# Prove that $\cos(x-y)+\cos(x+y)\cos(2x)>0$ for any $0<x,y<\frac{\pi}{2}$

Prove that the following inequality holds for any $x,y\in\mathbb{R}$ that satisfy $0<x,y<\frac{\pi}{2}$: $\cos(x-y)+\cos(x+y)\cos(2x)>0$

I've tried to use several trigonometric identities to prove it but wasn't able to reach something meaningful, Also, by plotting the graph of the function on a computer, It seems that by trying to use derivatives in order to find where the function is increasing/decreasing or in order to find its minimum/maximum points is not going to help in proving this inequality.

Thanks for any hint/help.

• Try $=cos(x-y)+cos(x+y)cos(x-y+x+y)$, not sure if you can do it in this way. – arberavdullahu Dec 20 '17 at 20:56

## 2 Answers

$$\cos \left( x-y \right) +\cos \left( x+y \right) \cos \left( 2x \right) >0\\ \cos { \left( x \right) \cos { \left( y \right) +\sin { \left( x \right) \sin { \left( y \right) +\left[ \cos { \left( x \right) \cos { \left( y \right) -\sin { \left( x \right) \sin { \left( y \right) } } } } \right] \cos { \left( 2x \right) = } } } } } \\ =\cos { \left( x \right) \cos { \left( y \right) \left[ 1+\cos { \left( 2x \right) } \right] +\sin { \left( x \right) \sin { \left( x \right) \left[ 1-\cos { \left( 2x \right) } \right] } = } } } \\ =\underset { >0 }{ \underbrace { \cos { \left( x \right) \cos { \left( y \right) \left[ 2\cos ^{ 2 }{ \left( x \right) } \right] } } } } +\underset { >0 }{ \underbrace { \sin { \left( x \right) \sin { \left( x \right) \left[ 2\sin ^{ 2 }{ \left( x \right) } \right] } } } } >0$$

$\cos 2x>-1$

On the other hand, $|x+y|>|x-y|$, so $\cos(x+y)<\cos(x-y)$. Furthermore, $|x-y|<\frac\pi2$, so $\cos(x-y)>0$.

Then $$\cos(x-y)+\cos2x\cos(x+y)>\cos(x-y)-\cos(x+y)>0$$

• you have a typo on the last line $- \to +$ – koftehor Dec 20 '17 at 21:42
• Thanks, @koftehor – ajotatxe Dec 20 '17 at 22:05