Question

The maximum and minimum values of $co{s}^6\theta +si{n}^6\theta$ are respectively

• A. $1$ and $1/4$
• B. $1$ and $0$
• C. $2$ and $0$
• D. $1$ and $1/2$

Answer 1

Answer: (A)

Let $f(\theta )={\sin }^6\theta +{\cos }^6\theta$
$\Rightarrow f(\theta )={\left({\sin }^2\theta \right)}^3+{\left({\cos }^2\theta \right)}^3$
$=\left({\sin }^2\theta +{\cos }^2\theta \right)$
$\left({\sin }^4\theta +{\cos }^4\theta -{\sin }^2\theta \cdot {\cos }^2\theta \right)$
$\left[\because a^3+b^3=(a+b)\left(a^2+b^2-ab\right)\right]$
$=1\cdot \left\{{\left({\sin }^2\theta +{\cos }^2\theta \right)}^2-3{\sin }^2\theta \cdot {\cos }^2\theta \right\}$
$=1\cdot \left(1-\frac{3}{4}\cdot 4{\sin }^2\theta \cdot {\cos }^2\theta \right)$
$=1-\frac{3}{4}(\sin 2\theta {)}^2(\because \sin 2A=2\sin A\cos A)$
$=1-\frac{3}{8}(1-\cos 4\theta )$
$=1-\frac{3}{8}+\frac{3}{8}\cos 4\theta$
$f(\theta )=\frac{5}{8}+\frac{3}{8}\cdot \cos 4\theta$
$\because -1\le \cos 4\theta \le 1$
$\Rightarrow \frac{-3}{8}\le \frac{3}{8}\cos 4\theta \le \frac{3}{8}$
$\Rightarrow \frac{5}{8}-\frac{3}{8}\le \frac{5}{8}+\frac{3}{8}\cos 4\theta \le \frac{5}{8}+\frac{3}{8}$
$\Rightarrow \frac{1}{4}\le f(\theta )\le 1$
[from Eq.
So, the maximum value is 1 and minimum value is
$\frac{1}{4}$