Prove that $x\sqrt{1-x^2} \leq \sin x \leq x$

Use the mean value theorem to prove that if $0 \leq x \leq 1$, then $$x\sqrt{1-x^2} \leq \sin x \leq x$$ The theorem guarantees the existence of a point, but not an inequality, so I don't know how to begin.

Solution 1:

All three of these functions are equal at $0$, and the inequality is obvious at $1$. If we had a point of equality between $0$ and $1$, then there would exist (for example) a point $c \in (0, 1)$ for which $\sin c = c$. In particular, this implies the existence of a point $d \in (0, c)$ such that

$$\sin' d = \frac{\sin c - \sin 0}{c - 0} = 1$$

by the Mean Value Theorem. But when is cosine equal to $1$?

Make a similar study for the lower inequality, and then consider why the intermediate value theorem means you're done.