How to prove that:
$$\cos{n\theta}=\cos^n{\theta}- \binom {n} {2}\cos^{n-2} \theta \cdot \sin^2 \theta+ \binom {n} {4}\cos^{n-4} \theta \cdot \sin^{4} \theta -\cdots$$
$$\sin n\theta = \binom {n} {1}\cos^{n-1} \theta \cdot \sin \theta - \binom {n} {3}\cos^{n-3} \theta \cdot \sin^3\theta +\cdots$$
I don't see how to start.
$\bullet \ \cos(\theta)+i \sin(\theta) = e^{i \theta}$
$\bullet\ \ (e^{i\theta})^n = (\cos(\theta)+i\sin(\theta))^n$
$\bullet \ \ (e^{i\theta})^n = e^{i(n\theta)} = \cos(n\theta)+i\sin(n\theta) = (\cos(\theta)+i\sin(\theta))^n $
Now just use binomial theorem
Here, there are identities
$$\sin(α+β)=\sin(α)\cos(β)+\cos(α)\sin(β)$$ $$\sin(α–β)=\sin(α)\cos(β)–\cos(α)\sin(β)$$ $$\cos(α+β)=\cos(α)\cos(β)–\sin(α)\sin(β)$$ $$\cos(α–β)=\cos(α)\cos(β)+\sin(α)\sin(β)$$ with induction proof with Binomial theorem
$$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}.$$
This is enough to prove the theorem. I managed to do it on a separate sheet.
