# Double Angle Formulas/Sine/Proof 3

## Theorem

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

## Proof

Consider a Isosceles Triangle $\triangle ABC$ with base $BC$, and head angle $\angle BAC = 2 \alpha$.

Draw a angle bisector to $\angle BAC$ and name it $AH$.

$\angle BAH = \angle CAH = \alpha$
$AH \perp BC$.
$\operatorname {Area} \left({\triangle BAH}\right) = \dfrac {BA \cdot AH \sin \alpha} 2$
$\operatorname {Area} \left({\triangle CAH}\right) = \dfrac {CA \cdot AH \sin \alpha} 2$

By definition of sine:

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

And so:

$\operatorname {Area} \left({\triangle BAH}\right) = \dfrac {BA \cdot CA \cos \alpha \sin \alpha} 2$
$\operatorname {Area} \left({\triangle CAH}\right) = \dfrac {CA \cdot BA \cos \alpha \sin \alpha} 2$

 $\displaystyle \operatorname {Area} \left({\triangle ABC}\right)$ $=$ $\displaystyle \operatorname {Area} \left({\triangle BAH}\right) + \operatorname {Area} \left({\triangle CAH}\right)$ $\displaystyle$ $=$ $\displaystyle \dfrac {BA \cdot CA \cos \alpha \sin \alpha} 2 + \dfrac {CA \cdot BA \cos \alpha \sin \alpha} 2$ $\displaystyle$ $=$ $\displaystyle BA CA \cos \alpha \sin \alpha$ $\displaystyle$ $=$ $\displaystyle \dfrac {BA \cdot CA \sin 2 \alpha} 2$ Area of Triangle in Terms of Two Sides and Angle ($\triangle ABC$)

And by cancelling out common terms:

$\sin 2 \alpha = 2 \cos \alpha \sin \alpha$

$\blacksquare$