Half Angle Formulas/Sine

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Theorem

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

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


Proof 1

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


We also have that:

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

$\blacksquare$


Proof 2

Define:

$u = \dfrac \theta 2$


Then:

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


We also have that:

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

$\blacksquare$


Also see


Sources