Answer
$\sin\theta=\frac{y}{r}=\frac{-4}{5}$
$\cos\theta=\frac{x}{r}=\frac{-3}{5}$
$\tan\theta=\frac{y}{x}=\frac{4}{3}$
$\cot\theta=\frac{x}{y}=\frac{3}{4}$
$\sec\theta=\frac{r}{x}=\frac{5}{-3}$
$\csc\theta=\frac{r}{y}=\frac{5}{-4}$
Work Step by Step
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
$\sin\theta=\frac{y}{r}=\frac{-4}{5}$
$\cos\theta=\frac{x}{r}=\frac{-3}{5}$
$\tan\theta=\frac{y}{x}=\frac{4}{3}$
$\cot\theta=\frac{x}{y}=\frac{3}{4}$
$\sec\theta=\frac{r}{x}=\frac{5}{-3}$
$\csc\theta=\frac{r}{y}=\frac{5}{-4}$
