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.

When $\theta = \left({2 k + 1}\right) \pi$, $\tan \dfrac \theta 2$ is undefined.


Proof

\(\displaystyle \tan \frac \theta 2\) \(=\) \(\displaystyle \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} }\) Half Angle Formula for Tangent
\(\displaystyle \) \(=\) \(\displaystyle \pm \sqrt {\frac {\left({1 - \cos \theta}\right) \left({1 + \cos \theta}\right)} {\left({1 + \cos \theta}\right)^2} }\) multiplying top and bottom by $\sqrt {1 + \cos \theta}$
\(\displaystyle \) \(=\) \(\displaystyle \pm \sqrt {\frac {1 - \cos^2 \theta} {\left({1 + \cos \theta}\right)^2} }\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle \pm \sqrt {\frac {\sin^2 \theta} {\left({1 + \cos \theta}\right)^2} }\) 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 $\theta = \left({2 k + 1}\right) \pi$, for $k \in \Z$.

At these points, $\sin \theta = 0$ as well.

Then:

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

So it follows that at $\theta = \left({2 k + 1}\right) \pi$, $\dfrac {\sin \theta} {1 + \cos \theta}$ 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 I and quadrant II: $\sin \theta > 0$
In quadrant III and quadrant 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 III.

Then from Bisection of Angle in Cartesian Plane: Corollary, $\theta$ is in quadrant I or quadrant II.

Therefore $\dfrac {\sin \theta} {1 + \cos \theta}$ is positive.


Let $\dfrac \theta 2$ be in quadrant II or quadrant IV.

Then from Bisection of Angle in Cartesian Plane: Corollary, $\theta$ is in quadrant III or quadrant IV.

Therefore $\dfrac {\sin \theta} {1 + \cos \theta}$ is negative.

$\blacksquare$

Also see


Sources