A geometry problem by Akshat Sharda

Using the identity $\displaystyle \prod_{k=1}^{n-1} \sin \left(\frac {k \pi}n \right) = \frac n{2^{n-1}}$ (see proof below), we have:

$\begin{aligned} \prod_{k=1}^6 \sin \left(\frac {k \pi}7 \right) & = \frac 7{2^6} \\ \sin \frac \pi 7 {\color{#3D99F6}{\sin \frac {2\pi} 7}} \sin \frac {3\pi} 7 {\color{#3D99F6}{\sin \frac {4\pi} 7}} \sin \frac {5\pi} 7 {\color{#3D99F6}{\sin \frac {6\pi} 7}} & = \frac 7{2^6} & \small \color{#3D99F6}{\text{As }\sin (\pi -x) = \sin x} \\ \sin \frac \pi 7 {\color{#3D99F6}{\sin \frac {5\pi} 7}} \sin \frac {3\pi} 7 {\color{#3D99F6}{\sin \frac {3\pi} 7}} \sin \frac {5\pi} 7 {\color{#3D99F6}{\sin \frac \pi 7}} & = \frac 7{2^6} \\ \sin^2 \frac \pi 7 \sin^2 \frac {3\pi} 7 \sin^2 \frac {5\pi} 7 & = \frac 7{2^6} \\ \implies \sin \frac \pi 7 \sin \frac {3\pi} 7 \sin \frac {5\pi} 7 & = \frac {\sqrt 7}{8} \end{aligned}$

$\implies a+b = 7+8 = \boxed{15}$

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Proof for Equation 24

$\begin{aligned} P & = \prod_{k=1}^{n-1} \sin \left(\frac {k \pi}n \right) \\ & = \prod_{k=1}^{n-1} \frac {i\left(e^{-\frac {k \pi}n i} - e^{\frac {k \pi}n i}\right)}2 \\ & = \frac {i^{n-1}}{2^{n-1}} \prod_{k=1}^{n-1} \frac {1 - e^{\frac {2 k \pi}n i}}{e^{\frac {k \pi}n i}} \\ & = \frac {i^{n-1}}{2^{n-1}} \frac {{\color{#3D99F6}\prod_{k=1}^{n-1} \left(1 - e^{\frac {2 k \pi}n i}\right)}}{\exp \left(\sum_{k=1}^{n-1} \frac {k \pi}n i \right)} & \small {\color{#3D99F6}\text{see Note}} \\ & = \frac {i^{n-1}}{2^{n-1}} \frac {{\color{#3D99F6}n}}{e^{\frac {(n-1)\pi i}2}} \\ & = \frac {i^{n-1}}{2^{n-1}} \frac n{i^{n-1}} \\ & = \frac n{2^{n-1}} \quad \blacksquare \end{aligned}$

Note that $e^{\frac {2 k \pi}n i} = \omega$ is the $n$ th root of unit, which satisfies the following equation.

$\begin{aligned} x^{n-1} + x^{n-2} + ... + x + 1 & = \prod_{k=1}^{n-1} (x-\omega^k) = 0 \end{aligned}$

$\begin{aligned} \implies \prod_{k=1}^{n-1} (x-\omega^k) & = \sum_{k=0}^{n-1}x^k \\ \prod_{k=1}^{n-1} (1-\omega^k) & = \sum_{k=0}^{n-1} 1 = n \end{aligned}$