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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.

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4 Answers 4

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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 $(4)$.

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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

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

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  • $\begingroup$ But I thought concavity of $\sin$ requires calculus .. :) $\endgroup$
    – r9m
    Commented Apr 18, 2020 at 18:10
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    $\begingroup$ 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. $\endgroup$
    – furfur
    Commented Apr 18, 2020 at 18:17
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    $\begingroup$ @r9m: the concavity of $\sin(x)$ follows from the formula for the sum of sines (as shown in my answer). $\endgroup$
    – robjohn
    Commented Apr 18, 2020 at 21:35
  • $\begingroup$ @robjohn That works! Nice proof! $\endgroup$
    – r9m
    Commented Apr 18, 2020 at 21:58
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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!

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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 $\pi$.

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  • $\begingroup$ 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 $\endgroup$
    – furfur
    Commented Apr 18, 2020 at 17:11
  • $\begingroup$ @furfur you may add that in your OP .. that you are not looking for calculus proofs. $\endgroup$
    – r9m
    Commented Apr 18, 2020 at 17:17

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