =$\frac{x}{y}$ =$\frac{-3}{-1}$ = $3$ $\sec\theta$ =$\frac{r}{x}$ =$\frac{\sqrt {10}}{-3}$ = -$\frac{\sqrt {10}}{3}$ $\csc\theta$ =$\frac{r}{y}$ =$\frac{\sqrt {10}}{-1}$= -$\sqrt {10}$