## Trigonometry (10th Edition)

$\sin\theta=\frac{y}{r}=\frac{-1}{3}$ $\cos\theta=\frac{x}{r}=\frac{\sqrt 8}{3}$ $\tan\theta=\frac{y}{x}=\frac{-1 }{\sqrt 8}=\frac{-\sqrt 2}{4}$ $\cot\theta=\frac{x}{y}=\frac{\sqrt 8}{-1}=-2\sqrt 2$ $\sec\theta=\frac{r}{x}=\frac{3}{\sqrt 8}=\frac{3\sqrt 2}{4}$ $\csc\theta=\frac{r}{y}=\frac{3}{-1}=-3$
1. x positive, because cos is positive; and y is negative because csc is negative y=-1 and r=3 2. Calculate x using distance formula $3=\sqrt {(x)^{2}+(1)^{2}}$ $9=1+x^{2}$ $x=\sqrt {8}$ 3. Plug those values to get final answer $\sin\theta=\frac{y}{r}=\frac{-1}{3}$ $\cos\theta=\frac{x}{r}=\frac{\sqrt 8}{3}$ $\tan\theta=\frac{y}{x}=\frac{-1 }{\sqrt 8}=\frac{-\sqrt 2}{4}$ $\cot\theta=\frac{x}{y}=\frac{\sqrt 8}{-1}=-2\sqrt 2$ $\sec\theta=\frac{r}{x}=\frac{3}{\sqrt 8}=\frac{3\sqrt 2}{4}$ $\csc\theta=\frac{r}{y}=\frac{3}{-1}=-3$