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Prove that ^6 + { ^6} + 3^2 ^2 = 1

Question

Prove that $sin^{6} θ + cos^{6} θ + 3 sin^{2} θ cos^{2} θ = 1$

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Prove that

sin^{6} θ + cos^{6} θ + 3 sin^{2} θ cos^{2} θ = 1

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Text solutionVerified

We know that

sin^{2} θ + cos^{2} θ = 1

Cubing both sides

(sin^{2} θ + cos^{2} θ)^{3} = 1^{3}

(sin^{2} θ)^{3} + (cos^{2} θ)^{3} + 3 sin^{2} θ cos^{2} θ (sin^{2} θ + cos^{2} θ) = 1

sin^{6} θ + cos^{6} θ + 3 sin^{2} θ cos^{2} θ (1) = 1

sin^{6} θ + cos^{6} θ + 3 sin^{2} θ cos^{2} θ = 1

Hence proved.

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Question Text	Prove that $sin^{6} θ + cos^{6} θ + 3 sin^{2} θ cos^{2} θ = 1$
Answer Type	Text solution:1
Upvotes	150