How do you evaluate $sin {\frac{π}{7}} sin {\frac{2 π}{7}} sin {\frac{3 π}{7}}$ ?

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Answer 1 · 2016-11-16T15:42:52

Let $x = sin (\frac{π}{7}) sin (\frac{2 π}{7}) sin (\frac{3 π}{7})$
and

$y = cos (\frac{π}{7}) cos (\frac{2 π}{7}) cos (\frac{3 π}{7})$

So

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

$= sin (\frac{2 π}{7}) sin (\frac{4 π}{7}) sin (\frac{6 π}{7})$

$= sin (\frac{2 π}{7}) sin (π - \frac{3 π}{7}) sin (π - \frac{π}{7})$

$= sin (\frac{2 π}{7}) sin (\frac{3 π}{7}) sin (\frac{π}{7}) = x$

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

Now let

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

So

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

Again

$8 x^{2} = 8 {sin}^{2} (\frac{π}{7}) {sin}^{2} (\frac{2 π}{7}) {sin}^{2} (\frac{3 π}{7})$

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

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

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

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

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

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

Now

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

$= \frac{1}{2} [2 cos (\frac{π}{7}) cos (\frac{3 π}{7}) - 2 cos (\frac{π}{7}) cos (\frac{2 π}{7}) - 2 cos (\frac{2 π}{7}) cos (\frac{3 π}{7})]$

$= \frac{1}{2} [cos (\frac{4 π}{7}) + cos (\frac{2 π}{7}) - cos (\frac{3 π}{7}) - cos (\frac{π}{7}) - cos (\frac{5 π}{7}) - cos (\frac{π}{7})]$

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

$= \frac{1}{2} [2 cos (\frac{2 π}{7}) - 2 cos (\frac{3 π}{7}) - 2 cos (\frac{π}{7})]$

$= b - a - c$

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

Hence we get

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

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

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

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

$\Rightarrow sin (\frac{π}{7}) sin (\frac{2 π}{7}) sin (\frac{3 π}{7}) = \frac{\sqrt{7}}{8}$

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