If $A+B+C+D+E = 540^\circ$ what is $\min (\cos A+\cos B+\cos C+\cos D+\cos E)$? Let each of $A, B, C, D, E$ be an angle that is less than $180^\circ$ and is greater than $0^\circ$. Note that each angle can be neither $0^\circ$ nor $180^\circ$.
If $A+B+C+D+E = 540^\circ,$ what is the minimum of the following function? $$\cos A+\cos B+\cos C+\cos D+\cos E$$
I suspect the minimum is achieved when $A=B=C=D=E$, but I can't prove it. I need your help.

What you're looking at is a constrained optimization problem. 
Put another way, you are being asked to minimize $\cos A+\cos B+\cos C+\cos D+\cos E$ given the constraint $A+B+C+D+E=540$. 
Using lagrange multipliers, we can rewrite this as the minimization of the function
$L(A,B,C,D,E,\lambda)=\cos A+\cos B+\cos C+\cos D+\cos E+\lambda(540-A-B-C-D-E)$
To minimize the function, you want to set each partial derivative of L ($\frac{\partial L}{\partial A},\frac{\partial L}{\partial B},...,\frac{\partial L}{\partial \lambda})$ to zero. This should get you the value for $\lambda$ as well as values for each angle. 
With that I think you can take the last step.
