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+3
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avatar+924 

Let $x_1,$ $x_2,$ $\dots,$ $x_9$ be real numbers such that \[\cos x_1 + \cos x_2 + \dots + \cos x_9 = 0.\]
Find the maximum value of $\cos 3x_1 + \cos 3x_2 + \dots + \cos 3x_9.$  

              

 Mar 31, 2020
edited by HELPMEEEEEEEEEEEEE  Apr 6, 2020
 #1
avatar+423 
+1

Hmm i can see your first part but not the second! Would you mind wrighting it in latex? TY!!cool

 Mar 31, 2020
 #3
avatar+924 
+1

I converted the question to an image, if you can answer this question that'll help a lot! Thanks!

HELPMEEEEEEEEEEEEE  Apr 6, 2020
 #2
avatar
-2

The maximum value is 3.

 Apr 1, 2020
 #4
avatar+21957 
+2

If you let  x1 = 0   x2 = 2pi/3   x3 = -2pi/3  x4 = 4pi/3  x5 = -4pi/3  x6 = 2pi  x7 = 8pi/3  x8 = -8pi/3  x9 = 4pi

the the sum of cos(3xi) becomes 9.

 Apr 7, 2020
 #5
avatar+924 
0

Thank you, that is correct!

HELPMEEEEEEEEEEEEE  Apr 7, 2020
 #6
avatar+111550 
0

How did you come by those values Geno?

Melody  Apr 8, 2020
 #7
avatar+924 
+1

Let three of the \(x_i\) be equal to 0, and let the remaining six be equal to \(\frac{2 \pi}{3}.\) Then

\(\cos x_1 + \cos x_2 + \dots + \cos x_9 = 3 \cos 0 + 6 \cos \frac{2 \pi}{3} = 3 + 6 \left( -\frac{1}{2} \right) = 0.\)
Also, 

\(\cos 3x_1 + \cos 3x_2 + \dots + \cos 3x_9 = 3 \cos 0 + 6 \cos 2 \pi = \boxed{9}.\)
Since \(\cos x \le 1\) for all \(x,\) this is clearly the maximum.

 

Answer from AOPS's Alcumus.

 

🔥🔥🔥

HELPMEEEEEEEEEEEEE  Apr 8, 2020
 #8
avatar+31296 
+1

More generally:

 

 Apr 9, 2020

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