# How do you use the point (7,-2) on the terminal side of the angle to evaluate the six trigonometric functions?

Nov 29, 2016

See explanation.

#### Explanation:

The trigonometric functions are defined as quotients of specified coordinates $x$ and $y$ and the distance between the point and the origin $r$. So first we have to calculate $r$:

$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{7}^{2} + {\left(- 2\right)}^{2}} = \sqrt{49 + 4} = \sqrt{53}$

Now we can calculate the functions according to their definitions:

$\sin \alpha = \frac{y}{r} = \frac{- 2}{\sqrt{53}} = \frac{- 2 \sqrt{53}}{53}$

$\cos \alpha = \frac{x}{r} = \frac{7}{\sqrt{53}} = \frac{7 \sqrt{53}}{\sqrt{53}}$

$\tan \alpha = \frac{y}{x} = \frac{- 2}{7} = - \frac{2}{7}$

$\cot \alpha = \frac{x}{y} = \frac{7}{- 2} = - \frac{7}{2}$

$\sec \alpha = \frac{r}{x} = \frac{\sqrt{53}}{7}$

$\csc \alpha = \frac{r}{y} = \frac{\sqrt{53}}{-} 2 = - \frac{\sqrt{53}}{2}$