plzz solve it as fast as possible
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plz marks as brainlist
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Prove that
8cos2(θ)+4cos4(θ)+cos6(θ)=4
Given
sin(θ)+sin^2(θ)+sin^3(θ)=1
sin(θ)+sin^3(θ)=1-sin^2(θ)
sin(θ)+sin^3(θ)=cos^2(θ)
sin (θ)[1+1-cos^2(θ)]=cos^2(θ)
sin (θ)[2-cos^2(θ)]=cos^2(θ)
sin (θ)=cos^2(θ)/[2-cos^2(θ)]
√(1-cos^2θ)=cos^2(θ)/[2-cos^2(θ)]
(1-cos^2θ)=cos^4(θ)/[4+cos^4θ-4cos^2(θ)]
4-4cos^2θ+cos^4θ-cos^6θ-4cos^2θ+4cos^4θ=cos^4
4-8cos^2θ-cos^6θ+4cos^4θ=0
cos^6θ-4cos^4θ+8cos^2θ=4
8cos2(θ)+4cos4(θ)+cos6(θ)=4
Given
sin(θ)+sin^2(θ)+sin^3(θ)=1
sin(θ)+sin^3(θ)=1-sin^2(θ)
sin(θ)+sin^3(θ)=cos^2(θ)
sin (θ)[1+1-cos^2(θ)]=cos^2(θ)
sin (θ)[2-cos^2(θ)]=cos^2(θ)
sin (θ)=cos^2(θ)/[2-cos^2(θ)]
√(1-cos^2θ)=cos^2(θ)/[2-cos^2(θ)]
(1-cos^2θ)=cos^4(θ)/[4+cos^4θ-4cos^2(θ)]
4-4cos^2θ+cos^4θ-cos^6θ-4cos^2θ+4cos^4θ=cos^4
4-8cos^2θ-cos^6θ+4cos^4θ=0
cos^6θ-4cos^4θ+8cos^2θ=4
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