Question

The maximum and minimum values of $cos^6θ + sin^6θ$ 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}-a b\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 2 A=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 \leq \cos 4 \theta \leq 1$
$\Rightarrow \frac{-3}{8} \leq \frac{3}{8} \cos 4 \theta \leq \frac{3}{8}$
$\Rightarrow \frac{5}{8}-\frac{3}{8} \leq \frac{5}{8}+\frac{3}{8} \cos 4 \theta \leq \frac{5}{8}+\frac{3}{8}$
$\Rightarrow \frac{1}{4} \leq f(\theta) \leq 1$
[from Eq.
So, the maximum value is 1 and minimum value is
$\frac{1}{4}$