Trig1.9

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Prove that \sin^6 \theta+\cos^6 \theta=1-3\sin^2 \theta\cdot\cos^2 \theta\,

LHS=\sin^6 \theta+\cos^6 \theta=(\sin^2 \theta)^3+(\cos^2 \theta)^3=(\sin^4 \theta+\cos^2 \theta)(\sin^2 \theta-\sin^2 \theta\cdot \cos^2 \theta+\cos^4 \theta)\,

(1)(\sin^4 \theta+\cos^4 \theta-\sin \theta \cos \theta)\,

[(\sin^2 \theta+\cos^2 \theta)^2-2\sin^2 \theta \cos^2 \theta-\sin^2 \theta\cos^2 \theta]\,

1-3\sin^2 \theta \cos^2 \theta\,=RHS


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