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

Question

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.

Answer 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.

Answer 2

Define $$F(x)=\sin{x}-x$$ $$F'(x)=\cos{x}-1\leq 0$$ If $0<b\in [0,1) $ then $[0,b] $ is a interval that satisfy, the value mean theorem, then

$$F(b)-F(0)<0$$ $$F(b)<0$$ $$\sin{b}<b$$ for all $b\in[0,1)$, the equality is have if $x=0,1$

Now you define $$H(x)=x\sqrt{1-x^2} -\sin x.... $$

Answer 3

Another way to prove this inequality is to notice that this can inequality can be rewritten as:

$$\frac{1}{2}\sin(2x)\le\sin(\sin(x))\le\sin(x)$$ for $x\in[0,\frac{\pi}{2}]$.

Let $f(x)=\frac{1}{2}\sin(2x)$, $g(x)=\sin(\sin(x))$, and $h(x)=\sin(x)$. Then $f'(x)=\cos(2x)$, $g'(x)=\cos(\sin(x))\cos(x)$, and $h'(x)=\cos(x)$. It follows quickly that:

$g'(x)=\cos(\sin(x))\cos(x)\le\cos(x)=h'(x)$ for any $x$ and we have $g(0)=h(0)$ so $g(x)\le h(x)$ for $x\in[0,\frac{\pi}{2}]$.

Since $\cos(x)$ is decreasing for $x\in[0,\frac{\pi}{2}]$ and the inequality proven in the previous paragraph establishes $\sin(x)\le x$ then for $x\in[0,\frac{\pi}{2}]$ we have:

$f'(x)=\cos(2x)=\cos^{2}(x)-\sin^{2}(x)\le\cos(x)\cos(x)\le\cos(\sin(x))\cos(x)=g'(x)$ for $x\in[0,\frac{\pi}{2}]$ and that $f(0)=g(0)$ so $f(x)\le g(x)$ for $x\in[0,\frac{\pi}{2}]$.