# How do you use a half-angle formula to simplify tan 195?

##### 1 Answer
Jun 15, 2015

$\tan \left({195}^{o}\right) = \tan \left({390}^{o} / 2\right)$

The identity you need is:

$\tan \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}}$
$+$ if quadrant I or III
$-$ if quadrant II or IV

It's derived from:

$\sin \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos x}{2}}$
$+$ if quadrant I or II
$-$ if quadrant III or IV

$\cos \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 + \cos x}{2}}$
$+$ if quadrant I or IV
$-$ if quadrant II or III

Divide the quadrant conditions to get the sign you need. $\tan {390}^{o} = \tan {30}^{o}$, so you're in quadrant I. Thus, no matter what, the sign of the answer is positive.

$\tan \left({390}^{o} / 2\right) = \sqrt{\frac{1 - \cos {390}^{o}}{1 + \cos {390}^{o}}}$

$= \sqrt{\frac{1 - \cos {30}^{o}}{1 + \cos {30}^{o}}}$

$= \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{\sqrt{3}}{2}}}$

Get common denominators:
$= \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{\frac{2 + \sqrt{3}}{2}}}$

Cancel:
$= \sqrt{\frac{2 - \sqrt{3}}{2 + \sqrt{3}}}$

Multiply by the conjugate of the denominator:
$= \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}} \cdot \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 - \sqrt{3}}}$

Simplify:
$= \frac{2 - \sqrt{3}}{\sqrt{{2}^{2} - {\left(\sqrt{3}\right)}^{2}}}$

$= \frac{2 - \sqrt{3}}{\sqrt{1}}$

$= 2 - \sqrt{3}$