We have to calculate the maximum value of 16sinθ−12sin2θ Let f(θ)=16sinθ−12sin2θ Differentiating, both sides, we get ⇒f′(θ)=16cosθ−12×2×sinθ×cosθ=16cosθ−24×sinθ×cosθ For maxima and minima, f′(θ)=0⇒16cosθ−24×sinθ×cosθ=0⇒8cosθ(2−3sinθ)=0⇒8cosθ=0 and (2−3sinθ)=0∴θ=90∘ and sinθ=32 or θ=sin−1(32) Now we have to find second derivative, ⇒f′′(θ)=−16sinθ−24[cos2θ−sin2θ]⇒f′′(θ)=−16sinθ−24[1−sin2θ−sin2θ]=−16sinθ−24[1−2sin2θ] When θ=.90∘⇒f(θ)∣θ=90∘=−16sin90∘−24[1−2sin290∘]=−16+24=8>0 So, function gives minimum value at θ=90∘ When θ=sin−1(32)⇒f′′(θ)∣θ=sin−1(32)=−16sin(sin−1(32))−24[1−2sin2(sin−1(32))]⇒f′′(θ)=−16(32)−24[1−2(32)2]=−(332)−(924)<0 So, function gives maximum value at θ=sin−1(32) or sinθ=32⇒f(θ)max=16(32)−12(32)2=(332)−(316)=316