# How do you find the exact value of tan 150 using the half angle identity?

##### 1 Answer
Jul 5, 2017

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

#### Explanation:

Call tan 150 = tan t -->
$\tan 2 t = \tan 300 = \tan \left(- 60 + 360\right) = \tan \left(- 60\right) = - \sqrt{3}$
Use trig identity:
$\tan 2 t = \frac{2 \tan t}{1 - {\tan}^{2} t}$
In this case:
$\frac{2 \tan t}{1 - {\tan}^{2} t} = - \sqrt{3}$.
Cross multiply -->
$- \sqrt{3} + \sqrt{3} {\tan}^{2} t = 2 \tan t$
$\sqrt{3} {\tan}^{2} t - 2 \tan t - \sqrt{3} = 0$.
Solve this quadratic equation for tan t.
$D = {d}^{2} = {b}^{2} - 4 a c = 4 + 12 = 16$ --> $d = \pm 4$
There are 2 real roots:
$\tan t = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{2}{2} \sqrt{3} \pm \frac{4}{2} \sqrt{3} = \frac{\sqrt{3}}{3} \pm 2 \frac{\sqrt{3}}{3}$
a. $\tan t = \tan 150 = \frac{3 \sqrt{3}}{3} = \sqrt{3}$ (rejected because tan 150 is negative)
b. $\tan t = \tan 150 = - \frac{\sqrt{3}}{3} = - 0.577$
Check by calculator.
$\tan 150 = - 0.577 = - \frac{\sqrt{3}}{3}$. Proved.