Kaspaueru2

2021-12-31

Basically, write $\mathrm{cos}4x$ as a polynomial in $\mathrm{sin}x$.
I've tried the double angles theorem and $\mathrm{cos}2x={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x$. I'm still having trouble right now though.

rodclassique4r

$\mathrm{cos}4x=1-2{\mathrm{sin}}^{2}2x$
$=1-2\left(1-{\mathrm{cos}}^{2}2x\right)$
$=1-2\left(1-{\left(1-2{\mathrm{sin}}^{2}x\right)}^{2}\right)$
$=1-2\left(4{\mathrm{sin}}^{2}x-4{\mathrm{sin}}^{4}x\right)$
$=1-8{\mathrm{sin}}^{2}x+8{\mathrm{sin}}^{4}x$

scomparve5j

If you prefer we can use complex numbers
Let $z=\mathrm{cos}x+i\mathrm{sin}x$. Using de Moivres

Vasquez

Double angle theorem once:
$\mathrm{cos}4x={\mathrm{cos}}^{2}2x+{\mathrm{sin}}^{2}2x$
Double angle theorem twice:
$=\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x{\right)}^{2}+\left(2\mathrm{sin}x\mathrm{cos}x{\right)}^{2}\right)$
Expand chicken soup and rice:
$={\mathrm{cos}}^{4}x-2{\mathrm{cos}}^{2}x{\mathrm{sin}}^{2}x+{\mathrm{sin}}^{4}x+4{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\right)$
To get it entirely in terms of ${\mathrm{sin}}^{k}x$ replace ${\mathrm{cos}}^{2}x$ with $1-{\mathrm{sin}}^{2}x$ (keeping in mind ${\mathrm{cos}}^{4}x=\left({\mathrm{cos}}^{2}x{\right)}^{2}\right)$
$=\left(1-{\mathrm{sin}}^{2}x{\right)}^{2}-2\left(1-{\mathrm{sin}}^{2}x\right){\mathrm{sin}}^{2}x+{\mathrm{sin}}^{4}x+4{\mathrm{sin}}^{2}x\left(1-{\mathrm{sin}}^{2}x\right)$

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