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Question

Prove that 2+2+2+2 cos 8θ=2 cos θ.

Or

Prove that (1+cosπ8)(1+cos3π8)(1+cos5π8)(1+cos7π8)=18

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Solution

LHS = 2+2+2+2cos 8θ

= 2+2+2(1+cos 8θ)

= 2+2+2.2cos2 4θ[ 2cos2 θ=1+cos 2θ]

= 2+2+2cos 4θ=2+2(1+cos 4θ)

= 2+2.2cos2 2θ[ 2cos2 θ=1+cos 2θ]

= 2+2cos 2θ

= 2(1+cos 2θ)

= 2.2cos2 θ=2cos θ=RHS Hence proved.

Or

LHS = (1+cosπ8)(1+cos3π8)(1+cos5π8)(1+cos7π8)

= (1+cosπ8)(1+cos3π8)[1+cos(π3π8)][1+cos(ππ8)]

= (1+cosπ8)(1+cos3π8)(1cosw3π8)(1cosπ8)

= (1+cosπ8)(1cosπ8)(1+cos3π8)(1cos3π8)

= (1cos2π8)(1cos23π8)=sin2π8×sin23π8

[ 1cos2 θ=sin2 θ]

= (1cos π/42)(1cos3π42)

[ sin2x=1cos 2x2]

= 14(112)(1+12)

= 14(112)=14×12=18=RHS Hence proved.


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