# How do you use the point (9,-5) on the terminal side of the angle to evaluate the six trigonometric functions?

Aug 16, 2016

Consider the following diagram:

The diagram is just to say that we know the two legs of the triangle, which lie in quadrant four. In other words, we can already state $\tan \alpha$ and $\cot \alpha$, but to find the other four trigonometric functions we need to know the length of the hypotenuse, or $r$.

We can do this by pythagorean theorem:

${r}^{2} = {\left(- 5\right)}^{2} + {9}^{2}$

$r = \sqrt{25 + 81}$

$r = \sqrt{106}$

Now, we can state the value of all 6 trigonometric ratios.

$\sin \alpha = \text{opposite"/"hypotenuse} = - \frac{5}{\sqrt{106}} = \frac{- 5 \sqrt{106}}{106}$

$\csc \alpha = \frac{1}{\text{opposite"/"hypotenuse}} = \frac{1}{- \frac{5}{\sqrt{106}}} = - \frac{\sqrt{106}}{5}$

$\cos \alpha = \text{adjacent/hypotenuse} = \frac{9}{\sqrt{106}} = \frac{9 \sqrt{106}}{106}$

$\sec \alpha = \frac{1}{\text{adjacent"/"hypotenuse}} = \frac{1}{\frac{9}{\sqrt{106}}} = \frac{\sqrt{106}}{9}$

$\tan \alpha = \text{opposite"/"adjacent} = - \frac{5}{9}$

$\cot \alpha = \frac{1}{\text{opposite"/"adjacent}} = - \frac{9}{5}$

Hopefully this helps!