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So I've the following example in my calculus book but I really don't understand the motivation behind each step.

The example in question:

If $x>2\pi$, then $\sin x \le 1<2\pi<x$. If $0<x\le2\pi$, then, by the Mean-Value-Theorem, there exists c in the open interval $(0,2\pi)$ such that

$\frac {\sin x}{x} =\frac {\sin x-\sin 0}{x-0}=\frac{d}{dx}\sin x|_{x=c}=\cos c<1$

Thus, $\sin x<x$ in this case too.

So I understand the concept of the MVT. But I can't really follow the authors reasoning to come to his conclusion?

could somebody provide more insight to why he first assumes $ x>2\pi$ then $0<x\le2\pi$?

Thank you in advance.

Aliasa Zarowny Pseudonymia
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  • Same source right? : https://math.stackexchange.com/questions/886840/using-mean-value-theorem-to-show-that-sin-x-lt-x –  Nov 29 '17 at 16:18
  • Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Jan 22 '18 at 21:37

3 Answers3

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Note that $\frac{d}{dx}(x-\sin x)=1-\cos x\geq 0$ implies the function $f(x)=x-\sin x$ is increasing. Also note that $f(0)=0$, hence $f(x)\geq 0$ for all $x\ge 0$.

QED
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I would do it this way: The inequality is obvious for $x>1.$ For each $0<x\le 1,$ we can use the MVT to see

$$\sin x = \sin x - \sin 0 = \cos c_x\cdot (x-0) = \cos c_x \cdot x$$

for some $c_x\in (0,1).$ Since $\cos u < 1$ for $0< u< 1,$ we're done.

zhw.
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For $sin x\leq x$ it suffices to use MVT:

$$\cos c =\frac{\sin x- \sin 0}{x-0}=\frac{\sin x}{x}\implies -1\leq \frac{\sin x}{x}\leq \implies \frac{\sin x}{x}\leq 1 \implies sin x\leq x$$

More in general:

$$ x - \frac{x^3}{3!} \leq \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} -\frac{x^7}{7!} + ...\leq x $$

For $x - \frac{x^3}{3!}\leq \sin(x) $ see here

Proof for $\sin(x) > x - \frac{x^3}{3!}$

user
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