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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
avatar+64 
-4

\(sqrt3/2\)

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 Jun 22, 2019
 #2
avatar+7683 
0

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
avatar+64 
-4

Try this: 

 

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

CuteDramione  Jun 22, 2019
 #4
avatar+64 
-4

3 sqrt3/2 whoops... :P I think..

CuteDramione  Jun 22, 2019
 #5
avatar+64 
-4

or sqrt3 not sure yet

CuteDramione  Jun 22, 2019
 #6
avatar+7683 
0

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
avatar+64 
-4

I cant type it here...

CuteDramione  Jun 22, 2019
 #8
avatar+7683 
0

Hmm ok. I will do the searching myself.

MaxWong  Jun 22, 2019
 #9
avatar+64 
-4

ok.!.!.!

CuteDramione  Jun 22, 2019
 #10
avatar+4296 
+1

Maximum Value: A=B=C=60 degrees, 1/2+1/2+1/2=3/2.

 Jun 22, 2019
edited by Guest  Jun 22, 2019

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