1. ## Infinitely differentiable

Suppose $f$ is infinitely differentiable on $\mathbb{R}$.

(a) If $f''(x) + f(x) = 0, f(0) = 0,$ and $f'(0) = 1,$ show that $f(x)$ = $sin x$.

2. Well, if $f(x) = \sin{x}$, then $f'(x) = \cos{x}$ and $f''(x) = -\sin{x}.$
So $f''(x)+f(x) = -\sin{x}+\sin{x} = 0$ and $f'(0) = \cos(0) = 1$.

3. This is the (one) definition of $\sin$. It is clear that $f^{(2n+1)}(0)=(-1)^n$, and all other derivatives of $f$ vanish. From this, we can construct a Taylor-like series definition of $f$, and see that it fits with the usual Taylor series of $\sin$:

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}$.

You have to be careful in using this statement, which may not preclude the analyticity of $f$.

4. Note that The CoffeeMachine's response shows that if f(x)= sin(x), then f satisfies the given conditions. To show the converse, which is the actual problem, you must also show uniqueness for solutions to such a differential equation. Once again, because the Original Poster shows no attempt at a solution, we do not know what he/she has available to do that.