Question

If $k=sin^{6}x+cos^{6}x,$ then find the minimum value of $4k$

Answer 1

$sin^{6}x+cos^{6}x=\left(sin^{2} x\right)^{3}+\left(cos^{2} x\right)^{3}$
$=\left(sin^{2} x + cos^{2} x\right)\left(sin^{4} x + cos^{4} x\right)\left(- sin^{2} x \cdot cos^{2} x\right)$
$=1\left[\left(sin^{2} x + cos^{2} x\right)^{2}\left(- 3 sin^{2} x cos^{2} x\right)\right]$
$=1-\frac{3}{4}sin^{2}2x$
$\Leftrightarrow \frac{1}{4}\leq 1-\frac{3}{4}sin^{2}2x\leq 1$
$\Rightarrow $ Minimum value of $sin^{6}x+cos^{6}x=\frac{1}{4}=0.25$
$\Rightarrow 4k=1$