# Show that sinΠ/14*sin3Π/14*sin5Π/14*sin7Π/14*sin9Π/14*sin11 Π/14*sin13Π/14=1/64

Akshay
185 Points
6 years ago
Hi,
Products are odd terms of pi/14 will be difficult to solve. So, convert it to product of even terms of pi/14.
Use sin x =cos(pi/2 -x) and cos x = sin(pi/2 + x),
sin(pi/14) = cos(6pi/14), sin(3pi/14)=cos(4pi/14), sin(5pi/14)=cos(2pi/14), sin(7pi/14)=1,
sin(9pi/14)=sin(pi/2 + 2pi/14)=cos(2pi/14), sin(11pi/14)=cos(4pi/14) and sin(13pi/14)=cos(6pi/14).

So, total product = [cos(2pi/14) * cos(4pi/14) * cos(6pi/14)] ^ 2.
Multiply and divide by sin2(2pi/14),
product = [sin(2pi/14) * cos(2pi/14) * cos(4pi/14) * cos(6pi/14) / sin(2pi/14)] ^ 2,
Use sin2x = 2*sin(x)*cos(x),
product = ¼ * [sin(4pi/14) * cos(4pi/14) * cos(6pi/14) / sin(2pi/14) ] ^ 2,
use sin(2x) identity again,
product = 1/16 * [sin(8pi/14) * cos(6pi/14) / sin(2pi/14) ] ^2,
Use [ 2 * sin a * cos b = sin(a+b) + sin(a-b) ],
product = 1/64 * [ (sin(14pi/14) + sin(2pi/14))/sin(2pi/14) ] ^2,
sin(14pi/14) =0 and sin(2pi/14) will get cancelled,
So,
Product = 1/64
Rishi Sharma
one year ago
Dear Student,

Products are odd terms of pi/14 will be difficult to solve. So, convert it to product of even terms of pi/14.
Use sin x =cos(pi/2 -x) and cos x = sin(pi/2 + x),
sin(pi/14) = cos(6pi/14), sin(3pi/14)=cos(4pi/14), sin(5pi/14)=cos(2pi/14), sin(7pi/14)=1,
sin(9pi/14)=sin(pi/2 + 2pi/14)=cos(2pi/14), sin(11pi/14)=cos(4pi/14) and sin(13pi/14)=cos(6pi/14).

So, total product = [cos(2pi/14) * cos(4pi/14) * cos(6pi/14)] ^ 2.
Multiply and divide by sin2(2pi/14),
product = [sin(2pi/14) * cos(2pi/14) * cos(4pi/14) * cos(6pi/14) / sin(2pi/14)] ^ 2,
Use sin2x = 2*sin(x)*cos(x),
product = ¼ * [sin(4pi/14) * cos(4pi/14) * cos(6pi/14) / sin(2pi/14) ] ^ 2,
use sin(2x) identity again,
product = 1/16 * [sin(8pi/14) * cos(6pi/14) / sin(2pi/14) ] ^2,
Use [ 2 * sin a * cos b = sin(a+b) + sin(a-b) ],
product = 1/64 * [ (sin(14pi/14) + sin(2pi/14))/sin(2pi/14) ] ^2,
sin(14pi/14) =0 and sin(2pi/14) will get cancelled,
So,
Product = 1/64

Thanks and Regards