An example of Trigonometric Optimization

Problem: If A, B, C are angles of a triangle, then find the maximum value of

$\cos{A}+\cos{B}+\cos{C}$ expressed as a reduced rational. (IITJEE 1984).

First Hint: Prove that the maximum must occur for an equilateral triangle.

Second Hint: For a fixed value of one of the three angles, show that the maximum must occur when the other two are equal.

Solution:

Here, the three angles A, B, C are constrained by the requirement that they are non-negative and add up to

$\pi$ . Let S be the set of all ordered triples of the form $(A,B,C)$ which satisfy these constraints, that is,

$S= \{ (A,B,C)= A \geq 0, B \geq 0, C \geq 0 and A+B+C=\pi\}$

and denote $\cos{A}+\cos{B}+\cos{C}$ by $f(A,B,C)$ . Then, the problem amounts to finding the maximum value of the function f over the set S.

Now, suppose $(A,B,C) \in S$ . We claim that unless $A=B$ , we can find some

$(A^{'}, B^{'}, C^{'}) \in S$ such that $f(A^{'},B^{'},C^{'})>f(A,B,C)$ . Indeed, we let $C^{'}=C$ and $A^{'}=B^{'}=\frac{A+B}{2}$ . Note that $A^{'}+B^{'}=A+B$ and hence,

$(A^{'},B^{'},C^{'}) \in S$ . Note that $\cos{\frac{A+B}{2}}=\cos{\frac{A^{'}+B^{'}}{2}} \neq 0$

(except in the degenerate case where $A=B=\pi/2$ ). Further, $\cos{\frac{A^{'}-B^{'}}{2}}=\cos{0}=1$ which is greater than $\cos{\frac{A-B}{2}}$ . This gives,