If $\sin(\pi/13)\sin(3\pi/13)\sin(4\pi/13)={a\over b}\sqrt{\frac{k-3\sqrt k}{2}}$, find $\frac{5a+b}{k}$.

Question

If $\sin(\pi/13)\sin(3\pi/13)\sin(4\pi/13)={a\over b}\sqrt{\frac{k-3\sqrt k}{2}}$, find $\frac{5a+b}{k}$.

Let $s=\sin(\pi/13)\sin(3\pi/13)\sin(4\pi/13)$ and $x=\sqrt k$. Then the above equation becomes

$$\frac{x(x-3)}{2}=\frac{s^2b^2}{a^2}\text{ i.e. }x^2-3x-\frac{2s^2b^2}{a^2}=0$$

Therefore $\displaystyle{x=\frac{3|a|\pm \sqrt{9a^2+8s^2b^2}}{2|a|}}$

I have to find the $s$. There is a problem here which evaluates $\sin(\pi/13)+\sin(3\pi/13)+\sin(4\pi/13)$.

So if I assume $\alpha=e^{i\pi/13}$. We get $s=\frac{(\alpha+\alpha^{12})(\alpha^3+\alpha^{10})(\alpha^4+\alpha^9)}{-8i}$.

I am not having any idea to proceed further. Can anyone help me with an idea to deal with this problem? Thanks for your help in advance.

Answer 1

If you use Newton-Girard formula $$\prod_{n=1}^6 \sin \left(\frac{n\pi }{13}\right)=\frac{\sqrt{13}}{64}$$ $$\frac{64}{\sqrt{13}}\sin \left(\frac{\pi }{13}\right) \sin \left(\frac{3 \pi }{13}\right) \sin \left(\frac{4 \pi }{13}\right)=\csc \left(\frac{2 \pi }{13}\right) \sec \left(\frac{\pi }{26}\right) \sec \left(\frac{3 \pi }{26}\right)$$ and the rhs is "just" $$4 \sqrt{2-\frac{6}{\sqrt{13}}}$$ which makes

$${a\over b}\sqrt{\frac{k-3\sqrt k}{2}}=\frac{1}{16} \sqrt{13 \left(2-\frac{6}{\sqrt{13}}\right)}=\frac{1}{256}\sqrt{26-6 \sqrt{13}}$$

Just finish !