# How do you find all six trigonometric function of theta if the point (sqrt3,-1) is on the terminal side of theta?

Aug 25, 2017

Given: $\left(x , y\right) = \left(\sqrt{3} , - 1\right)$

Compute r:

#r = sqrt(x^2+y^2)

$r = \sqrt{{\left(\sqrt{3}\right)}^{2} + {\left(- 1\right)}^{2}}$

$r = \sqrt{4}$

$r = 2$

$\sin \left(\theta\right) = \frac{y}{r}$

$\sin \left(\theta\right) = - \frac{1}{2} \text{ [1]}$

$\cos \left(\theta\right) = \frac{x}{r}$

$\cos \left(\theta\right) = \frac{\sqrt{3}}{2} \text{ [2]}$

$\csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right)$

$\csc \left(\theta\right) = - 2 \text{ [3]}$

$\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right)$

$\sec \left(\theta\right) = \frac{2 \sqrt{3}}{3} \text{ [4]}$

$\tan \left(\theta\right) = \sin \left(\theta\right) \sec \left(\theta\right)$

$\tan \left(\theta\right) = - \frac{\sqrt{3}}{3} \text{ [5]}$

$\cot \left(\theta\right) = \frac{1}{\tan} \left(\theta\right)$

$\cot \left(\theta\right) = - \sqrt{3} \text{ [6]}$