Prove that: cos2pi/7 .cos4pi/7 cos8pi/7=1/8
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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
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