Proving $\sin(x) < x$ for $0<x<2\pi$

Question

Prove that $\sin(x) < x$ when $0<x<2\pi.$

I have been struggling on this problem for quite some time and I do not understand some parts of the problem. I am supposed to use rolles theorem and Mean value theorem

First using the mean value theorem I got $\cos(x) = \dfrac {\sin(x)}x$ and since $1 ≥ \cos x ≥ -1$ , $1 ≥ \dfrac {\sin(x)}x$ which is $x ≥ \sin x$ for all $x ≥ 0$.

Here the first issue is that I didn't know how to change $≥$ to $>$.

The second part is proving when $x<2\pi$ and this part I have no idea.

I know that $2\pi > 1$ , and $1 ≥ \sin x$ and my thought process ends here.

Answer 1

$x-\sin\, x=\int_0^{x}[1-\cos\, t] dt \geq 0$ and equality can hold only of the non-negative continuous function $1-\cos\, t$ is identically $0$ from $0$ to $x$. This is not true for any $x>0$ so strict inequality holds.

Answer 2

Using $f : [0, \infty) : x \mapsto x - \sin(x)$ and the mean value theorem, we can solve this problem. Note that $$f'(x) = 1 - \cos(x) \ge 0.$$ The mean value theorem tells us that $f$ is therefore non-decreasing (prove the contrapositive!). So, for all $x \ge 0$, we have $$x - \sin(x) = f(x) \ge f(0) = 0 \implies x \ge \sin(x)$$ as you deduced already. The only thing we need to do is show $f(x) > 0$ for $x > 0$. Note that, in order for this to be true, since $f$ is non-decreasing, we would need to have $f(x) = 0$ on some interval $[0, \lambda]$; as soon as $f(\lambda) = 0$ for some $\lambda > 0$, then $$x \in [0, \lambda] \implies 0 = f(0) \le f(x) \le f(\lambda) = 0 \implies f(x) = 0.$$ But, this is not the case. This would imply that $f'(x) = 0$ for all $x \in (0, \lambda)$, which is not the case as $\cos$ is not constant locally to the right of $0$.

Answer 3

The inequality obviously holds for $x>1$.

Then for $0<x\le1$,

$$\cos x<1$$ and by integration from $0$

$$\sin x<x.$$

Answer 4

Let $f(x) = \sin x, x \in \Bbb R.$ Let $x \in (0,1)$ then using Lagrange's MVT on the interval $[0,x]$ we get $$\frac {f(x)-f(0)} {x-0} = f'(c)$$ where $c \in (0,x) \subset (0,1).$Therefore we have $$\frac {\sin x} {x} = \cos c < 1,$$ since $0<c<1$ and $\cos x$ is strictly decreasing on $\left (0, {\pi} \right )$ and hence on $(0,1).$ Hence for all $x \in (0,1)$ we have $\sin x < x.$ Also for $x=1$ we have $\sin x = \sin 1 < \sin \left (\frac {\pi} {2} \right ) = 1,$ since $1 < \frac {\pi} {2}$ and $\sin x$ is strictly increasing on $\left (0,\frac {\pi} {2} \right ).$ Also for $x > 1$ we have $\sin x \leq 1 < x.$ So we have $\sin x < x$ for all $x > 0.$