Question

If 12 sin θ − 9sin² θ attains its maximum value at θ = α, then write the value of sin α.

Answer 1

$$\begin{gathered} \mathrm{Let}f\left(\theta \right)=12\sin \theta -9{\sin }^2\theta \\ =-\left(9{\sin }^2\theta -12\sin \theta \right) \\ =-\left[{\left(3\sin \theta \right)}^2-2.3\sin \theta .2+2^2-4\right] \\ =-\left[{\left(3\sin \theta -2\right)}^2-4\right] \\ =4-{\left(3\sin \theta -2\right)}^2 \\ \mathrm{Minimum}\mathrm{value}\mathrm{of}{\left(3\sin \theta -2\right)}^2\mathrm{is}0. \\ \mathrm{Therefore},\mathrm{maximum}\mathrm{value}\mathrm{of}f\left(\theta \right)=4-{\left(3\sin \theta -2\right)}^2\mathrm{is}4. \\ \mathrm{We}\mathrm{are}\mathrm{given}\mathrm{that}12\sin \theta -9{\sin }^2\theta \mathrm{will}\mathrm{attain}\mathrm{its}\mathrm{maximum}\mathrm{value}\mathrm{at}\theta =\alpha . \\ \therefore 12\sin \alpha -9{\sin }^2\alpha =4 \\ \Rightarrow -9{\sin }^2\alpha +12\sin \alpha -4=0 \\ \Rightarrow 9{\sin }^2\alpha -12\sin \alpha +4=0 \\ \Rightarrow 9{\sin }^2\alpha -6\sin \alpha -6\sin \alpha +4=0 \\ \Rightarrow 3\sin \alpha \left(3\sin \alpha -2\right)-2\left(3\sin \alpha -2\right)=0 \\ \Rightarrow \left(3\sin \alpha -2\right)\left(3\sin \alpha -2\right)=0 \\ \therefore \sin \alpha =\frac{2}{3} \end{gathered}$$