## Calculus, 10th Edition (Anton)

First we will find the derivative using the Chain Rule: $$g'(x)=(f(x)\sin x)'=f'(x)\sin x+f(x)(\sin x)'=f'(x)\sin x+f(x)\cos x$$ Now we will evaluate $g'(x)$ for $x=0$: $$g'(0)=f'(0)\sin0+f(0)\cos0=f'(0)\cdot0+f(0)\cdot1=f(0)$$ So, the statement is true.