# Double Angle Formulas/Sine/Proof 3

## Theorem

$\sin 2 \theta = 2 \sin \theta \cos \theta$

## Proof

Consider an isosceles triangle $\triangle ABC$ with base $BC$ and apex $\angle BAC = 2 \alpha$.

Construct the angle bisector to $\angle BAC$ and name it $AH$:

$\angle BAH = \angle CAH = \alpha$
$AH \perp BC$
 $\displaystyle \map \Area {\triangle BAH}$ $=$ $\displaystyle \dfrac {BA \cdot AH \sin \alpha} 2$ $\displaystyle \map \Area {\triangle CAH}$ $=$ $\displaystyle \dfrac {CA \cdot AH \sin \alpha} 2$

By definition of sine:

 $\displaystyle AH$ $=$ $\displaystyle CA \cos \alpha$ $\displaystyle AH$ $=$ $\displaystyle BA \cos \alpha$

and so:

 $\displaystyle \map \Area {\triangle BAH}$ $=$ $\displaystyle \dfrac {BA \cdot CA \cos \alpha \sin \alpha} 2$ $\displaystyle \map \Area {\triangle CAH}$ $=$ $\displaystyle \dfrac {CA \cdot BA \cos \alpha \sin \alpha} 2$ $\displaystyle \map \Area {\triangle ABC}$ $=$ $\displaystyle \map \Area {\triangle BAH} + \map \Area {\triangle CAH}$ $\displaystyle$ $=$ $\displaystyle \frac {BA \cdot CA \cos \alpha \sin \alpha} 2 + \frac {CA \cdot BA \cos \alpha \sin \alpha} 2$ $\displaystyle$ $=$ $\displaystyle BA \cdot CA \cos \alpha \sin \alpha$ $\displaystyle$ $=$ $\displaystyle \frac {BA \cdot CA \sin 2 \alpha} 2$ Area of Triangle in Terms of Two Sides and Angle ($\triangle ABC$) $\displaystyle \leadsto \ \$ $\displaystyle \sin 2 \alpha$ $=$ $\displaystyle 2 \cos \alpha \sin \alpha$ dividing both sides by $\dfrac {BA \cdot CA} 2$

$\blacksquare$