The point (-12,-5) is on the terminal side of an angle in standard position, how do you determine the exact values of the six trigonometric functions of the angle?

May 2, 2017

See explanation.

Explanation:

To calculate the values of trigonometric functions you have to calculate the distance between the point ant the origin:

$r = \sqrt{{\left(- 12 - 0\right)}^{2} + {\left(- 5 - 0\right)}^{2}} = \sqrt{144 + 25} = \sqrt{169} = 13$

Now we can calculate the trigonometric functions:

$\sin \alpha = \frac{y}{r} = \frac{- 5}{13} = - \frac{5}{13}$

$\cos \alpha = \frac{x}{r} = \frac{- 12}{13} = - \frac{12}{13}$

$\tan \alpha = \frac{y}{x} = \frac{- 5}{- 12} = \frac{5}{12}$

$\cot \alpha = \frac{x}{y} = \frac{- 12}{- 5} = \frac{12}{5} = 2 \frac{2}{5}$

$\sec \alpha = \frac{r}{x} = \frac{13}{- 12} = - 1 \frac{1}{12}$

$\csc \alpha = \frac{r}{y} = \frac{13}{- 5} = - 2 \frac{3}{5}$