Question

Cos(π/7)cos(2π/7)cos(3π/7)cos(4π/7)cos(5π/7)cos(6π/7)

Answer 1

Cos(π/7)cos(2π/7)cos(3π/7)cos(4π/7)cos(5π/7)cos(6π/7)

we know , cos(π - α) = -cosα

so, cos(6π/7) = cos(π - π/7) = -cos(π/7)

cos(5π/7) = cos(π - 2π/7) = -cos(2π/7)

and cos(3π/7) = cos(π - 4π/7) = -cos(4π/7)

so, Cos(π/7)cos(2π/7)cos(3π/7)cos(4π/7)cos(5π/7)cos(6π/7) = -cos²(π/7).cos²(2π/7).cos²(4π/7)

= -[cos(π/7).cos(2π/7).cos(4π/7)]² ....(1)

now use the application, cosβ.cos2β.cos4β = sin8β/8sinβ

let β = π/7

then, cos(π/7).cos(2π/7).cos(4π/7)

= cosβ.cos2β.cos4β

= sin8β/8sinβ

= sin(8π/7)/8sin(π/7)

= sin(π + π/7)/8sin(π/7)

= -sin(π/7)/8sin(π/7)

= -1/8 putting it in equation (1),

so, -[cos(π/7).cos(2π/7).cos(4π/7)]²

= - [ -1/8 ]²

= -1/64

hence, Cos(π/7)cos(2π/7)cos(3π/7)cos(4π/7)cos(5π/7)cos(6π/7) = -1/64