Half Angle Formulas/Tangent/Corollary 2

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Theorem

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

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

\(\displaystyle \tan \frac \theta 2\) \(=\) \(\displaystyle \pm \sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} }\) $\quad$ Half Angle Formula for Tangent $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \pm \sqrt {\frac {\left({1 - \cos \theta}\right)^2} {\left({1 + \cos \theta}\right) \left({1 - \cos \theta}\right)} }\) $\quad$ multiplying both numerator and denominator by $\sqrt {1 - \cos \theta}$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \pm \sqrt {\frac {\left({1 - \cos \theta}\right)^2} {1 - \cos^2 \theta} }\) $\quad$ Difference of Two Squares $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \pm \sqrt {\frac {\left({1 - \cos \theta}\right)^2} {\sin^2 \theta} }\) $\quad$ Sum of Squares of Sine and Cosine $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \pm \frac {1 - \cos \theta} {\sin \theta}\) $\quad$ $\quad$


When $\sin \theta = 0$, the above is undefined.

This happens when $\theta = k \pi$ for $k \in \Z$.


When $\theta = \left({2 k + 1}\right) \pi$, the value of $1 - \cos \theta$ is $2$.

Thus at $\theta = \left({2 k + 1}\right) \pi$, the value of $\tan \dfrac \theta 2$ is undefined.

$\Box$


When $\theta = 2 k \pi$, the value of $\cos \theta = 1$ and so $1 - \cos \theta$ is $0$.

Then:

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

Thus $\tan \dfrac \theta 2$ is defined at $\theta = 2 k \pi$, and equals $0$.

$\Box$


At all other values of $\theta$, $1 - \cos \theta > 0$.

Therefore the sign of $\dfrac {1 - \cos \theta} {\sin \theta}$ 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 $\dfrac {1 - \cos \theta} {\sin \theta}$.


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

$\blacksquare$

Also see


Sources