Find the formula of : $\cos (\theta x) = ?$ I was thinking about how $\cos (a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)$ and that $\cos(2\theta) = \cos^2(\theta) - sin^2(\theta)$. Is there a pattern for $\cos(\theta),\cos(2\theta),\cos(3\theta),\cos(4\theta) ...$? So basically what does $\cos(\theta x)= ?$ (like the binomial theorem I was thinking)
Thanks
 A: If $x$ is strictly considered to be a natural number, then there is using imaginary numbers in its derivation.
Euler's formula states that $e^{i\alpha} = \cos(\alpha) + i\sin(\alpha)$. Now, consider $\alpha = \theta n$ for some natural number n. 
The function $\Re(z)$ is defined as the real part of an imaginary number $z$, and so $\Re(a+ib) = a$.
Then $\cos(\alpha) = \Re(\cos(\alpha) + i\sin(\alpha)) = \Re(e^{i\alpha}) = \Re((e^{i\theta})^n) = \Re([cos(\theta) + i\sin(\theta)]^n)$. Now, we can use the binomial theorem to expand $[\cos(\theta)+i\sin(\theta)]^n$, and then isolate the real terms. Remember, $i^2 = -1$.
To make things easier, let $u = \cos(\theta)$ and $\omega = \sin(\theta)$.
$(u+i\omega)^n = \sum_{k=0}^{n}\binom{n}{k}u^{n-k}(i\omega)^{k}$
$(u+i\omega)^n = u^n + nu^{n-1}(i\omega) - \binom{n}{2}u^{n-2}\omega^2 + \cdots + \binom{n}{n-1}u(i\omega)^{n-1} + (i\omega)^n$
Then, isolating the real terms, and replacing the sines and cosines back, we get:
$$cos(\theta n) = \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor}\binom{n}{2k} \cos^{n-2k}(\theta) \sin^{2k}(\theta)(-1)^k$$
Where $\lfloor x\rfloor$ is the floor function of $x$, and $\binom{p}{r}$ is the binomial coefficient $pCr$. 
A: You are asking about cos(n\Theta) so i can send you to DeMoivre formulas, you can study the subject starting by 
https://en.wikipedia.org/wiki/De_Moivre%27s_formula
A: 
The addition formulas are enough to derive the expressions of $\cos(n\theta),\sin(n\theta)$ for any natural $n$.
This is compactly achieved using the complex notation $\text{cis}(\theta)=\cos(\theta)+i\sin(\theta)$, and you can write the recurrence relation
$$\begin{align}\text{cis}(n\theta+\theta)&=\cos(n\theta+\theta)+i\sin(n\theta+\theta)\\
&=(\cos(n\theta)\cos(\theta)-\sin(n\theta)\sin(\theta))+i(\sin(n\theta)\cos(\theta)+\cos(n\theta)\sin(\theta))\\
&=\text{cis}(n\theta)\text{cis}(\theta).\end{align}$$
With $\text{cis}(0\theta)=1$, this simply gives
$$\text{cis}(n\theta)=\text{cis}^n(\theta).$$
Now you can expand the RHS using the Binomial formula, and split in real and imaginary parts.

For example,
$$\begin{align}\cos(3\theta)+i\sin(3\theta)
&=\cos^3(\theta)+i3\cos^2(\theta)\sin(\theta)+i^23\cos(\theta)\sin^2(\theta))+i^3\sin^3(\theta)\\
&=(\cos^3(\theta)-3\cos(\theta)\sin^2(\theta))+i(3\cos^2(\theta)\sin(\theta)-\sin^3(\theta)).\end{align}$$
For the cosine, you add terms starting from $\cos^n$, with alternating signs and every other Binomial coefficient, trading a $\cos^2$ for a $\sin^2$ every time. For the sine, you start from $n\cos^{n-1}\sin$ and proceed similarly.
This also shows you that you can also derive "cosine-only" and "sine-only" formulas formulas when the parity allows it, leading to the Chebyshev polynomials.
