Half Angle Formula for Tangent/Corollary 2

Theorem

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

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

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

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

Thus at $\theta = \left({2 k + 1}\right) \pi$, 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 I and quadrant II: $\sin \theta > 0$


 * In quadrant III and quadrant 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 I or quadrant III.

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

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

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

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

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

Also see

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