Question

The maximum and minimum values of ${\cos }^6\theta +{\sin }^6\theta$ are respectively

• A. $1$ and $\frac{1}{4}$
• B. $1$ and $0$
• C. $2$ and $0$
• D. $1$ and $\frac{1}{2}$

Answer 1

Answer: (A)

$1$ and $\frac{1}{4}$

Let $f\left(\theta \right)={\cos }^6\theta +{\sin }^6\theta$

$\Rightarrow$ $f\left(\theta \right)={\left({\sin }^2\theta \right)}^3+{\left({\cos }^2\theta \right)}^3=({\sin }^2\theta +{\cos }^2\theta )({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta )$

$[\because a^3+b^3=(a+b\left)\right(a^2+b^2-ab\left)\right]$
$=1.{\{({\sin }^2\theta +{\cos }^2\theta )}^2-3{\sin }^2\theta {\cos }^2\theta \}$

$=1.(1-\frac{3}{4}.4{\sin }^2\theta {\cos }^2\theta )$

$=1-\frac{3}{4}{\left(\sin 2\theta \right)}^2(\because \sin 2A=2\sin A\cos A)$

$=1-\frac{3}{8}(1-\cos 4\theta )$=1-\cfrac { 3 }{ 8 } +\cfrac { 3 }{ 8 } \cos { 4\theta } $

$f\left(\theta \right)=\frac{5}{8}+\frac{3}{8}\cos 4\theta ...\left(i\right)$

$-1\le \cos 4\theta \le 1$

$\Rightarrow \frac{-3}{8}\le \frac{3}{8}\cos 4\theta \le \frac{3}{8}$

$\Rightarrow \frac{5}{8}-\frac{3}{8}\le \frac{5}{8}+\frac{3}{8}\cos 4\theta \le \frac{5}{8}+\frac{3}{8}$

$\Rightarrow \frac{1}{4}\le f\left(\theta \right)\le 1$