Question

If $P = \frac{1}{2} \sin^2 \theta + \frac{1}{3} \cos^2 \, \theta$, then

• A. $\frac{1}{3} \le P \le \frac{1}{2}$
• B. $P \ge \frac{1}{2}$
• C. $2 \le P \le 3$
• D. $\frac{- \sqrt{13}}{6} \le P \le \frac{\sqrt{13}}{6}$

Answer 1

Answer: (A)

$P = \frac{1}{2} \sin^{2} \theta + \frac{1}{3} \cos^{2} \theta $
$= \frac{1}{3} \left(\sin^{2} \theta + \cos^{2} \theta\right) + \left(\frac{1}{2} - \frac{1}{3}\right) \sin^{2} \theta$
$ = \frac{1}{3} + \frac{1}{6} \sin^{2} \theta \ge \frac{1}{3} $
and again $P = \frac{1}{2} \sin^{2} \theta + \frac{1}{3} \sin^{2} \theta$
$ = \frac{1}{2} \left(\sin^{2} \theta + \cos^{2} \theta\right) - \left(\frac{1}{2} - \frac{1}{3}\right) \cos^{2} \theta$
$ = \frac{1}{2} - \frac{1}{6} \cos^{2} \theta \le \frac{1}{2} $
Alternatively,
$\min \left(a \sin^{2} \theta + b \cos^{2} \theta\right) = \min \left\{a, b\right\}$
and $ \max\left(a \sin^{2} \theta + b \cos^{2} \theta\right) = \max\left\{a,b\right\}$