How do you use the half angle identify to find the exact value of tan22.5^circ?

Nov 16, 2016

The half angle is, tan22.5º=(-1+sqrt2)/2

Explanation:

We start from the definition of $\tan \theta$

$\tan \theta = \sin \frac{\theta}{\cos} \theta = \frac{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{{\cos}^{2} \left(\frac{\theta}{2}\right) - {\sin}^{2} \left(\frac{\theta}{2}\right)}$

Dividing by ${\cos}^{2} \left(\frac{\theta}{2}\right)$

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

$= \frac{2 \tan \left(\frac{\theta}{2}\right)}{1 - {\tan}^{2} \left(\frac{\theta}{2}\right)}$

Let $\tan \left(\frac{\theta}{2}\right) = t$

We need $\tan 22.5 = \tan \left(\frac{\theta}{2}\right)$

tantheta=tan45º=1

Therefore
$1 = \frac{2 t}{1 - {t}^{2}}$

$1 - {t}^{2} = 2 t$

${t}^{2} + 2 t - 1 = 0$

We solve this quadratic equation with

$\Delta = 4 + 4 = 8$

$\Delta > 0$, therefore 2 real roots

$t = \frac{- 2 \pm \left(\sqrt{8}\right)}{2} = \left(- 1 \pm \sqrt{2}\right)$

We keep the positive root

t=(-1+sqrt2)/2 = tan22.5º