Half Angle Formula for Tangent/Corollary 1
Theorem
- $\tan \dfrac \theta 2 = \dfrac {\sin \theta} {1 + \cos \theta}$
where $\tan$ denotes tangent, $\sin$ denotes sine and $\cos$ denotes cosine.
Where $\theta = \paren {2 k + 1} \pi$, $\tan \dfrac \theta 2$ is undefined.
Proof
\(\ds \tan \frac \theta 2\) | \(=\) | \(\ds \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} }\) | Half Angle Formula for Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \pm \sqrt {\frac {\paren {1 - \cos \theta} \paren {1 + \cos \theta} } {\paren {1 + \cos \theta}^2} }\) | multiplying top and bottom by $\sqrt {1 + \cos \theta}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \pm \sqrt {\frac {1 - \cos^2 \theta} {\paren {1 + \cos \theta}^2} }\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \pm \sqrt {\frac {\sin^2 \theta} {\paren {1 + \cos \theta}^2} }\) | Sum of Squares of Sine and Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds \pm \frac {\sin \theta} {1 + \cos \theta}\) |
Since $\cos \theta \ge -1$, it follows that $\cos \theta + 1 \ge 0$.
When $\cos \theta = -1$ it follows that $1 + \cos \theta = 0$.
This happens when $\theta = \paren {2 k + 1} \pi$, for $k \in \Z$.
At these points, $\sin \theta = 0$ as well.
Then:
\(\ds \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\sin \theta} {1 + \cos \theta}\) | \(=\) | \(\ds \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\dfrac \d {\d \theta} \sin \theta} {\map {\dfrac \d {\d \theta} } {1 + \cos \theta} }\) | L'Hôpital's Rule | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\cos \theta} {\sin \theta}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 0\) |
So it follows that at $\theta = \paren {2 k + 1} \pi$, $\dfrac {\sin \theta} {1 + \cos \theta}$ is undefined.
$\Box$
At all other values of $\theta$, $\cos \theta + 1 > 0$.
Therefore the sign of $\dfrac {\sin \theta} {1 + \cos \theta}$ 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 $\dfrac {\sin \theta} {1 + \cos \theta}$.
Let $\dfrac \theta 2$ be in quadrant 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 $\dfrac {\sin \theta} {1 + \cos \theta}$ 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 $\dfrac {\sin \theta} {1 + \cos \theta}$ is negative.
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.43$