Half Angle Formulas

Theorem

 * $(1): \quad \sin \dfrac \theta 2 = \pm \sqrt {\dfrac {1 - \cos \theta} {2}}$
 * $(2): \quad \cos \dfrac \theta 2 = \pm \sqrt {\dfrac {1 + \cos \theta} {2}}$
 * $(3): \quad \tan \dfrac \theta 2 = \dfrac {\sin \theta} {1 + \cos \theta} = \dfrac {1 - \cos \theta} {\sin \theta}$

Proof
Define:


 * $u = \dfrac \theta 2$

Since $\cos \theta \ge -1$, it follows that $\cos \theta + 1 \ge 0$.

We also have:
 * $\sin \theta > 0$ if $\theta$ is in the first or second quadrant and:
 * $\sin \theta < 0$ if $\theta$ is in the third or fourth quadrant.

But $\displaystyle \tan \frac \theta 2 \ge 0$ if $\theta$ is in the first or second quadrant (because $\tan\theta > 0$ if $\theta$ is in the first quadrant), and $\displaystyle \tan \frac \theta 2 < 0$ if $\theta$ is in the third or fourth quadrant (because $\tan \theta < 0$ if $\theta$ is in the second quadrant).

Thus, $\displaystyle \tan \frac \theta 2$ and $\sin \theta$ have the same sign, so we can drop the $\pm$, and we obtain:
 * $\displaystyle \tan \frac \theta 2 = \frac {\sin \theta} {1 + \cos \theta}$

If we had proceeded by writing $\displaystyle \tan \frac \theta 2 = \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta}} = \pm \sqrt {\frac {\left({1 - \cos \theta}\right)^2} {\left({1 + \cos \theta}\right) \left({1 - \cos \theta}\right)} }$, we would have ended up with:
 * $\displaystyle \tan \frac \theta 2 = \frac {1 - \cos \theta} {\sin \theta}$

Note
Technically, we should also check the boundaries between the first and fourth quadrants and the second and third quadrants.

If $\theta = \pi + 2k \pi, k \in \Z$, then $\displaystyle \tan\frac{\theta}{2}$ is undefined, $\displaystyle \frac{\sin\theta}{1 + \cos \theta}$ is undefined, $\displaystyle \frac{1 - \cos \theta}{\sin \theta}$ is undefined.

If $\theta = 2k \pi, k \in \Z$, then $\displaystyle \tan\frac{\theta}{2} = 0$, $\displaystyle \frac{\sin\theta}{1+\cos\theta} = 0$, and $\displaystyle \frac{1-\cos\theta}{\sin\theta}$ is undefined (although by L'Hôpital's Rule, $\displaystyle \lim_{\theta \to 0}\frac{1-\cos\theta}{\sin\theta} = 0$).

Thus, $\displaystyle \frac {1 - \cos \theta} {\sin \theta}$ is not a perfect formula for $\displaystyle \tan \frac \theta 2$.