[Math] Prove that $\sin 2\alpha=2 \sin \alpha \cos\alpha$

Question

In this triangle

$AD=AC=1$, $BC=a$, $BAC=2\alpha$

I thought $\sin 2\alpha=a$, but I don't know how to continue.

Answer 1

drop a perpendicular from $A$ to $DC$ and call the foot of the perpendicular $E.$ then $$DE = EC = \cos \alpha, DC = 2 \cos \alpha.$$ now look at the right triangle $CBD$ with hypotenuse $2\cos \alpha$ and the $\angle BDC = \alpha,$ therefore the opposite side $BC$ is $$ BC = 2\cos \alpha \sin \alpha.$$ you can also look at the right angle triangle $ABC$ with hypotenuse $AC = 1, \angle BAC = 2\alpha$ which give the opposite side $$BC = \sin 2\alpha $$

therefore $$ BC = \sin 2\alpha = 2\sin \alpha \cos \alpha $$