# The point (-4,10) is on the terminal side of an angle in standard position, how do you determine the exact values of the six trigonometric functions of the angle?

Dec 14, 2016

#### Explanation:

Let $x = - 4$

Let $y = 10$

Let $r =$ the length of a line segment drawn from the origin to the point:

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

$r = \sqrt{{\left(- 4\right)}^{2} + {10}^{2}}$

$r = \sqrt{116} = 2 \sqrt{29}$

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

$\sin \left(\theta\right) = \frac{10}{2 \sqrt{29}}$

$\sin \left(\theta\right) = \frac{5 \sqrt{29}}{29}$

$\csc \left(\theta\right) = \frac{1}{\sin} \left(\theta\right) = \frac{\sqrt{29}}{5}$

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

$\cos \left(\theta\right) = - \frac{4}{2 \sqrt{29}}$

$\cos \left(\theta\right) = - \frac{2 \sqrt{29}}{29}$

$\sec \left(\theta\right) = \frac{1}{\cos} \left(\theta\right) = - \frac{\sqrt{29}}{2}$

$\tan \left(\theta\right) = \frac{y}{x}$

$\tan \left(\theta\right) = \frac{10}{-} 4$

$\tan \left(\theta\right) = - 2.5$

$\cot \left(\theta\right) = \frac{1}{\tan} \left(\theta\right) = - \frac{2}{5}$