+934 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 $\cos 3x_1 + \cos 3x_2 + \dots + \cos 3x_9.$
+422 Hmm i can see your first part but not the second! Would you mind wrighting it in latex? TY!!
+934 I converted the question to an image, if you can answer this question that'll help a lot! Thanks!
+23252 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.
+934 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.
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