Half Angle Formulas

Theorem

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

Proof
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 quadrants and $\sin\theta < 0$ if $\theta$ is in the third or fourth quadrants

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

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

If, when we had $\displaystyle \tan\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$, we had instead multiplied by $\sqrt{1-\cos\theta}$, we end 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 $\displaystyle \tan\frac{\theta}{2}$.