# The terminal side of theta in standard position contains (-6,8), how do you find the exact values of the six trigonometric functions of theta?

Feb 28, 2018

See explanation.

#### Explanation:

If the point in the angle's terminal side is $P = \left(x , y\right)$ then the trigonometric functions can be calculated as:

## $\csc \alpha = \frac{r}{y}$

where $r = \sqrt{{x}^{2} + {y}^{2}}$

For the given point we have:

$r = \sqrt{{\left(- 6\right)}^{2} + {8}^{2}} = \sqrt{36 + 64} = 10$

So the functions are:

$\sin \alpha = \frac{8}{10} = \frac{4}{5}$

$\cos \alpha = \frac{- 6}{10} = - \frac{3}{5}$

$\tan \alpha = \frac{8}{- 6} = - \frac{4}{3}$

$\cot \alpha = \frac{- 6}{8} = - \frac{3}{4}$

$\sec \alpha = \frac{10}{- 6} = - \frac{5}{3} = - 1 \frac{2}{3}$

$\csc \alpha = \frac{10}{8} = \frac{5}{4} = 1 \frac{1}{4}$