Asked • 03/23/19

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.

Tom K.

If all 5 are 108 degrees, 5 cos(108 degrees) = 5 * (1-sqrt(5))/4 = -1.54508 If one is 0 degrees and 4 are 135 degrees, cos(0) + 4 * cos(135 degrees) = 1 + 4 * (- sqrt(2)/2) = 1 - 2 * sqrt(2) = -1.82843. (If all 5 are between 90 and 180 degrees, the function is convex, so the minimum is where you suggest.)
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03/23/19

1 Expert Answer

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Daniel M. answered • 05/22/20

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