4

From the article Products of Sines, we have $\sin 15^\circ\sin75^\circ=\sin 18^\circ\sin54^\circ=\frac{1}{4}$. We can rewrite this as $\sin \frac{\pi}{12}\sin\frac{5\pi}{12}=\sin \frac{\pi}{10}\sin\frac{3\pi}{10}=\frac{1}{4}$. Is there any good method to get $x,y$ such that $\sin x\sin y=\frac{1}{4}$ or more generally to get $x_i$ such that $$\prod_{i=1}^n\sin x_i=k$$ where $k$ is a rational number?

Later
  • 722
  • 2
  • 8
  • 24
pipi
  • 2,461
  • I know of one such identity $$\prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$$ but I don't know if there is a similar one in general. – Alexander Vlasev Oct 24 '12 at 05:40
  • 2
    A slightly more general identity than the one cited by Aleks is $$ 2,\sin \left( n\theta \right) =\prod _{k=0}^{n-1}2,\sin \left( \theta+{\frac {k\pi }{n}} \right) $$ See e.g. https://groups.google.com/forum/?hl=en&fromgroups=#!topic/sci.math/gFBgDRypbxs – Robert Israel Oct 24 '12 at 05:49
  • (note that if you divide by $\theta$ and take the limit as $\theta \to 0$, you get @AleksVlasev's identity) – Robert Israel Oct 24 '12 at 05:56
  • 1
    In particular taking $\theta = \pi/(2 n)$ we get $$2^{1-n} = \prod_{k=0}^{n-1} \sin\left(\left(k+\frac{1}{2}\right) \frac{\pi}{n}\right)$$ – Robert Israel Oct 24 '12 at 06:06
  • By combining these, we can get any rational as a quotient of products of sines of rational multiples of $\pi$. – Robert Israel Oct 24 '12 at 06:26
  • 4
    I wrote a paper in which I found all rational products of three and of four sines of rational angles (the case of two had already been done). Rational products of sines of rational angles, Aequationes Math. 45 (1993) 70-82, MR 93m:11140. Maybe this link works: http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002039893&IDDOC=178038 – Gerry Myerson Oct 24 '12 at 06:32

0 Answers0