Answer
See proof
Work Step by Step
Consider the function $f(x)= \sin x$, which is continuous and differentiable on $R$.
Let $a$ be a number such that $0\lt a\lt 2\pi$.
Then f is continuous on $[0,a]$ and differentiable on $(0,a)$.
By the Mean Value Theorem, there is a number c in $(0,a)$ such that
$f(a)-f(0)$ = $f'(c)(a-0)$
therefore
$\sin a-0 = \cos c(a)$
$\sin a = \cos c(a)$
Now $\cos c\lt 1$ for $0 \lt c \lt 2\pi$
so $\sin a \lt 1(a) = a$
As we took $a$ to be an arbitrary number in $(0,2\pi)$
it follows that $\sin x \lt x$ for all $x$ satisfying $0\lt x \lt 2\pi$.
