Question

the maximum value of 12 sin theta -9 sin^2 theta=

Answer 1

Answer:

$12 \: sin \alpha - 9 {sin}^{2} \alpha = - 9( {sin}^{2} \alpha - \frac{4}{3} \: sin \alpha ) = - 9( {sin}^{2} \alpha - 2sin \alpha \times \frac{2}{3} + ( \frac{2}{3} ) {}^{2} - \frac{4}{9} ) =$

$- 9(sin \alpha - \frac{2}{3} ) {}^{2} + 4......(i)$

Now,

$- 1 \leqslant sin \alpha \leqslant 1$

$- 1 - \frac{2}{ 3} \leqslant sin \alpha - \frac{2}{3} \leqslant 1 - \frac{2}{3}$

$0\leqslant (sin \alpha - \frac{2}{3}) {}^{2} \leqslant \frac{25}{9}$

$- 9 \times \frac{25}{9} \leqslant - 9(sin \alpha - \frac{2}{3}) {}^{2} \leqslant 0$

$- 25 + 4\leqslant - 9(sin \alpha - \frac{2}{3}) {}^{2} + 4\leqslant 0 + 4$

$-21\leqslant - 9(sin \alpha - \frac{2}{3}) {}^{2} + 4\leqslant 4$

Maximum Value = 4

Minimum Value = –21