Basically, write \cos 4x as a polynomial in \sin x. I've

Question

Basically, write \cos 4x as a polynomial in \sin x. I've

Kaspaueru2 2021-12-31 Answered

Basically, write

\cos 4 x

as a polynomial in

\sin x

.
I've tried the double angles theorem and

\cos 2 x = \cos^{2} x - \sin^{2} x

. I'm still having trouble right now though.

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Expert Answer

rodclassique4r

Answered 2022-01-01 Author has 37 answers

\cos 4 x = 1 - 2 \sin^{2} 2 x

= 1 - 2 (1 - \cos^{2} 2 x)

= 1 - 2 (1 - {(1 - 2 \sin^{2} x)}^{2})

= 1 - 2 (4 \sin^{2} x - 4 \sin^{4} x)

= 1 - 8 \sin^{2} x + 8 \sin^{4} x

This is helpful 0

scomparve5j

Answered 2022-01-02 Author has 38 answers

If you prefer we can use complex numbers
Let

z = \cos x + i \sin x

. Using de Moivres

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Vasquez

Answered 2022-01-08 Author has 460 answers

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