While it's trivial to prove $\prod_{k=0}^{n-1}\cos(2^kx)=\frac{\sin(2^nx)}{2^n\sin x}$, Wikipedia refers to a "similar" identity $\sin\tfrac{\pi}{9}\sin\tfrac{2\pi}{9}\sin\tfrac{4\pi}{9}=\frac{\sqrt{3}}{8}$. How does this generalize to a result for $\prod_{k=0}^{n-1}\sin(2^kx)$? Failing that, how do we prove this special case?
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I can show you a proof of the similar identity, but I don't know a general formula for $\prod_{k=0}^{n-1}\sin(2^kx)$.
According to Showing $\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}} = \frac{\sqrt{13}}{64}$, we have that $$\prod _{k=1}^{n-1}\,\sin \left({\frac {k\pi }{n}} \right)=\frac{n}{2^{n-1}} .$$ For $n=9$, by symmetry, we have that $$\left(\sin \left({\frac {\pi }{9}}\right) \sin \left({\frac {2\pi }{9}}\right)\sin \left({\frac {4\pi }{9}}\right)\right)^2\underbrace{\sin^2\left({\frac {\pi }{3}}\right)}_{3/4}=\prod _{k=1}^{8}\,\sin \left({\frac {k\pi }{9}} \right)=\frac{9}{256} $$ and it follows that $$\sin \left({\frac {\pi }{9}}\right) \sin \left({\frac {2\pi }{9}}\right)\sin \left({\frac {4\pi }{9}}\right)=\sqrt{\frac{3}{64}}=\frac{\sqrt{3}}{8}.$$ In a similar way, for $n=7$, we show that $$\sin \left({\frac {\pi }{7}}\right) \sin \left({\frac {2\pi }{7}}\right)\sin \left({\frac {4\pi }{7}}\right)=\frac{\sqrt7}{8}.$$
Robert Z
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