# How do I prove the minimum for given trigonometric expression of two angles

by Parallelism Alert   Last Updated July 17, 2019 10:20 AM - source

Given the expression defined for $$x,y \in {[}0,2\pi{)}$$ \begin{align} E(x,y)&=\cos(2x)+\cos(2y)+\cos(2x+2y) \\ &+4\cos{x}+4\cos y+4\cos(x+y)\\ &-2\cos(x-y)-2\cos(2x+y)-2\cos(2y+x) \end{align} Wolfram-Alpha tells me that the minimum of this multivariable expression is $$-\frac{27}{2}$$, for $$x=y=\frac{2\pi}{3}$$.

How can this be proved (easily)?

I've tried using multivariable calculus but the calculations are really "messy". Using formulae that transform sums of trigonometric functions to products did not help either, in my case.

Tags :

It may not be very simple but i'll try.

First observe that it's a continuous function and it should have minimum and maximum on a closed interval. Next, observe that the function is symmetrical relative to $$x$$ and $$y$$ so if there is a minimum inside of the interval it will be achieved at some point $$(a,a)$$ because partial derivatives will be symmetrical as well. So we just need to find minimum of $$f(x)=E(x,x)=\cos 4x-4\cos 3x+6\cos 2x+8\cos x-2$$ Using multiple angle formulas, we can write this function as a polynomial of $$\cos x$$: $$f(x)=8\cos^4x-16\cos^3x+4\cos^2x+20\cos x-7$$ Let's find derivative $$f'(x)=-\sin x \cdot (32\cos^3x-48\cos^2x+8\cos x+20)$$ or $$f'(x)=-4\sin x \cdot(2\cos x+1)(4\cos^2x-8\cos x+5)$$.

We have two critical points to check: $$x=0$$ and $$\cos x=-0.5$$ ((plus the boundaries).

$$f(0)=f(2\pi)=9$$, when $$x_m=\frac{2\pi}{3}$$ we have our minimum $$f(x_m)=-\frac{27}{2}$$

Vasya
July 16, 2019 13:30 PM