# How do you evaluate sin{pi/7} sin{(2pi)/7} sin{(3pi)/7}?

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Nov 17, 2016

Let $x = \sin \left(\frac{\pi}{7}\right) \sin \left(\frac{2 \pi}{7}\right) \sin \left(\frac{3 \pi}{7}\right)$
and

$y = \cos \left(\frac{\pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \cos \left(\frac{3 \pi}{7}\right)$

So

$8 x y = 2 \sin \left(\frac{\pi}{7}\right) \cos \left(\frac{\pi}{7}\right) \times 2 \sin \left(\frac{2 \pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \times 2 \sin \left(\frac{3 \pi}{7}\right) \cos \left(\frac{3 \pi}{7}\right)$

$= \sin \left(\frac{2 \pi}{7}\right) \sin \left(\frac{4 \pi}{7}\right) \sin \left(\frac{6 \pi}{7}\right)$

$= \sin \left(\frac{2 \pi}{7}\right) \sin \left(\pi - \frac{3 \pi}{7}\right) \sin \left(\pi - \frac{\pi}{7}\right)$

$= \sin \left(\frac{2 \pi}{7}\right) \sin \left(\frac{3 \pi}{7}\right) \sin \left(\frac{\pi}{7}\right) = x$

Hence $y = \frac{1}{8}$

Now let

$\cos \left(\frac{\pi}{7}\right) = a , \cos \left(\frac{2 \pi}{7}\right) = b , \cos \left(\frac{3 \pi}{7}\right) = c$

So

$y = \cos \left(\frac{\pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \cos \left(\frac{3 \pi}{7}\right) = a b c = \frac{1}{8}$

Again

$8 {x}^{2} = 8 {\sin}^{2} \left(\frac{\pi}{7}\right) {\sin}^{2} \left(\frac{2 \pi}{7}\right) {\sin}^{2} \left(\frac{3 \pi}{7}\right)$

$= \left[1 - \cos \left(\frac{2 \pi}{7}\right)\right] \left[1 - \cos \left(\frac{4 \pi}{7}\right)\right] \left[1 - \cos \left(\frac{6 \pi}{7}\right)\right]$

$= \left[1 - \cos \left(\frac{2 \pi}{7}\right)\right] \left[1 + \cos \left(\frac{3 \pi}{7}\right)\right] \left[1 + \cos \left(\frac{\pi}{7}\right)\right]$

$= \left(1 - b\right) \left(1 + c\right) \left(1 + a\right)$

$= \left(1 - b\right) \left(1 + c\right) \left(1 + a\right)$

$= 1 + a - b + c + a c - a b - b c - a b c$

$= 1 + a - b + c + a c - a b - b c - \frac{1}{8}$

$\implies 8 {x}^{2} = \frac{7}{8} + a - b + c + a c - a b - b c$

Now

$a c - a b - b c = \frac{1}{2} \left(2 a c - 2 a b - 2 b c\right)$

=1/2[2cos(pi/7)cos((3pi)/7) -2cos(pi/7)cos((2pi)/7)-2cos((2pi)/7)cos((3pi)/7)]

=1/2[cos((4pi)/7)+cos((2pi)/7) -cos((3pi)/7)-cos(pi/7)-cos((5pi)/7)-cos(pi/7)]

=1/2[-cos((3pi)/7)+cos((2pi)/7) -cos((3pi)/7)+cos((2pi)/7)-2cos(pi/7)]

=1/2[2cos((2pi)/7) -2cos((3pi)/7)-2cos(pi/7)]

$= b - a - c$

$\implies a c - a b - b c + a - b + c = 0$

Hence we get

$\implies 8 {x}^{2} = \frac{7}{8} + a - b + c + a c - a b - b c$

$\implies 8 {x}^{2} = \frac{7}{8} + 0$

$\implies {x}^{2} = \frac{7}{64}$

$\implies x = \frac{\sqrt{7}}{8}$

$\implies \sin \left(\frac{\pi}{7}\right) \sin \left(\frac{2 \pi}{7}\right) \sin \left(\frac{3 \pi}{7}\right) = \frac{\sqrt{7}}{8}$

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