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I was trying to compute: $$P=\sin(2\pi/7) \sin(4\pi/7) \sin(8\pi/7)$$ So I managed to analyse the equation $\sin(7x)=0$, thus $x = k\pi/7$, when $k$ is an integer.

The problem arrives when the equation $\sin(7x) = 7\sin(x) - 56\sin^{3}(x) + 112\sin^{5}(x) - 64\sin^{7}(x)=0$ is derived, because I can't exactly know what are its roots.

Assuming $\sin(x) ≠ 0$, the equation becomes $$7\ - 56\sin^{2}(x) + 112\sin^{4}(x) - 64\sin^{6}(x)= 7 - 56t^{2} + 112t^{4} - 64t^{6} = 0$$ Where $t=\sin(x)$.

But, to solve my problem, how can I set the values for the integer parameter $k$, since I've already looked in Wolfram, and the roots of that equation in $t$ are, apparently, $± \,\sin(2\pi/7),\,± \,\sin(4\pi/7),\,± \,\sin(8\pi/7)$, so some good values for $k$ are $±\,1,\,±\,2$ and $\,±\,3$, since we have three pairs of roots with opposite signs, and, by setting that, I successfully get the desired result, $P=-\,\frac{\sqrt7}{8}$. But, I don't know how to set those "good" values for $k$, can you help me?

Another question, that has to do with this one, is it correct to affirm that the Vieta's relation does not hold for trigonometric polynomials like the one for $\frac{\sin(7x)}{\sin(x)}=7\ - 56\sin^{2}(x) + 112\sin^{4}(x) - 64\sin^{6}(x)$, because the sign of the roots would not satisfy it?

Thanks in advance!

GuySa
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user405919
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1 Answers1

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You have $$\frac{\sin 7x}{\sin x}=P(\sin x)$$ where $$P(t)=7-56t^2+112t^4-64t^6.$$ The distinct roots of $P(t)$ are $\pm\sin(2\pi/7)$, $\pm\sin(4\pi/7)$ and $\pm\sin(6\pi/7)$. But $P(t)=Q(t^2)$ where $$Q(u)=7-56u+112u^2-64u^3.$$ The distinct roots of $Q(u)$ are $\sin^2(2\pi/7)$, $\sin^2(4\pi/7)$ and $\sin^2(6\pi/7)$. By Vieta, $$\sin^2(2\pi/7)\sin^2(4\pi/7)\sin^2(6\pi/7)=\frac7{64}.$$ Taking square roots: $$\sin(2\pi/7)\sin(4\pi/7)\sin(6\pi/7)=\frac{\sqrt7}{8}$$ as these sines are all positive. Finally $$\sin(2\pi/7)\sin(4\pi/7)\sin(8\pi/7)=-\frac{\sqrt7}{8}$$ as $\sin(8\pi/7)=-\sin(6\pi/7)$.

Angina Seng
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