# The point (-1,1) 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?

May 25, 2018

#### Explanation:

Given: terminal side point $\left(- 1 , 1\right)$

The point $\left(- 1 , 1\right)$ is in the 2nd quadrant.

It has the $x$ value $= - 1$ and the $y$-value $= 1$. Drop the point to the $x$-axis and form a reference triangle.

Using the Pythagorean Theorem $r = \sqrt{{\left(- 1\right)}^{2} + {1}^{2}} = \sqrt{2}$

$\sin \theta = \frac{y}{r} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

$\cos \theta = \frac{x}{r} = - \frac{1}{\sqrt{2}} = - \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = - \frac{\sqrt{2}}{2}$

$\tan \theta = \frac{y}{x} = \frac{1}{-} 1 = - 1$

$\csc \theta = \frac{1}{\sin \theta} = \frac{r}{y} = \frac{\sqrt{2}}{1} = \sqrt{2}$

$\sec \theta = \frac{1}{\cos \theta} = \frac{r}{x} = \frac{\sqrt{2}}{-} 1 = - \sqrt{2}$

$\cot \theta = \frac{1}{\tan \theta} = \frac{x}{y} = - \frac{1}{1} = - 1$