Question

Prove that cos 2π/7.cos4π/7.cos8π/7= 1/8​

Answer 1

Step-by-step explanation :

i)

Let P = cos(2π/7) * cos(4π/7) * cos(8π/7)

ii) Multiply and divide by 8*sin(2π/7) & grouping

==> P = {1/8*sin(2π/7)}*[{2*sin(2π/7)*cos(2π/7)}...

==> P = {1/8*sin(2π/7)}[2sin(4π/7)cos(4π/7)*{2co...

[Application of 2sinAcosA = sin(2A)]

==> P = {1/8*sin(2π/7)}[2sin(8π/7)cos(8π/7)}]

==> P = {1/8*sin(2π/7)}*{sin(16π/7) = sin(2π/7)/8*sin(2π/7)

[Since sin(16π/7) = sin(2π + 2π/7) = sin(2π/7)]

Thus P simplifies to 1/8

Hence it is proved that: cos(2π/7) * cos(4π/7) * cos(8π/7) = 1/8

Answer 2

Answer:

Hope it helps you Please mark me as brainliest