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If A = cos^2θ + sin^4θ, then prove that for all values of θ, 3/4≤A≤1. |
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Answer» A=cos^2θ+sin^4θ ⟹A=1−sin^2θ+sin^4θ ⟹A=1+(sin^4θ−2⋅sin^2θ⋅1/2+1/4)−1/4 ⟹A=(sinθ−1/2)^2+3/4 We know, 0≤sin^2θ≤1,∀θ∈R. ∴0≤(sin2θ−1/2)2≤1/4 ⟹0+3/4≤(sinθ−1/2)2+3/4≤14+3/4 ∴3/4≤A≤1.(∀θ∈R)† |
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