# How do you find the exact value of the six trigonometric functions of the angle whose terminal side passes through (-10/3, -2/3)?

Feb 27, 2018

As detailed below.

#### Explanation:

r = sqrt(x^2 + y^2) = sqrt((-10/3)^2 + (-2/3)^2) ) = sqrt(104/9)

$\sin \theta = \frac{y}{r} = \frac{- \frac{2}{3}}{\sqrt{\frac{104}{9}}} = - \frac{1}{\sqrt{26}}$

$\csc \theta = \frac{1}{\sin} \theta = - \sqrt{26}$

$\cos \theta = \frac{x}{r} = \frac{- \frac{10}{3}}{\sqrt{\frac{104}{9}}} = - \frac{5}{\sqrt{26}}$

$\sec \theta = \frac{1}{\cos} \theta = - \frac{\sqrt{26}}{5}$

$\tan \theta = \frac{y}{x} = - \frac{\frac{2}{3}}{-} \left(\frac{10}{3}\right) = \frac{1}{5}$

$\cot \theta = \frac{1}{\tan} \theta = 5$