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 $\max \left(a{\sin }^2\theta +b{\cos }^2\theta \right)=\max \left\{a,b\right\}$