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Basically, write $\cos4x$ as a polynomial in $\sin x$.

I've tried the double angles theorem and $\cos2x=\cos^2x-\sin^2x$. I'm still having trouble right now though.

Please help! Thanks!

Juan Castaño
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riorpewruiowe
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    Try the double angle formula twice (or more). E.g. What is $ \cos 4x$ equal to? – Calvin Lin Jan 18 '21 at 06:46
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    I think the better thing to do is this : equality as polynomials implies that at a particular value of $x$, equality holds. You can try to substitute values of $x$ to help you. For example, note that putting $-x$ in place of $x$ is helpful. Putting $x=0$ is helpful. Of course, this works only if you know that it is a polynomial in $\sin x$, otherwise substituting values won't help. – Sarvesh Ravichandran Iyer Jan 18 '21 at 06:46
  • Do double angles twice. You will get $\cos 4x$ as an equation of powers of $\sin x$ and $\cos x$. If you replace the $\cos^2 x$ with $1-\sin^2 x$ you will probably get something very straighforward. – fleablood Jan 18 '21 at 06:50

3 Answers3

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$$\begin{align}\cos4x&=1-2\sin^22x\\&=1-2(1-\cos^22x)\\&=1-2(1-(1-2\sin^2x)^2)\\&=1-2(4\sin^2x-4\sin^4x)\\&=1-8\sin^2x+8\sin^4x\end{align}$$

19aksh
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Parcly Taxel
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If you prefer we can use complex numbers (yes!).

Let $z=\cos x+i\sin x$. Using de Moivre's theorem: $$\begin{align} z^4&=\cos 4x+i\sin 4x=(\cos x+i\sin x)^4\\&=\cos^4x+4\cos x(i\sin x)+6\cos^2x(i\sin x)^2+4\cos x(i\sin x)^3+(i\sin x)^4\end{align}$$ Since we're only concerned with the real part of this expression, $\cos 4x$, we see that $$\cos4x=\cos^4x-6\cos^2x\sin^2x+\sin^4x$$ and now use $\cos^2x=1-\sin^2x$ repeatedly.

A-Level Student
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Double angle theorem once:

$\cos 4x = \cos^2 2x + \sin^2 2x$.

Double angle theorem twice:

$= (\cos^2 x -\sin^2 x)^2 + (2\sin x\cos x)^2$.

Expand chicken soup and rice:

$= \cos^4 x- 2\cos^2 x\sin^2 x + \sin^4 x +4\sin^2 x\cos^2 x$

Can you finish?

To get it entirely in terms of $\sin^k x$ replace $\cos^2 x$ with $1-\sin^2 x$ (keeping in mind $\cos^4 x = (\cos^2 x)^2$)

Can you finish?

$= (1 - \sin^2 x)^2 - 2(1-\sin^2 x) \sin^2 x + \sin^4 x +4\sin^2 x(1-\sin^2 x) $

fleablood
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