find extreme values of $\cos(x)+\cos(y)+\cos(z)$ when $x+y+z=\pi$

Question

How can I find the maximum and minimum of $\cos(x)+\cos(y)+\cos(z)$ if $x,y,z\geq0$ such that they are vertices of a triangle with $x+y+z=\pi$. I don't know how to start, but I feel like the Lagrange multipliers are a good place to start.

Answer 1

The method of Lagrange multipliers is indeed a good way to start the analysis. We are finding the extrema of $\cos x+\cos y+\cos z$ subject to the constraint $x+y+z=w$. The Lagrangian is $\cos x+\cos y+\cos z-(x+y+z-\pi)$.

By what I will assume is a routine calculation for you, the process yields $\sin x=\sin y=\sin z$. This has two solutions. The obvious one is $x=y=z=\frac{\pi}{3}$. But maybe two of the angles, say $x$ and $y$, are equal, and $z=\pi-x$. That yields $x=0$.

The maximum is when all the angles are equal. The minimum, for "real" triangles, is not attained.

Answer 2

i don't know about the ten proofs but here is one way to do this.

let $$f = \cos x +\cos y + \cos z = \cos x + \cos y - \cos(x+y), 0 \le x, y \le \pi.$$ we will look at the unconstrained optimization of $f$. at a local maximum, we need $$f_x = -\sin x + \sin(x+y) = 0, f_y = -\sin y + \sin(x+y)= 0 $$ now, $$\sin x = \sin y, 0 \le x, y \le \pi \implies x = y, \sin x = \sin 2x$ \to x = 0, \pi, x = \pi/3.$$

$\begin{array}{|c|c|c|} \hline x & 0 & \pi/3 \\ \hline y&0 & \pi/3\\ \hline \cos x + \cos y - \cos(x+y) & 1 &3/2 \\\hline \end{array}$

on all the the boundaries $f = 0,$ therefore the global maximum is $3/2.$

Answer 3

There are about $10$ proofs out there for this problem. But my favorite is: $\cos x + \cos y + \cos z = 1+\dfrac{r}{R}$. But by Euler's inequality: $R \geq 2r$. The result follows.