f(θ)=4{sin4(27π−θ)+sin4(11π+θ)}−2{sin2(23π−θ)+sin2(9π−θ)}
f(θ)=4(cos4(θ)+sin4(θ))−2(cos6θ+sin6θ)
f(θ)=4(1−2sin2θcos2θ)−2(1−3sin2θcos2θ)
f(θ)=2−2sin2θcos2θ
f(θ)=2−2sin2(2θ)
α=f(θ)max=2
β=f(θ)min=23
⇒α+2β=5
Let α and β respectively be the maximum and the minimum values of the function f(θ)=4(sin4(27π−θ)+sin4(11π+θ))−2(sin6(23π−θ)+sin6(9π−θ)),θ∈R. Then α+2β is equal to :
Held on 23 Jan 2026 · Verified 6 Jul 2026.
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