Answer
a) No.
b) Yes.
Work Step by Step
a)$$\sin 60^{\circ } = \sin ( 2 \cdot 30^{\circ })=\frac{\sqrt{3}}{2}, \quad \sin 30^{\circ }= \frac{1}{2} \\ \Rightarrow \quad \sin ( 2 \cdot 30^{\circ })= \frac{\sqrt{3}}{2} \neq 1=2(\frac{1}{2})= 2 \sin 30^{\circ }$$
b)$$\sin 60^{\circ } = \sin ( 2 \cdot 30^{\circ })=\frac{\sqrt{3}}{2}, \quad \sin 30^{\circ }= \frac{1}{2}, \quad \cos 30^{\circ }=\frac{\sqrt{3}}{2} \\ \Rightarrow \quad \sin ( 2 \cdot 30^{\circ })= \frac{\sqrt{3}}{2} =2(\frac{1}{2})(\frac{\sqrt{3}}{2})= 2 \sin 30^{\circ } \cos 30^{\circ }$$
