Given a positive angle theta in standard position with the terminal side passing through (1, -3). How do you find the six trigonometric functions of theta?

Nov 7, 2016

Please see the explanation for steps leading to all 6 functions.

Explanation:

Given: $x = 1 , \mathmr{and} y = - 3$

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

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

$r = \sqrt{10}$

$x = r \cos \left(\theta\right) \mathmr{and} y = r \sin \left(\theta\right)$

$1 = \sqrt{10} \cos \left(\theta\right)$

$\cos \left(\theta\right) = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$

$- 3 = \sqrt{10} \sin \left(\theta\right)$

$\sin \left(\theta\right) = - \frac{3}{\sqrt{10}} = - 3 \frac{\sqrt{10}}{10}$

$\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right) = \frac{- \frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}} = - 3$

$\cot \left(\theta\right) = \frac{1}{\tan} \left(\theta\right) = - \frac{1}{3}$

$\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right) = \sqrt{10}$

$\csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right) = - \frac{\sqrt{10}}{3}$