for x ≠ kπ, k∈Z show that: sin8x / 8sinx = cosx.cos2x.cos4x deduce the value of E = cosπ / 7.cos2π / 7.cos4π / 7 ?

for #x ≠ kπ, k∈Z# show that:
#(sin8x) / (8sinx) = cosx xxcos2x xxcos4x#
deduce the value of #E = cosπ / 7xx(cos2π) / 7xx(cos4π) / 7#

1 Answer
Jan 27, 2018

The product should have been written as #(\cos(\pi/7))×(\cos((2\pi)/7))×(\cos((4\pi)/7))# and that is #-1/8#.

Explanation:

Following Eshan S's answer we provevthe general trigonometric identity with the double angle formula for sines:

#\sin(8x)=2\sin(4x)\cos(4x)#

#=2(2\sin(2x)\cos(2x))\cos(4x)#

#=2(2(2\sin(x)\cos(x))\cos(2x))\cos(4x)#

#=8\sin(x)\cos(x)\cos(2x)\cos(4x)#

Then putting #x=\pi/7#:

#(sin((8\pi)/7))/(8\sin(\pi/7))=\cos(\pi/7)×cos((2\pi)/7)×cos((4\pi)/7)#

But also #(8\pi)/7=\pi+\pi/7# so #sin((8\pi)/7)=-\sin(\pi/7)#. Therefore:

#\cos(\pi/7)×cos((2\pi)/7)×cos((4\pi)/7)=(-\sin(\pi/7))/(8\sin(\pi/7))=-1/8#.