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Multiple Choice Questions
1. $$sin^{6}θ + cos^{6} θ$$ is equal to
(A): $$1$$
(B): $$1 - 3 sin^{2} θ cos^{2} θ$$
(C): $$1 - 3 sin θ cos θ$$
(D): $$1 + 3 sin^{2} θ cos^{2} θ$$
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By: anil on 05 May 2019 01.53 am
$$sin^6 θ + cos^6 θ = (sin^2θ)^3 + (cos^2θ)^3 $$
Consider
$$(sin^2 θ + cos^2 θ)^3 = sin^6 θ + cos^6 θ + 3 sin^4 θ cos^2 θ + 3 sin^2 θ cos^4 θ $$
$$1 = sin^6 θ + cos^6 θ + 3 sin^4 θ cos^2 θ + 3 sin^2 θ cos^4 θ $$
$$1= sin^6 θ + cos^6 θ + 3 sin^2 θ cos^2 θ (sin^ 2 θ + cos^2 θ) $$
$$1 - 3 sin^2 θ cos^2 θ (sin^ 2 θ + cos^2 θ) = sin^6 θ + cos^6 θ $$
$$1 - 3 sin^2 θ cos^2 θ = sin^6 θ + cos^6 θ $$
Hence Option B is the correct answer.
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SSC CGL 2014 Tier 1 26 Oct shift 4
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Consider
$$(sin^2 θ + cos^2 θ)^3 = sin^6 θ + cos^6 θ + 3 sin^4 θ cos^2 θ + 3 sin^2 θ cos^4 θ $$
$$1 = sin^6 θ + cos^6 θ + 3 sin^4 θ cos^2 θ + 3 sin^2 θ cos^4 θ $$
$$1= sin^6 θ + cos^6 θ + 3 sin^2 θ cos^2 θ (sin^ 2 θ + cos^2 θ) $$
$$1 - 3 sin^2 θ cos^2 θ (sin^ 2 θ + cos^2 θ) = sin^6 θ + cos^6 θ $$
$$1 - 3 sin^2 θ cos^2 θ = sin^6 θ + cos^6 θ $$
Hence Option B is the correct answer.