# Half Angle Formulas/Tangent

## Theorem

 $\ds \tan \frac \theta 2$ $=$ $\ds +\sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} }$ for $\dfrac \theta 2$ in quadrant $\text I$ or quadrant $\text {III}$ $\ds \tan \frac \theta 2$ $=$ $\ds -\sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} }$ for $\dfrac \theta 2$ in quadrant $\text {II}$ or quadrant $\text {IV}$

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

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

### Corollary 1

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

### Corollary 2

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

### Corollary 3

$\tan \dfrac \theta 2 = \csc \theta - \cot \theta$

## Proof

 $\ds \tan \frac \theta 2$ $=$ $\ds \frac {\sin \frac \theta 2} {\cos \frac \theta 2}$ Tangent is Sine divided by Cosine $\ds$ $=$ $\ds \frac {\pm \sqrt {\frac {1 - \cos \theta} 2} } {\pm \sqrt {\frac {1 + \cos \theta} 2} }$ Half Angle Formula for Sine and Half Angle Formula for Cosine $\ds$ $=$ $\ds \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} }$

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 $\text I$:

 $\ds \tan \frac \theta 2$ $>$ $\ds 0$ $\ds \leadsto \ \$ $\ds \tan \frac \theta 2$ $=$ $\ds +\sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} }$

### 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 $\text {II}$:

 $\ds \tan \frac \theta 2$ $<$ $\ds 0$ $\ds \leadsto \ \$ $\ds \tan \frac \theta 2$ $=$ $\ds -\sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} }$

### 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 $\text {III}$:

 $\ds \tan \frac \theta 2$ $>$ $\ds 0$ $\ds \leadsto \ \$ $\ds \tan \frac \theta 2$ $=$ $\ds + \sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} }$

### 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 $\text {IV}$:

 $\ds \tan \frac \theta 2$ $<$ $\ds 0$ $\ds \leadsto \ \$ $\ds \tan \frac \theta 2$ $=$ $\ds -\sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} }$

$\blacksquare$