Algebra 2 (1st Edition)

$\dfrac{1}{2} \sin \dfrac{2x}{3}=\sin \dfrac{x}{3} \cos \dfrac{x}{3}$
We need to verify that $\dfrac{1}{2} \sin \dfrac{2x}{3}=\sin \dfrac{x}{3} \cos \dfrac{x}{3}$ We need to recall that: $\sin 2 \theta =2 \sin \theta \cos \theta$ $\dfrac{1}{2} ( 2 \sin \dfrac{x}{3}\cos \dfrac{x}{3}) =\sin \dfrac{x}{3} \cos \dfrac{x}{3}$ Thus, it has been proven that $\sin \dfrac{x}{3} \cos \dfrac{x}{3}=\sin \dfrac{x}{3} \cos \dfrac{x}{3}$