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# Got a question

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In $$\triangle ABC$$, what is the maximum value of $$\cos A + \cos B + \cos C$$?

I have solved the counterpart of this question with sin instead of cos. I used Jensen's inequality to get the result that the maximum occurs at A = B = C = pi/3. However, as the concavity of y = cos x changes at x = pi/2, Jensen's inequality does not work anymore. Can someone help?

Jun 22, 2019
edited by MaxWong  Jun 22, 2019

#1
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$$sqrt3/2$$

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Jun 22, 2019
#2
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Nope. I am sure it could be greater.

For example. in a 90-45-45 triangle, cos A + cos B + cos C = 2 sqrt (2), which is already greater than sqrt(3)/2.

MaxWong  Jun 22, 2019
#3
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Try this:

If A is fixed then B+C is fixed, and the product is greatest when B=C

CuteDramione  Jun 22, 2019
#4
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3 sqrt3/2 whoops... :P I think..

CuteDramione  Jun 22, 2019
#5
+73
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or sqrt3 not sure yet

CuteDramione  Jun 22, 2019
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If A is fixed, max value occurs when B = C,

If B is fixed, max value occurs when A = C.

If C is fixed, max value occurs when A = B.

So the maximum will occur when A = B = C

Max value = 3 cos (pi / 3) = 3/2

I got the answer, but why does the max value occur when B = C if A is fixed?

MaxWong  Jun 22, 2019
#7
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I cant type it here...

CuteDramione  Jun 22, 2019
#8
+9665
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Hmm ok. I will do the searching myself.

MaxWong  Jun 22, 2019
#9
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ok.!.!.!

CuteDramione  Jun 22, 2019
#10
+4613
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Maximum Value: A=B=C=60 degrees, 1/2+1/2+1/2=3/2.

Jun 22, 2019
edited by Guest  Jun 22, 2019