We use, a^3+b^3=(a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2) to get,
(costheta)^6+(sintheta)^6=cos^6theta+sin^6theta(cosθ)6+(sinθ)6=cos6θ+sin6θ
=(cos^2theta+sin^2theta)(cos^4theta-cos^2thetasin^2theta+sin^4theta)=(cos2θ+sin2θ)(cos4θ−cos2θsin2θ+sin4θ)
=(1){(cos^2theta+sin^2theta)^2-2sin^2thetacos^2theta-cos^2thetasin^2theta}=(1){(cos2θ+sin2θ)2−2sin2θcos2θ−cos2θsin2θ}
=1-3cos^2thetasin^2theta=1−3cos2θsin2θ
=1-3/4(2sinthetacostheta)^2=1−34(2sinθcosθ)2
=1-3/4(sin2theta)^2=1−34(sin2θ)2
=1-3/4(sin^2(2theta))=1−34(sin2(2θ))
Recall that, 1-cos2A=2sin^2A1−cos2A=2sin2A.
Hence, the Exp.=1-3/8{2sin^2(2theta)}=1−38{2sin2(2θ)}
=1-3/8(1-cos4theta)=1−38(1−cos4θ)
=1-3/8+3/8cos4theta=1−38+38cos4θ
=5/8+3/8cos4theta=58+38cos4θ
