# Prove trigonometric inequality $\sin x\leq 1-\left(\frac{2x}{\pi}-1\right)^2$

I was working on a trigonometric inequality and after some manipulations I needed to prove that:

$$\sin x\leq 1-\left(\dfrac{2x}{\pi}-1\right)^2, \enspace \forall x\in \left[0,\pi\right).$$

My idea was to move the square on a side and then square root and prove what we got. But I failed. Please help me solve this! Thank you! Please don't use calculus for the proof.

Concavity of Sine

For $$x,y\in[0,\pi]$$, \begin{align} \frac{\sin(x)+\sin(y)}2 &=\sin\left(\frac{x+y}2\right)\cos\left(\frac{x-y}2\right)\\[6pt] &\le\sin\left(\frac{x+y}2\right)\tag1 \end{align} Since $$\sin(x)$$ is continuous, $$(1)$$ shows that $$\sin(x)$$ is concave on $$[0,\pi]$$.

The Inequality

Note that for $$x=0$$ and $$x=\frac\pi4$$, $$\sin(x)=\frac{2\sqrt2x}\pi$$. Thus, since $$\sin(x)$$ is concave on $$\left[0,\frac\pi4\right]$$, we have $$\sin(x)\ge\frac{2\sqrt2x}\pi\tag2$$ for $$x\in\left[0,\frac\pi4\right]$$. Thus, for $$\frac x2\in\left[0,\frac\pi4\right]$$, that is, $$x\in\left[0,\frac\pi2\right]$$, \begin{align} \cos(x) &=1-2\sin^2\left(\frac x2\right)\tag3\\ &\le1-\frac{4x^2}{\pi^2}\tag4 \end{align} where step $$(4)$$ is simply an application of $$(2)$$. Since $$(4)$$ is even, it is true for $$x\in\left[-\frac\pi2,\frac\pi2\right]$$. Therefore, for $$\frac\pi2-x\in\left[-\frac\pi2,\frac\pi2\right]$$, that is, $$x\in[0,\pi]$$, \begin{align} \sin(x) &=\cos\left(\frac\pi2-x\right)\tag5\\ &\le1-\frac{4\left(\frac\pi2-x\right)^2}{\pi^2}\tag6\\ &=1-\left(\frac{2x}\pi-1\right)^2\tag7 \end{align} where $$(6)$$ is an application of $$(4)$$.

Note that $$f(t)=\frac\pi{\sqrt2}\sin \frac t2,\> t\in [0,\pi/2]$$ is a concave function with $$f(0)=0$$ and $$f(\frac\pi2)= \frac\pi2$$, which implies $$f(t)\ge t$$, i.e.

$$\frac\pi{\sqrt2}\sin \frac t2- t\ge 0 \implies \sin^2\frac t2 \ge (\frac{\sqrt2 t}\pi)^2 \implies 1-\cos t \ge (\frac{2t}\pi)^2$$

Substitute $$x= \frac\pi2+t, \> x\in [0, \pi]$$, to obtain

$$\sin x\leq 1-\left(\dfrac{2x}{\pi}-1\right)^2$$

• But I thought concavity of $\sin$ requires calculus .. :)
– r9m
Commented Apr 18, 2020 at 18:10
• The concavity of $\sin$ Is well-known even before calculus. It can be proved simply using Jensen’s inequality reciprocal which does not require calculus. Commented Apr 18, 2020 at 18:17
• @r9m: the concavity of $\sin(x)$ follows from the formula for the sum of sines (as shown in my answer).
– robjohn
Commented Apr 18, 2020 at 21:35
• @robjohn That works! Nice proof!
– r9m
Commented Apr 18, 2020 at 21:58

We need to prove that $$2\cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\geq\left(\frac{2x}{\pi}-1\right)^2$$ or $$\left(\sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)-\frac{2x}{\pi}+1\right)\left(\sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)+\frac{2x}{\pi}-1\right)\geq0.$$ Now, show that $$\sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)-\frac{2x}{\pi}+1\geq0$$ and $$\sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)+\frac{2x}{\pi}-1\geq0,$$ which we can prove by using one derivative only:

Let $$f(x)=\sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)-\frac{2x}{\pi}+1.$$

Thus, $$f'(x)=\frac{1}{\sqrt2}\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)-\frac{2}{\pi}<0,$$ which says $$f(x)>f(\pi)=0.$$ Let $$g(x)=\sqrt2\cos\left(\frac{\pi}{4}-\frac{x}{2}\right)+\frac{2x}{\pi}-1.$$

Thus, $$g'(x)=\frac{1}{\sqrt2}\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)+\frac{2}{\pi}>0,$$ which gives $$g(x)\geq g(0)=0$$ and we are done!

I would study the function $$f(x)=\sin x+\frac{4x^2}{\pi^2}-\frac{4x}{\pi}$$ should not be a problem to prove that it is negative between $$0$$ and $$\pi$$.

• Yeah.. sadly I'm not allowed to use derivatives to prove it. I'm learning for a contest and they are not allowed on my level Commented Apr 18, 2020 at 17:11
• @furfur you may add that in your OP .. that you are not looking for calculus proofs.
– r9m
Commented Apr 18, 2020 at 17:17