Prove $|\cos(x+\sin 2x)|\leq |\cos x|$ holds for all $x \in \mathbb{R}$.

Question

Prove $$|\cos(x+\sin 2x)|\leq |\cos x|$$ holds for all $x \in \mathbb{R}$.

This is true, by WA graphing, but seems to be hard to prove.

Answer 1

We need $$0\ge\sin^2(x+\sin2x)-\sin^2x$$

$=\sin(2x+\sin2x)\sin(\sin2x)=f(x)$(say)

Observe that $f(x)=f(-x)$

Now if $x>0,0<\sin2x\le1<\dfrac\pi2,\sin(\sin2x)>0$

We need to establish $0\le2x+\sin2x\le\pi$

The left inequality is obvious

So, we need $$\sin2x\le\pi-2x$$

Set $\pi-2x=z\implies\sin2z\le2z$

Use How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

Answer 2

Note that \begin{eqnarray} f(x)&=&\cos^2x-\cos^2(x+\sin(2x)\\ &=&\frac{1+\cos(2x)}{2}-\frac{1+\cos(2x+2\sin(2x))}{2}\\ &=&\frac{\cos(2x)-\cos(2x+2\sin(2x)}{2}\\ &=&\sin(2x+\sin(2x))\sin(\sin(2x)). \end{eqnarray} From this, one can see that $f(x)$ is a periodic function with period $\pi$.

Case 1: $x\in[0,\frac{\pi}{2}-\frac12]$. Then $\sin(2x)\in[0,1]$ and $$ 0\le2x+\sin(2x)\le\pi-1+1=\pi$$ and hence $\sin(2x+\sin(2x))\in[0,1]$. So $f(x)\ge0$.

Case 2: Similarly for $x\in[\frac{\pi}2+\frac12,\pi]$, $f(x)\ge 0$.

Case 3: $x\in[\frac{\pi}{2}-\frac12,\frac{\pi}{2}+\frac12]$. Let $t=x-\frac{\pi}{2}$ and then $t\in[-\frac12,\frac12]$. So $$ f(x)=\sin(2t-\sin(2t))\sin(\sin(2t)) $$ is an even function. One only can consider $t\in[0,\frac12]$. Since $$ 0\le 2t-\sin(2t)\le 2t\le1 $$ one has $f(x)\ge0$.