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If +^{2} +^{3} =1, then ^{6} -4^{4} +8 ^{2} =4.

Question

If $sin θ + sin^{2} θ + sin^{3} θ = 1$ , then $cos^{6} θ - 4 cos^{4} θ + 8 cos^{2} θ = 4$ .

True

False

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Jack Discussed

sin θ + sin^{2} θ + sin^{3} θ = 1

, then

cos^{6} θ - 4 cos^{4} θ + 8 cos^{2} θ = 4

5 mins ago

Discuss this question LIVE

5 mins ago

Text solutionVerified

Given

sin θ + sin^{2} θ + sin^{3} θ = 1

⟹ sin θ (1 + sin^{2} θ) = 1 - sin^{2} θ

⟹ sin θ (2 - cos^{2} θ) = cos^{2} θ

Squaring on both sides

⟹ sin^{2} θ (2 - cos^{2} θ)^{2} = cos^{4} θ

⟹ (1 - cos^{2} θ) (4 + cos^{4} θ - 4 cos^{2} θ) = cos^{4} θ

⟹ 4 + cos^{4} θ - 4 cos^{2} θ - 4 cos^{2} θ - cos^{6} θ + 4 cos^{4} θ = cos^{4} θ

⟹ cos^{6} θ - 4 cos^{4} θ + 8 cos^{2} θ = 4

So the relation is

True

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Question Text	If $sin θ + sin^{2} θ + sin^{3} θ = 1$ , then $cos^{6} θ - 4 cos^{4} θ + 8 cos^{2} θ = 4$ .
Answer Type	Text solution:1
Upvotes	150