"Show that between any two roots of $\sin(x)$, there is at least one root of $\cos(x)$".
I want to confirm that I'm using the right strategy for this proof. First we need to show that the sine function, call it $f(x)$ is differentiable everywhere, and therefore continuous everywhere. The key property of $f$ at this point is that $$f(x+h)-f(x)=f(x)f'(h)+f'(x)f(h)-f(x)$$
, and it can be shown that the limit is precisely $f'(x)=\cos(x)$. Let $a,b \in \Bbb{R}$ such that $a,b \in [a,b]$ and $f(a)=f(b)=0$. Now we can use Rolle's Theorem and claim that there is a point $c\in (a,b)$, such that $$f'(c)=0$$
and since $c \in (a,b)$, we have that $a<c<b$. Is there something wrong with this proof?