# How do you use the unit circle to find the values of the six trigonometric functions for 150 degrees?

Jun 17, 2018

$\cos 150 = - \frac{\sqrt{3}}{2}$

$\sin 150 = \frac{1}{2}$

$\tan 150 = - \frac{\sqrt{3}}{3}$

$\cot 150 = - \sqrt{3}$

$\sec 150 = \frac{- 2 \sqrt{3}}{3}$

$\csc 150 = 2$

#### Explanation:

We know the reference angle is ${30}^{\circ}$, and from the unit circle, we know the coordinates for ${30}^{\circ}$ are

$\left(\frac{\sqrt{3}}{2} , \frac{1}{2}\right)$

Our angle, ${150}^{\circ}$ is in the second quadrant, where cosine is negative and sine is positive. Unit circle coordinates are given by

$\left(\cos \theta , \sin \theta\right)$

This means the coordinates for ${150}^{\circ}$ are

$\left(- \frac{\sqrt{3}}{2} , \frac{1}{2}\right)$

We know:

$\textcolor{b l u e}{\cos 150 = - \frac{\sqrt{3}}{2}}$

$\textcolor{\mathrm{da} r k b l u e}{\sin 150 = \frac{1}{2}}$

$\textcolor{\lim e}{\tan \theta} = \frac{\textcolor{\mathrm{da} r k b l u e}{\sin \theta}}{\textcolor{b l u e}{\cos \theta}}$

And from our definitions of trig functions:

$\textcolor{red}{\cot \theta} = \frac{1}{\textcolor{\lim e}{\tan \theta}}$

$\textcolor{\mathrm{da} r k v i o \le t}{\sec \theta} = \frac{1}{\textcolor{b l u e}{\cos \theta}}$

$\textcolor{\mathmr{and} a n \ge}{\csc \theta} = \frac{1}{\textcolor{\mathrm{da} r k b l u e}{\sin \theta}}$

After plugging in the appropriate values (and rationalizing the denominator when necessary), we get

$\textcolor{\lim e}{\tan 150 = - \frac{\sqrt{3}}{3}}$

$\textcolor{red}{\cot 150 = - \sqrt{3}}$

$\textcolor{\mathrm{da} r k v i o \le t}{\sec 150 = \frac{- 2 \sqrt{3}}{3}}$

$\textcolor{\mathmr{and} a n \ge}{\csc 150 = 2}$

Hope this helps!