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

trigonometry

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.