# Showing $\sin(x) < x$ for all $x>0$ using the mean value theorem

I want to show that $\sin(x) < x$ for all $x>0$, using the mean value theorem.

Since the sine is bounded above by $1$, it's obviously true for $x > 1$. Consider $x \in ]0,1]$. Let $f(x)=\sin(x)$. Choose $a=0$ and $x>0$, then there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with

$$f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow (\sin(x))'(x_0)= \frac{\sin(x)-\sin(a)}{x} \Leftrightarrow \cos(x_0)=\frac{\sin(x)}{x}$$

Since $1\geq x_0>0 \Rightarrow \cos(x_0) < 1$,

$$\Rightarrow 1 > \cos(x_0)=\frac{\sin(x)}{x} \Rightarrow x > \sin(x)$$

Is my proof correct?

• Seems okay, but what about $sin(1)$? You excluded it in your interval. – Alex Jan 25 '14 at 23:53
• It's also true, i've fixed it. – fear.xD Jan 25 '14 at 23:57
• I'd write $\sin'(x_0)$ rather than $(\sin(x))'(x_0)$. (I would not, however, write $(\sin(x_0))'$, since that would be $0$.) – Michael Hardy Jan 26 '14 at 0:06
• Related question: Prove $\sin(x)< x$ when $x>0$ using LMVT – Martin Sleziak Jul 10 '15 at 7:32
• @weakmathematician : From the context it is clear that $y'$ means $\dfrac {dy} {dx}$ and not $\dfrac {dy} {dx_0}.$ The number $x_0$ does not change as $x$ changes, or in other words, it is a constant. So $\sin x_0$ does not change as $x$ changes. Thus you have $\dfrac d{dx} \sin x=\cos x$ and $\dfrac d{dx} \sin x_0 = 0.$ That last fact can be expressed by saying $(\sin x_0)'=0.$ But $\sin' x_0$ means the derivative of the sine function, evaluated at $x_0,$ so it is $\cos x_0. \qquad$ – Michael Hardy Jun 6 '17 at 13:54

Generally, the way one proves that if $f'>0$ everywhere then $f$ is increasing is by using the mean value theorem in this same way. One could apply that idea to $f(x)=x-\sin x$. Since $f(0)=0$ and $f'(x)=1-\cos x>0$ (for $0<x<2\pi$), one concludes that $f(x)$ increases as $x$ increases from $0$; hence $f(x)>0$ for those values of $x$. That is a bit different from your way, but either works.