# The point (-4,1) 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?

Jun 11, 2018

$\sin t = \frac{1}{\sqrt{17}}$
$\cos t = - \frac{4}{\sqrt{17}}$

#### Explanation:

Point (x = -4, y = 1) on the terminal side is in the Quadrant 2.
Call t the angle (arc):
$\tan t = \frac{y}{x} = - \frac{1}{4}$
${\cos}^{2} t = \frac{1}{1 + {\tan}^{2} t} = \frac{1}{1 + \frac{1}{16}} = \frac{16}{17}$
$\cos t = \pm \frac{4}{\sqrt{17}}$
Since t lies in Quadrant 2 --> cos t is negative.
$\cos t = - \frac{4}{\sqrt{17}}$
${\sin}^{2} t = 1 - {\cos}^{2} t = 1 - \frac{16}{17} = \frac{1}{17}$
$\sin t = \pm \frac{1}{\sqrt{17}}$
Since t lies in Quadrant 2 --> sin t is positive
$\sin t = \frac{1}{\sqrt{17}}$