Question

cos2π÷7.cos4π÷7.cos8π÷7=1÷8​

Answer 1

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)}*{2cos(4π/7)}*{2cos(8π/7)}

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

[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:

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