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 $\theta = \paren {2 k + 1} \pi$, $\tan \dfrac \theta 2$ is undefined.


Proof

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


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

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


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

Thus at $\theta = \paren {2 k + 1} \pi$, 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:

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

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 $\text I$ 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 $\dfrac {1 - \cos \theta} {\sin \theta}$.


Let $\dfrac \theta 2$ be in quadrant $\text I$ or quadrant $\text {III}$.

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

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

$\blacksquare$


Also see


Sources