Question

If $A={\cos }^2\theta +{\sin }^2\theta$ for all values of $\theta$ then

• A. $1\le A\le 2$
• B. $\frac{13}{16}\le A\le 1$
• C. $\frac{3}{4}\le A\le \frac{13}{16}$
• D. $\frac{3}{4}\le A\le 1$

Answer 1

Answer: (D)

$\frac{3}{4}\le A\le 1$

$A={\cos }^2\theta +{\sin }^4\theta$
$={\cos }^2\theta +{\sin }^2\theta \times {\sin }^2\theta$
Since ${\sin }^2\theta \le 1$
$\Rightarrow A\le ({\cos }^2\theta +{\sin }^2\theta )$
$\Rightarrow A\le 1$
Also $A={\cos }^2\theta +{\sin }^4\theta$
$=(1-{\sin }^2\theta )+{\sin }^4\theta$
$\displaystyle =\left ( \sin ^{2}\theta -\frac{1}{2} \right )^{2}+\frac{3}{4} \geq \frac{3}{4}\because \left ( \sin ^{2}\theta -\frac{1}{2} \right )^{2}\geq 0$
Thus $\frac{3}{4}\le A\le 1$