Answer
a. $\sin 36^\circ$
b. $\sin 6\theta$
Work Step by Step
a. Use the formula $\sin 2x=2\sin x\cos x$ with $x=18^\circ$. Then $2\sin 18^\circ\cos 18^\circ=\sin(2*18^\circ)=\sin 36^\circ$.
b. Use the formula $\sin 2x=2\sin x\cos x$ with $x=3\theta$. Then $2\sin 3\theta\cos 3\theta=\sin(2*3\theta)=\sin 6\theta$.
