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The expression can be reduced as $$|\cos \frac{2\pi}{7} -\cos \frac {\pi}{7} +\cos \frac{4\pi}{7} -\cos \frac{3\pi}{7}+\cdots|$$ $$=|2\sin \frac{\pi}{14}|| \sin \frac{3\pi}{14} + \sin \frac{7\pi}{14} +\sin \frac {11\pi}{14}|$$ $$ =2\sin \frac{\pi}{14} | 1 + 2\sin \frac{\pi}{2} \cos \frac{8\pi}{14}|$$ $$=2[\sin \frac {9\pi}{14} + \sin \frac {\pi}{14}]$$

I don’t think it’s reducible any further, but the given answer is 1. Where am I going wrong?

user26857
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Aditya
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  • $\cos 2 \pi/7 + \cos 4 \pi/7 + \cos 6 \pi/7 = -\frac{1}{2}$ has been proven many times, for example here. Since $\cos 8 \pi/7 = \cos 6 \pi/7$ and so on, you can just use the link. – Toby Mak Feb 18 '21 at 05:26
  • @TobyMak so there is no way to solve other than using complex numbers? I think there should be one using angle formulas – Aditya Feb 18 '21 at 07:29

1 Answers1

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The error is that $\sin \frac{3\pi}{14} + \sin \frac{7\pi}{14} +\sin \frac {11\pi}{14} \ne 1 + 2\sin \frac{\pi}{2} \cos \frac{8\pi}{14}$.

$\frac{3+11}{2} = 7$ instead of $8$, thus $\sin \frac{3 \pi}{14} + \sin \frac{11 \pi}{14} = 2 \sin \frac{7 \pi}{14} \cos \frac{4 \pi}{14}$, so the whole thing should equal $1 + 2 \cos \frac{2 \pi}{7}$ instead.

Toby Mak
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  • Ok, so it is $2\sin \frac{\pi}{14} (1+ 2\cos \frac{4\pi}{ 14})$. But I am still not able to reduce it to 1 – Aditya Feb 18 '21 at 05:12
  • If $x = 2 \pi / 14$, then inside the brackets is $1 + 2 \cos 2x = 1 + 2 (2 \cos^2 x - 1)$. If you have $\sin \pi/14$ and $\cos \pi/14$ that will be easier: try that first. – Toby Mak Feb 18 '21 at 05:15
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    Of course if you know complex numbers, you have $|(1 + \omega + \cdots + \omega^6) - 1| = |\frac{\omega^7 - 1}{\omega - 1} - 1| = 1$ where $\omega$ is a seventh root of unity, but I assume you're not allowed to use this. – Toby Mak Feb 18 '21 at 05:18
  • There must be some way using product-to-sum formulas again. – Toby Mak Feb 18 '21 at 05:23
  • I totally expect the answer to be found using product to sum formula. Complex numbers shouldn’t really be involved here, at least for my level. – Aditya Feb 18 '21 at 07:30