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What is $\cos\frac{\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}+\cos\frac{7\pi}{13}+\cos\frac{9\pi}{13}+\cos\frac{11\pi}{13}$

I feel like this question shouldn't be too difficult but I'm getting stuck on it for some reason. I paired the first and last terms, second and second-to-last terms, etc, and used the sum of cosines formula to get

$\cos\frac{\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}+\cos\frac{7\pi}{13}+\cos\frac{9\pi}{13}+\cos\frac{11\pi}{13}=2\cos\frac{6\pi}{13}(\cos\frac{5\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{\pi}{13})$

But I don't know where to go from here.

Bernard
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Silverleaf1
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4 Answers4

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You have the high school formula: $$ 1+\cos\theta+\cos 2\theta+\dots+\cos n\theta=\frac{\sin\frac{(n +1)\theta}{2}}{\sin \frac{\theta}{2}}\,\cos\frac{n\theta}{2}.$$

Bernard
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$$\cos\frac{\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}+\cos\frac{7\pi}{13}+\cos\frac{9\pi}{13}+\cos\frac{11\pi}{13}=$$ $$=\tfrac{2\sin\frac{\pi}{13}\cos\frac{\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{3\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{5\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{7\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{9\pi}{13}+2\sin\frac{\pi}{13}\cos\frac{11\pi}{13}}{2\sin\frac{\pi}{13}}=$$ $$=\tfrac{\sin\frac{2\pi}{13}+\sin\frac{4\pi}{13}-\sin\frac{2\pi}{13}+\sin\frac{6\pi}{13}-\sin\frac{4\pi}{13}+\sin\frac{8\pi}{13}-\sin\frac{6\pi}{13}+\sin\frac{10\pi}{13}-\sin\frac{8\pi}{13}+\sin\frac{12\pi}{13}-\sin\frac{10\pi}{13}}{2\sin\frac{\pi}{13}}=\frac{1}{2}.$$

Michael Rozenberg
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Let $a = \frac \pi{13}$ and note that,

$$ 2\sin a \cos a = \sin 2a$$ $$ 2\sin a \cos 3a = \sin 4a -\sin 2a $$ $$ 2\sin a \cos 5a = \sin 6a -\sin 4a $$

$$...$$ $$ 2\sin a \cos 11a = \sin 12a -\sin 10a $$

Add up both sides and divide by $2\sin a$,

$$\cos\frac{\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}+\cos\frac{7\pi}{13}+\cos\frac{9\pi}{13}+\cos\frac{11\pi}{13}= \frac {\sin 12a }{2\sin a} = \frac12$$

where $\sin12a = \sin a$ is used in the last step.

Quanto
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HINT.-$$(\cos\frac{\pi}{13}++\cos\frac{11\pi}{13})+(\cos\frac{3\pi}{13}+\cos\frac{9\pi}{13})+(\cos\frac{5\pi}{13}+\cos\frac{7\pi}{13})$$ $$2\cos6X(\cos5X+\cos3X+\cos X)$$ where $X=\dfrac{\pi}{13}$ $$2\cos6X(2\cos4X\cos X+\cos X)=2\cos6X\cos X(2\cos4X+1)$$ $$2\cos\dfrac{6\pi}{13}\cos\frac{\pi}{13}(2\cos\frac{4\pi}{13}+1)=\frac12$$

Piquito
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