I saw this expansion formula for the cosine in a Digital Signal Processing book (Proakis)

$$\cos(kx)=\sum^k_{i=0}\beta_i(\cos^i(x))$$

This expansion is new to me, how can I determine the coefficients $\beta_i$ of the expansion?

As an example, the book stated that

$$

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

$$

This is obvious by means of trigonometric identities however, using the expansion formula means

$$

\beta_2=2\\

\beta_1=0\\

\beta_0=-1

$$

Edit:

I found the answer to the complication. This is because I forgot the $0$ in $\beta_0\cos^0(x)$. However my question on the determining the coefficients without trigonometric identities still exists

## Best Answer

One way is to consider De'Moivre's formula

$(\cos(x)+i\sin(x))^k= \cos(kx)+i\sin(kx)$. Then we have

$\cos(kx)= \Re((\cos(x)+i\sin(x))^k)=\Re(\sum_{j=0}^k \binom{k}{j}\cos(x)^j \sin(x)^{k-j}i^{k-j})$

If $k$ is even then we should only retain the even values of $j$

$\cos(kx) = \sum_{l=0}^{k/2} \binom{k}{2l}\cos^{2l}(x) \sin^{k-2l}(x)(-1)^{k/2-l} =\sum_{l=0}^{k/2} \binom{k}{2l}\cos^{2l}(x) (\cos^2(x)-1)^{k/2-l}$

So your example is $k=2$ and we get

$\cos(2x)=\sum_{l=0}^{1}\binom{2}{2l}\cos^{2l}(x)(\cos^2(x)-1)^{1-l}=\cos^2(x)-1 +\cos^2(x)=2\cos^2(x)-1$.

You can get the formula for odd $k$ as well, just now only consider the terms where $j$ is odd so that the powers of $i$ are even.