Half Angle Formulas/Tangent/Corollary 1

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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

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


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:

\(\displaystyle \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\sin \theta} {1 + \cos \theta}\) \(=\) \(\displaystyle \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\dfrac \d {\d \theta} \sin \theta} {\map {\dfrac \d {\d \theta} } {1 + \cos \theta} }\) L'Hôpital's Rule
\(\displaystyle \) \(=\) \(\displaystyle \lim_{x \mathop \to \paren {2 k + 1} \pi^+} \frac {\cos \theta} {\sin \theta}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 0\)

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


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