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# HELP ME!! Precalc/Trig

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

Mar 31, 2020
#3
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I converted the question to an image, if you can answer this question that'll help a lot! Thanks!

HELPMEEEEEEEEEEEEE  Apr 6, 2020
#2
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The maximum value is 3.

Apr 1, 2020
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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
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Thank you, that is correct!

HELPMEEEEEEEEEEEEE  Apr 7, 2020
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How did you come by those values Geno?

Melody  Apr 8, 2020
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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.

🔥🔥🔥

HELPMEEEEEEEEEEEEE  Apr 8, 2020
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More generally:

Apr 9, 2020