2

When considering “Langley's Adventitious Angles”, I noticed that $$\sin40^\circ\sin50^\circ=\sin30^\circ\sin80^\circ.$$ What other distinct rational multiples of $\pi$ radians, say $0<\alpha, \beta, \gamma, \delta<\frac12\pi$, satisfy such an equation, namely $$\sin\alpha\sin\beta=\sin\gamma\sin\delta,$$ if any?

Edit:$\;$ As J.G. points out in a comment, $$\sin x\sin(\tfrac12\pi-x)=\sin x\cos x=\tfrac12\sin2x=\sin\tfrac16\pi\sin2x$$ provides a more general template for many solutions. So we are looking for solutions not of this type.

John Bentin
  • 20,004
  • 6
    The above example is a special case of $\sin x\sin(\pi/2-x)=\sin x\cos x=\tfrac12\sin2x=\sin(\pi/6)\sin2x$, which for your restriction requires only $0<x<\pi/4$. It's worth editing your question to ask whether there are any solutions other than that. – J.G. Sep 25 '22 at 16:40
  • 2
    An example not of that type is $\sin(15^o) \sin(75^o) = \sin(18^o) \sin(54^o)$. – Robert Israel Sep 25 '22 at 17:01

1 Answers1

2

The question specified that the angles are distinct, to exclude the trivial solutions where $\{\alpha,\beta\}=\{\gamma,\delta\}$. However, if this restriction is weakened to admit nontrivial solutions where just three of the angles are distinct, it turns out that there are three such solutions. These are included in the solutions listed below (see solutions 10,11,15). Apart from the family of solutions given by $$\{\{\alpha,\beta\},\{\gamma,\delta\}\}=\{\{r\pi,(\tfrac12-r)\pi\},\{\tfrac16\pi,2r\pi\}\}\quad(r\in\Bbb Q;\,0<r<\tfrac14),$$ which were pointed out by J.G., there are altogether just fifteen “sporadic” solutions; namely $\pi$ multiplied by the pairs of rational-number pairs tabulated as $$\begin{array}{c|cc|cc}\\ \hline 1&1/21 &8/21 &1/14 &3/14 \\ 2&1/14 &5/14 &2/21& 5/21\\ 3&4/21 &10/21 &3/14 &5/14\\ 4&1/20& 9/20& 1/15& 4/15\\ 5&2/15& 7/15& 3/20& 7/20\\ 6&1/30 &3/10& 1/15& 2/15\\ 7&1/15& 7/15& 1/10& 7/30\\ 8&1/10& 13/30& 2/15& 4/15\\ 9&4/15 &7/15& 3/10& 11/30\\ 10&1/30&11/30& 1/10& 1/10\\ 11&7/30 &13/30& 3/10 &3/10\\ 12&1/15& 4/15 &1/10& 1/6\\ 13&2/15 &8/15& 1/6 &3/10\\ 14&1/12& 5/12& 1/10& 3/10\\ 15&1/10& 3/10 &1/6 &1/6\\ \hline \end{array}$$ This comprehensive set of solutions (of which #14 was found by Robert Israel) is due to Gerry Myerson, and stated as Theorem 4 in his 2005 paper.

John Bentin
  • 20,004