# How do you use a half-angle formula to simplify tan 105?

Jun 11, 2015

A derivation for $\tan \left(\frac{x}{2}\right)$ is at the bottom.

$\tan \left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}}$

depending on the quadrant of ${105}^{o}$ (quadrant II, thus the horizontal axis $< 0$ on the unit circle).

$\tan {105}^{o} = \tan \left(\frac{7 \pi}{12}\right) = \tan \left(\frac{1}{2} \cdot \frac{7 \pi}{6}\right) = - \sqrt{\frac{1 - \cos \left(\frac{7 \pi}{6}\right)}{1 + \cos \left(\frac{7 \pi}{6}\right)}}$

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

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

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

$= - \sqrt{{\left(2 + \sqrt{3}\right)}^{2} / \left(\left(2 - \sqrt{3}\right) \left(2 + \sqrt{3}\right)\right)}$

$= - \frac{\sqrt{{\left(2 + \sqrt{3}\right)}^{2}}}{1}$

And since we already specified the quadrant, there's no need for $\pm$ (and $2 + \sqrt{3} > 0$ of course).

$= - \left(2 + \sqrt{3}\right)$

$= - 2 - \sqrt{3}$

You can derive the half-angle formula if you don't remember it.
${\sin}^{2} \left(x\right) = \frac{1 - \cos \left(2 x\right)}{2}$

Similarly:

${\sin}^{2} \left(\frac{x}{2}\right) = \frac{1 - \cos \left(x\right)}{2}$

Thus:

$| \sin \left(\frac{x}{2}\right) | = \sqrt{\frac{1 - \cos x}{2}}$

Similarly:

${\cos}^{2} \left(x\right) = \frac{1 + \cos \left(2 x\right)}{2}$

${\cos}^{2} \left(\frac{x}{2}\right) = \frac{1 + \cos x}{2}$

$| \cos \left(\frac{x}{2}\right) | = \sqrt{\frac{1 + \cos x}{2}}$

Now we can divide them.

$| \sin \frac{\frac{x}{2}}{\cos \left(\frac{x}{2}\right)} | = | \tan \left(\frac{x}{2}\right) | = \frac{\sqrt{\frac{1 - \cos x}{2}}}{\sqrt{\frac{1 + \cos x}{2}}}$

$= \sqrt{\frac{1 - \cos x}{1 + \cos x}}$

Jun 12, 2015

Simplify tan 105
Use trig identity:$\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a . \tan b}$

#### Explanation:

tan 105 = tan (45 + 60)

tan 45 = 1 ; tan 60 = sqrt3

$\tan 105 = \frac{1 + \sqrt{3}}{1 - \sqrt{3}}$