jelentetvq

2022-01-28

Simplification of ${\mathrm{cos}}^{4}\left(x\right)+{\mathrm{sin}}^{4}\left(x\right)$
${\left(\mathrm{sin}x\right)}^{4}+{\left(\mathrm{cos}x\right)}^{4}=\frac{{\left(1-\mathrm{cos}2x\right)}^{2}}{4}+\frac{{\left(1+\mathrm{cos}2x\right)}^{2}}{4}$
$=\frac{1-2\mathrm{cos}2x+{\left(\mathrm{cos}2x\right)}^{2}+1+2\mathrm{cos}2x+{\left(\mathrm{cos}2x\right)}^{2}}{4}$
$=\frac{1+{\left(\mathrm{cos}2x\right)}^{2}}{2}$
its correct?

Alfred Mueller

A faster way:
${\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x={\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)}^{2}-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x=1-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x$
$=1-\frac{1}{2}{\left(2\mathrm{sin}x\mathrm{cos}x\right)}^{2}=1-\frac{1}{2}{\mathrm{sin}}^{2}2x$
$=1-\frac{1}{2}\frac{1-\mathrm{cos}4x}{2}=\frac{3+\mathrm{cos}4x}{4}$

Aiden Cooper

${\left(\frac{{e}^{ix}-{e}^{-ix}}{2i}\right)}^{4}+{\left(\frac{{e}^{ix}+{e}^{-ix}}{2}\right)}^{4}=\frac{1}{{2}^{4}}\left({\left({e}^{ix}-{e}^{-ix}\right)}^{4}+{\left({e}^{ix}+{e}^{-ix}\right)}^{4}\right)$
but
${\left(a+b\right)}^{4}+{\left(a-b\right)}^{4}=2\left({a}^{4}+6{a}^{2}{b}^{2}+{b}^{4}\right)$
hence
${\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x=\frac{1}{{2}^{3}}\left(2\mathrm{cos}\left(4x\right)+6\right)=\frac{1}{4}\left(\mathrm{cos}\left(4x\right)+3\right)$

Do you have a similar question?