 Alex Cervantes

2022-01-28

How do I show that ${\mathrm{cos}}^{4}x=\frac{1}{8}\mathrm{cos}\left(4x\right)+\frac{1}{2}\mathrm{cos}\left(2x\right)+\frac{3}{8}$
I know how to prove that
${\mathrm{cos}}^{2}x=\frac{1}{2}+\frac{1}{2}\mathrm{cos}\left(2x\right)$
by substituting $\mathrm{cos}2x$ with $2{\mathrm{cos}}^{2}x-1$ according to the double angle identity
$\mathrm{cos}\left(2x\right)=2{\mathrm{cos}}^{2}x-1$
However, how do I do that for ${\mathrm{cos}}^{4}x$?
Do I do it by writing ${\mathrm{cos}}^{4}x$ as
${\mathrm{cos}}^{2}\left(x\right)\cdot {\mathrm{cos}}^{2}\left(x\right)$
and thus get it by squaring the LHS of
${\mathrm{cos}}^{2}x=\frac{1}{2}+\frac{1}{2}\mathrm{cos}\left(2x\right)$
Im not sure how to proceed. plusmarcacw

Hint: By squaring we get
${\mathrm{cos}}^{4}\left(x\right)=\frac{1}{4}\left({\mathrm{cos}}^{2}\left(2x\right)+2\mathrm{cos}\left(2x\right)+1\right)$
and then
$\mathrm{cos}\left(4x\right)=2{\mathrm{cos}}^{2}\left(2x\right)-1$ Mazzuranavf

Since
${\mathrm{cos}}^{2}x=\frac{1}{2}+\frac{1}{2}\mathrm{cos}\left(2x\right)$
we have
${\mathrm{cos}}^{2}\left(2x\right)=\frac{1}{2}+\frac{1}{2}\mathrm{cos}\left(4x\right)$
and thus, if we square first equation we get
${\mathrm{cos}}^{4}x=\frac{1}{4}+\frac{1}{2}\mathrm{cos}\left(2x\right)+\frac{1}{4}{\mathrm{cos}}^{2}\left(2x\right)=$
$=\frac{1}{4}+\frac{1}{2}\mathrm{cos}\left(2x\right)+\frac{1}{4}\left(\frac{1}{2}+\frac{1}{2}\mathrm{cos}\left(4x\right)\right)$
$=\frac{3}{8}+\frac{1}{2}\mathrm{cos}\left(2x\right)+\frac{1}{8}\mathrm{cos}\left(4x\right)$

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