Prove that for any $xin$ the half$-$open interval $(0,\frac{}{2}]$ one has

$(\frac{sinx}{x}{)}^3>cosx$ Solution.

$On(0,\frac{}{2}],$

$sinx>x-\frac{x^3}{6}$

$(\frac{sinx}{x}{)}^3>cosxif(1-\frac{x^2}{6}{)}^3>1-\frac{x^2}{2}+\frac{x^4}{24}$

$q$ iff $\frac{x^4}{12}-\frac{x^6}{216}>\frac{x^4}{24}$

iff $\frac{x^4}{24}>\frac{x^6}{216}$

iff $3>x.$

But $x\frac{}{2}$~~ the desired result follows.

Please explain all steps $in$ this solution