Half Angle Formulas/Tangent

Theorem
where $\tan$ denotes tangent and $\cos$ denotes cosine.

When $\theta = \paren {2 k + 1} \pi$, $\tan \dfrac \theta 2$ is undefined.

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

When $\cos \theta = -1$ it follows that $\cos \theta + 1 = 0$ and then $\tan \dfrac \theta 2$ is undefined.

This happens when $\theta = \paren {2 k + 1} \pi$.

We have that:
 * $\tan \dfrac \theta 2 = \dfrac {\sin \dfrac \theta 2} {\cos \dfrac \theta 2}$

Quadrant $\text I$
In quadrant $\text I$, we have:
 * Sine in First Quadrant: $\sin \dfrac \theta 2 > 0$
 * Cosine in first Quadrant: $\cos \dfrac \theta 2 > 0$

and so in quadrant $\text I$:

Quadrant $\text {II}$
In quadrant $\text {II}$, we have:
 * Sine in Second Quadrant: $\sin \dfrac \theta 2 > 0$
 * Cosine in Second Quadrant: $\cos \dfrac \theta 2 < 0$

and so in quadrant $\text {II}$:

Quadrant $\text {III}$
In quadrant $\text {III}$, we have:
 * Sine in Third Quadrant: $\sin \dfrac \theta 2 < 0$
 * Cosine in Third Quadrant: $\cos \dfrac \theta 2 < 0$

and so in quadrant $\text {III}$:

Quadrant $\text {IV}$
In quadrant $\text {IV}$, we have:
 * Sine in Fourth Quadrant: $\sin \dfrac \theta 2 < 0$
 * Cosine in Fourth Quadrant: $\cos \dfrac \theta 2 > 0$

and so in quadrant $\text {IV}$:

Also see

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