Vilharjevw7

2022-01-27

Prove that $4{\mathrm{cos}}^{4}x-2\mathrm{cos}2x-\frac{1}{2}\mathrm{cos}4x$ is independent of x
I don't know how to proceed. $\mathrm{cos}4x=2{\mathrm{cos}}^{2}\left(2x\right)-1$

rakije2v

Expert

Yes, $\mathrm{cos}4x=2{\mathrm{cos}}^{2}\left(2x\right)-1$ is useful, also ${\mathrm{cos}}^{2}x=\frac{1+\mathrm{cos}2x}{2}$
Now
$4{\mathrm{cos}}^{4}x-2\mathrm{cos}2x-\frac{1}{2}\mathrm{cos}4x=4{\left(\frac{1+\mathrm{cos}2x}{2}\right)}^{2}-2\mathrm{cos}2x-\frac{1}{2}\left(2{\mathrm{cos}}^{2}\left(2x\right)-1\right)$
then let $\mathrm{cos}2x=k$.

Ydaxq

Expert

$4{\mathrm{cos}}^{4}x-2\mathrm{cos}2x-\frac{1}{2}\mathrm{cos}4x={\left(1+\mathrm{cos}2x\right)}^{2}-2\mathrm{cos}2x-\frac{1}{2}\left(2{\mathrm{cos}}^{22}x-1\right)=\frac{3}{2}.$