Question

Prove that cos pi/65.cos 2pi/65.cos 4pi/65.cos 8pi/65.cos 16pi/65.cos 32 pi/65 = 1/64?

Answer 1

Answer:

Proved

Cos(π/65).Cos(2π/65).Cos(4π/65).Cos(8π/65).Cos(16π/65).Cos(32π/65) = 1/64

Step-by-step explanation:

Prove that cos pi/65.cos 2pi/65.cos 4pi/65.cos 8pi/65.cos 16pi/65.cos 32 pi/65 = 1/64?

LHS =

= Cos(π/65).Cos(2π/65).Cos(4π/65).Cos(8π/65).Cos(16π/65).Cos(32π/65)

Multiply & Divide by Sin(π/65)

=(1/Sin(π/65)) (Sin(π/65) . Cos(π/65).Cos(2π/65).Cos(4π/65).Cos(8π/65).Cos(16π/65).Cos(32π/65))

Using SinθCosθ = Sin(2θ) / 2

= (1/2Sin(π/65)) (Sin(2π/65).Cos(2π/65).Cos(4π/65).Cos(8π/65).Cos(16π/65).Cos(32π/65))

=  (1/4Sin(π/65)) (Sin(4π/65).Cos(4π/65).Cos(8π/65).Cos(16π/65).Cos(32π/65))

= (1/8Sin(π/65)) (Sin(8π/65).Cos(8π/65).Cos(16π/65).Cos(32π/65))

= (1/16Sin(π/65)) (Sin(16π/65).Cos(16π/65).Cos(32π/65))

= (1/32Sin(π/65)) (Sin(32π/65).Cos(32π/65))

=  (1/64Sin(π/65)) (Sin(64π/65))

= (1/64)( (Sin(64π/65) /Sin(π/65) )

Sinθ = Sin(π-θ)

Sin(π/65) = Sin(π - π/65) = Sin(64π/65)

= (1/64) ( (Sin(64π/65) /Sin(64π/65) )

= 1/64

= RHS

QED

Answer 2

Answer:

dear this is your answer

Step-by-step explanation:

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