Question

${\sin }^{6}x+{\cos }^{6}x$ lies between

• A. $\frac{1}{4}$ and $1$
• B. $\frac{1}{4}$ and $2$
• C. $0$ and $1$
• D. None of these

Answer 1

The correct option is A $\frac{1}{4}$ and $1$

We have ${\sin }^{6}x+{\cos }^{6}x={\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3$

$={({\sin }^{2}x+{\cos }^{2}x)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)$

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

$\Rightarrow$ Maximum value of ${\sin }^{6}x+{\cos }^{6}x$ is $1-\frac{3}{4}\times 0=1$ and minimum values is $1-\frac{3}{4}\times 1=\frac{1}{4}$