# Proof for $\frac{2}{\pi}x \lt \sin{x}$ for $x \in (0,\frac{\pi}2)$

The following is part of exercise 6.26.21 from Tom Apostol's Calculus Volume 1. I wonder if my proof is correct and if there is a simpler alternative proof.

Prove the following by examining the sign of the derivative of an appropriate function: $$\frac{2}{\pi}x \lt \sin{x} \qquad \text{if} \qquad 0 \lt x \lt \frac{\pi}{2} \tag{1}\label{1}$$

Let $f(x)=\sin{x}-\frac{2}{\pi}x$, $0 \le x \le \frac{\pi}{2}$ then

$$f(0)=f\left(\frac{\pi}{2}\right)=0 \tag{2}\label{2}$$

and

$$f''(x)=-\sin{x} \lt 0 \tag{3}\label{3}$$

From $\eqref{2}$ and $\eqref{3}$ we know that $f$ has a maximum at exactly one point, this together with $\eqref{2}$ proves $\eqref{1}$.

• seems correct to me...you can also say by (2) f is strictly concave hence $f(x)=f(1\cdot x+0\cdot y)> 1\cdot0+0\cdot 0=0$ – math635 Dec 28 '15 at 1:05
• This is the first part of Jordan's inequality. – user258700 Dec 28 '15 at 1:06
• You should consider $f(x)=\sin x-2x/\pi$ for $0\le x\le\pi/2$, just to be picky. – egreg Dec 28 '15 at 1:07
• @math635 thanks, was interesting to see how the concavity proves the point by its very definition. Btw, as a comparison, I think my "maximum at exactly one point" is not sufficient as that wording would still allow a minimum at another point, so as you and others pointed out I should've simply used the concavity property. – Imre Deák Dec 28 '15 at 10:50
• @math635, one note is that in $f(x)=f(\lambda x_0+(1-\lambda)y_0) \gt \lambda f(x_0)+(1-\lambda )f(y_0)$ it's not clear to me how you chose $\lambda, x_0, y_0$. I see how it works if we let $\lambda \in (0,1) \; \text, \; x0=0 \; \text, \; y_0=\frac{\pi}2$. – Imre Deák Dec 28 '15 at 12:30

Your proof is fine: the concavity of the sine function over $\left(0,\frac{\pi}{2}\right)$ gives the wanted inequality in a straightforward way. Anyway, if you like to kill flies with hydrogen bombs, you may consider that: $$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2 \pi^2}\right)$$ hence if $x\in\left(0,\frac{\pi}{2}\right)$ we have: $$\frac{\sin x}{x}> \prod_{n\geq 1}\left(1-\frac{1}{4n^2}\right) = \prod_{n\geq 1}\frac{2n-1}{2n}\cdot\frac{2n+1}{2n}$$ where the RHS is the reciprocal of the Wallis product, i.e. $\frac{2}{\pi}$ as wanted.
Take $f(x)=\frac{\sin x}{x}$ then $f$ is decreasing as its derivative is negative.
• That proves that $\sin{x} \lt x$, but not (1)? – Imre Deák Dec 28 '15 at 1:38
• Its decreasing down to $\frac{\pi}{2}$ so this is a lower bound. – Rene Schipperus Dec 28 '15 at 1:40
• Ok, I understand now if you meant $\frac{2}{\pi}$. – Imre Deák Dec 28 '15 at 2:16