Question

If $k=si{n}^{6}x+co{s}^{6}x,$ then find the minimum value of $4k$

Answer 1

Answer: 1

$si{n}^{6}x+co{s}^{6}x={\left(si{n}^{2}x\right)}^3+{\left(co{s}^{2}x\right)}^3$
$=\left(si{n}^{2}x+co{s}^{2}x\right)\left(si{n}^{4}x+co{s}^{4}x\right)\left(-si{n}^{2}x\cdot co{s}^{2}x\right)$
$=1\left[{\left(si{n}^{2}x+co{s}^{2}x\right)}^2\left(-3si{n}^{2}xco{s}^{2}x\right)\right]$
$=1-\frac{3}{4}si{n}^{2}2x$
$\iff \frac{1}{4}\le 1-\frac{3}{4}si{n}^{2}2x\le 1$
$\Rightarrow$ Minimum value of $si{n}^{6}x+co{s}^{6}x=\frac{1}{4}=0.25$
$\Rightarrow 4k=1$