Question

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`

Answer 1

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`.