Prove that Cosπ11Cos2π11Cos3π11Cos4π11Cos5π11=132

C=Cosπ11Cos2π11Cos3π11Cos4π11Cos5π11and S=Sinπ11Sin2π11Sin3π11Sin4π11Sin5π11now C×S=Cosπ11Sinπ11Cos2π11Sin2π11Cos3π11Sin3π11Cos4π11Sin4π11Cos5π11Sin5π11=122Cosπ11Sinπ11122Cos2π11Sin2π11122Cos3π11Sin3π11122Cos4π11Sin4π11122Cos5π11Sin5π11=132Sin2π11Sin4π11Sin6π11Sin8π11Sin10π11=132Sin2π11Sin4π11Sin6π11Sin8π11Sin10π11

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Prove that
$Cos \frac{π}{11} Cos \frac{2 π}{11} Cos \frac{3 π}{11} Cos \frac{4 π}{11} Cos \frac{5 π}{11} = \frac{1}{32}$

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Detailed Solution

$C = Cos \frac{π}{11} Cos \frac{2 π}{11} Cos \frac{3 π}{11} Cos \frac{4 π}{11} Cos \frac{5 π}{11}$

and $S = Sin \frac{π}{11} Sin \frac{2 π}{11} Sin \frac{3 π}{11} Sin \frac{4 π}{11} Sin \frac{5 π}{11}$

now $C \times S = Cos \frac{π}{11} Sin \frac{π}{11} Cos \frac{2 π}{11} Sin \frac{2 π}{11} Cos \frac{3 π}{11} Sin \frac{3 π}{11} Cos \frac{4 π}{11} Si n \frac{4 π}{11} Cos \frac{5 π}{11} Sin \frac{5 π}{11}$

$= \frac{1}{2} (2 Cos \frac{π}{11} Sin \frac{π}{11}) \frac{1}{2} (2 Cos \frac{2 π}{11} Sin \frac{2 π}{11}) \frac{1}{2} (2 Cos \frac{3 π}{11} Sin \frac{3 π}{11}) \frac{1}{2} (2 Cos \frac{4 π}{11} Sin \frac{4 π}{11}) \frac{1}{2} (2 Cos \frac{5 π}{11} Sin \frac{5 π}{11})$