Half Angle Formula for Tangent/Corollary 1

Theorem

 * $\tan \dfrac \theta 2 = \dfrac {\sin \theta} {1 + \cos \theta}$

where $\tan$ denotes tangent, $\sin$ denotes sine and $\cos$ denotes cosine.

When $\theta = \left({2 k + 1}\right) \pi$, $\tan \dfrac \theta 2$ is undefined.

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 quadrant and:
 * $\sin \theta < 0$ if $\theta$ is in the third or fourth quadrant.

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

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

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

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 $\tan \dfrac \theta 2$ is undefined, $\dfrac{\sin \theta} {1 + \cos \theta}$ is undefined, $\dfrac{1 - \cos \theta} {\sin \theta}$ is undefined.

If $\theta = 2k \pi, k \in \Z$, then $\tan \dfrac \theta 2 = 0$, $\dfrac {\sin\theta} {1+\cos\theta} = 0$

Also see

 * Half Angle Formula for Sine
 * Half Angle Formula for Cosine