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)<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

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