Question

If $A={\sin }^8\theta +{\cos }^{14}\theta$, then for all values of $\theta$

• A. $A>1$
• B. $A\ge 1$
• C. $A<1$
• D. $A\le 1$

Answer 1

Answer: (D)

$A\le 1$

Since, ${\sin }^8\theta \le {\sin }^2\theta$
${\cos }^{14}\theta \le {\cos }^2\theta$
$\Rightarrow {\sin }^8\theta +{\cos }^{14}\theta \le {\sin }^2\theta +{\cos }^2\theta =1$
$\Rightarrow A\le 1$
Also, A > 0 as $A=0$ if $\sin \theta =\cos \theta =0$ which is not possible.