# How do you use the point (6,-11) on the terminal side of the angle to evaluate the six trigonometric functions?

Oct 13, 2016

see below

#### Explanation:

Since every ordered pair in the unit circle has coordinates $\left(\cos \theta , \sin \theta\right)$ then the point $\left(6 , - 11\right)$ is in quadrant IV
and hence,
$\cos \theta = 6 , \sin \theta = - 11 , \tan \theta = \sin \frac{\theta}{\cos} \theta = - \frac{11}{6} ,$

$\sec \theta = \frac{1}{\cos} \theta = \frac{1}{6} , \csc \theta = \frac{1}{\sin} \theta = \frac{1}{-} 11$

$\cot \theta = \cos \frac{\theta}{\sin} \theta = \frac{6}{-} 11$