Half Angle Formulas/Sine

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Theorem

\(\ds \sin \frac \theta 2\) \(=\) \(\ds +\sqrt {\frac {1 - \cos \theta} 2}\) for $\dfrac \theta 2$ in quadrant $\text I$ or quadrant $\text {II}$
\(\ds \sin \frac \theta 2\) \(=\) \(\ds -\sqrt {\dfrac {1 - \cos \theta} 2}\) for $\dfrac \theta 2$ in quadrant $\text {III}$ or quadrant $\text {IV}$

where $\sin$ denotes sine and $\cos$ denotes cosine.


Proof 1

\(\ds \cos \theta\) \(=\) \(\ds 1 - 2 \ \sin^2 \frac \theta 2\) Double Angle Formula for Cosine: Corollary $2$
\(\ds \leadsto \ \ \) \(\ds 2 \ \sin^2 \frac \theta 2\) \(=\) \(\ds 1 - \cos \theta\)
\(\ds \leadsto \ \ \) \(\ds \sin \frac \theta 2\) \(=\) \(\ds \pm \sqrt {\frac {1 - \cos \theta} 2}\)


We also have that:

In quadrant $\text I$ and quadrant $\text {II}$, $\sin \theta > 0$
In quadrant $\text {III}$ and quadrant $\text {IV}$, $\sin \theta < 0$.

$\blacksquare$


Proof 2

Define:

$u = \dfrac \theta 2$


Then:

\(\ds \sin^2 u\) \(=\) \(\ds \frac {1 - \cos 2 u} 2\) Power Reduction Formulas
\(\ds \leadsto \ \ \) \(\ds \sin \frac \theta 2\) \(=\) \(\ds \pm \sqrt {\frac {1 - \cos \theta} 2}\)


We also have that:

In quadrant $\text I$, and quadrant $\text {II}$, $\sin \theta > 0$
In quadrant $\text {III}$, and quadrant $\text {IV}$, $\sin \theta < 0$.

$\blacksquare$


Also see


Sources

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