Half Angle Formula for Tangent/Corollary 2/Proof 1

Proof
When $\sin \theta = 0$, the above is undefined.

This happens when $\theta = k \pi$ for $k \in \Z$.

When $\theta = \paren {2 k + 1} \pi$, the value of $1 - \cos \theta$ is $2$.

Thus at $\theta = \paren {2 k + 1} \pi$, the value of $\tan \dfrac \theta 2$ is undefined.

When $\theta = 2 k \pi$, the value of $\cos \theta = 1$ and so $1 - \cos \theta$ is $0$.

Then:

Thus $\tan \dfrac \theta 2$ is defined at $\theta = 2 k \pi$, and equals $0$.

At all other values of $\theta$, $1 - \cos \theta > 0$.

Therefore the sign of $\dfrac {1 - \cos \theta} {\sin \theta}$ is equal to the sign of $\sin \theta$.

We recall:


 * In quadrant $\text I$ and quadrant $\text {II}$: $\sin \theta > 0$


 * In quadrant $\text {III}$ and quadrant $\text {IV}$: $\sin \theta < 0$

Thus it follows that the same applies to $\dfrac {1 - \cos \theta} {\sin \theta}$.

Let $\dfrac \theta 2$ be in quadrant $\text I$ or quadrant $\text {III}$.

Then from Bisection of Angle in Cartesian Plane: Corollary, $\theta$ is in quadrant $\text I$ or quadrant $\text {II}$.

Therefore $\dfrac {1 - \cos \theta} {\sin \theta}$ is positive.

Let $\dfrac \theta 2$ be in quadrant $\text {II}$ or quadrant $\text {IV}$.

Then from Bisection of Angle in Cartesian Plane: Corollary, $\theta$ is in quadrant $\text {III}$ or quadrant $\text {IV}$.

Therefore $\dfrac {1 - \cos \theta} {\sin \theta}$ is negative.