## 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,

$f(A,B,C)=\cos{A}+\cos{B}+\cos{C}=2\cos{\frac{A+B}{2}}\cos{\frac{A-B}{2}}+\cos{C}$

which is less than

$2\cos{\frac{A^{'}+B^{'}}{2}}\cos{\frac{A^{'}-B^{'}}{2}}+\cos{C^{'}}$

which equals $\cos{A^{'}}+\cos{B^{'}}+\cos{C^{'}}=f(A^{'},B^{'},C^{'})$.

By a similar reasoning, unless $B=C$, we can find some $(A^{''}, B^{''}, C^{''}) \in S$ such that

$f(A,B,C). A similar assertion holds if $C \neq A$. It then follows that when

$f(A,B,C)$ is maximum, A, B, C must be all equal and hence, each equals $\pi/3$

(since $A+B+C=\pi/3$). But, $f(\pi/3,\pi/3,\pi/3)=1/2+1/2+1/2=3/2$. So, the maximum value of

$\cos{A}+\cos{B}+\cos{C}=3/2$.

More later,

Nalin Pithwa