How do you derive the half angle formula for sin,cos and tan?

Jun 12, 2015

You can start from ${\sin}^{2} \left(x\right)$.

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

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

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

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

Therefore:
$\tan \left(\frac{x}{2}\right) = \sin \frac{\frac{x}{2}}{\cos} \left(\frac{x}{2}\right) = \left[\frac{\pm \sqrt{\frac{1 - \cos x}{2}}}{\pm \sqrt{\frac{1 + \cos x}{2}}}\right]$

$= \pm \sqrt{\frac{1 - \cos x}{1 + \cos x}}$
$+$ if in quadrant I or III
$-$ if in quadrant II or IV

In quadrant I, the division gives $\left(\frac{+}{+}\right) = +$

In quadrant III, the division gives $\left(\frac{-}{-}\right) = +$

In quadrant II, the division gives $\left(\frac{+}{-}\right) = -$

In quadrant IV, the division gives $\left(\frac{-}{+}\right) = -$