Double Angle Formulas/Sine/Proof 3

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 $AH$.
 * $\angle BAH = \angle CAH = \alpha$

From Angler Bisector and Altitude coincide iff triangle is isosceles:
 * $AH \perp BC$.

From Area of Triangle in Terms of Two Sides and Angle:
 * $\operatorname {Area} \left({\triangle BAH}\right) = \dfrac {BA \cdot AH \sin \alpha} 2$
 * $\operatorname {Area} \left({\triangle CAH}\right) = \dfrac {CA \cdot AH \sin \alpha} 2$

By definition of sine:
 * $AH = CA \cos \alpha$
 * $AH = BA \cos \alpha$

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

And by cancelling out common terms:
 * $\sin 2 \alpha = 2 \cos \alpha \sin \alpha $