# Half Angle Formulas/Sine

## 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, $\sin \theta < 0$.

$\blacksquare$