Half Angle Formulas/Tangent
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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 $\text {IV}$:
| \(\ds \tan \frac \theta 2\) | \(<\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \tan \frac \theta 2\) | \(=\) | \(\ds -\sqrt {\dfrac {1 - \cos \theta} {1 + \cos \theta} }\) |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.43$