# Half Angle Formulas/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 $\theta = \paren {2 k + 1} \pi$, $\tan \dfrac \theta 2$ is undefined.

## Proof

 $\displaystyle \tan \frac \theta 2$ $=$ $\displaystyle \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} }$ Half Angle Formula for Tangent $\displaystyle$ $=$ $\displaystyle \pm \sqrt {\frac {\paren {1 - \cos \theta}^2} {\paren {1 + \cos \theta} \paren {1 - \cos \theta} } }$ multiplying both numerator and denominator by $\sqrt {1 - \cos \theta}$ $\displaystyle$ $=$ $\displaystyle \pm \sqrt {\frac {\paren {1 - \cos \theta}^2} {1 - \cos^2 \theta} }$ Difference of Two Squares $\displaystyle$ $=$ $\displaystyle \pm \sqrt {\frac {\paren {1 - \cos \theta}^2} {\sin^2 \theta} }$ Sum of Squares of Sine and Cosine $\displaystyle$ $=$ $\displaystyle \pm \frac {1 - \cos \theta} {\sin \theta}$

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.

$\Box$

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

Then:

 $\displaystyle \lim_{x \mathop \to 0^+} \frac {1 - \cos \theta} {\sin \theta}$ $=$ $\displaystyle \lim_{x \mathop \to 0^+} \frac {\map {\dfrac \d {\d \theta} } {1 - \cos \theta} } {\dfrac \d {\d \theta} \sin \theta}$ L'Hôpital's Rule $\displaystyle$ $=$ $\displaystyle \lim_{x \mathop \to 0^+} \frac {\sin \theta} {\cos \theta}$ $\displaystyle$ $=$ $\displaystyle \frac 0 1$ $\displaystyle$ $=$ $\displaystyle 0$

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

$\Box$

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.

$\blacksquare$