## Trigonometry (10th Edition)

$\sin\theta=\frac{y}{r}=\frac{-4}{5}$ $\cos\theta=\frac{x}{r}=\frac{-3}{5}$ $\tan\theta=\frac{y}{x}=\frac{4}{3}$ $\cot\theta=\frac{x}{y}=\frac{3}{4}$ $\sec\theta=\frac{r}{x}=\frac{5}{-3}$ $\csc\theta=\frac{r}{y}=\frac{5}{-4}$
1. The angle is in the III quadrant, therefore, x and y are negative. 2.Lets find the y, by using the x, r and distance formula $r=\sqrt {(-3)^{2}+(y)^{2}} =5$ $25=9+y^{2}$ $y^{2}=16$ $y=-4$ 3. Insert the values to find trig functions $\sin\theta=\frac{y}{r}=\frac{-4}{5}$ $\cos\theta=\frac{x}{r}=\frac{-3}{5}$ $\tan\theta=\frac{y}{x}=\frac{4}{3}$ $\cot\theta=\frac{x}{y}=\frac{3}{4}$ $\sec\theta=\frac{r}{x}=\frac{5}{-3}$ $\csc\theta=\frac{r}{y}=\frac{5}{-4}$