- First: solving (sin^2theta+cos^2theta)^3=(1)^3(sin2θ+cos2θ)3=(1)3
(sin^2theta+cos^2theta)(sin^4theta+2sin^2theta*cos^2theta+cos^4theta)=1(sin2θ+cos2θ)(sin4θ+2sin2θ⋅cos2θ+cos4θ)=1
(sin^6theta+2sin^4theta*cos^2theta+sin^2theta*cos^4theta)+(cos^2theta*sin^4theta+2sin^2theta*cos^4theta+cos^6theta)=1(sin6θ+2sin4θ⋅cos2θ+sin2θ⋅cos4θ)+(cos2θ⋅sin4θ+2sin2θ⋅cos4θ+cos6θ)=1
(sin^6theta+cos^6theta)+[(2sin^4theta*cos^2theta+2sin^2theta*cos^4theta)+(sin^2theta*cos^4theta+cos^2theta*sin^4theta)]=1(sin6θ+cos6θ)+[(2sin4θ⋅cos2θ+2sin2θ⋅cos4θ)+(sin2θ⋅cos4θ+cos2θ⋅sin4θ)]=1
(sin^6theta+cos^6theta)+[sin^2theta*cos^2theta*((2sin^2theta+2cos^2theta)+(cos^2theta+sin^2theta))]=1(sin6θ+cos6θ)+[sin2θ⋅cos2θ⋅((2sin2θ+2cos2θ)+(cos2θ+sin2θ))]=1
(sin^6theta+cos^6theta)+[sin^2theta*cos^2theta*((2)+(1))]=1(sin6θ+cos6θ)+[sin2θ⋅cos2θ⋅((2)+(1))]=1
(sin^6theta+cos^6theta)+3[sin^2theta*cos^2theta]=1(sin6θ+cos6θ)+3[sin2θ⋅cos2θ]=1
:.sin^6theta+cos^6theta=1-3[sin^2theta*cos^2theta]
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Using a the same method starting with (sin^2theta+cos^2theta)^2=(1)^2
you get:
sin^4theta+cos^4theta=1-2[sin^2theta*cos^2theta]
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The left side of the equation =
2(sin^6theta+cos^6theta)-3(sin^4theta+cos^4theta)+1
=2(1-3sin^2theta*cos^2theta)-3(1-2sin^2theta*cos^2theta)+1
=2cancel(-6sin^2theta*cos^2theta)-3+cancel(6sin^2theta*cos^2theta)+1
=2-3+1 =0