Proving a $\cos(2nx)$ identity using induction

Question

Prove that $\cos(2nx)=∑_{k=0}^n (-1)^k \dbinom{2n}{2k} \cos^{2(n-k)}(x)\cdot \sin^{2k}(x):=p(n)$

I'd start using induction, with $n=1$ we have: $$cos(2x)=\cos^2(x)-\sin^2(x)$$ True. Now assume $p(n)$ is true, we proof that $p(n+1)$ is true. $$p(n+1)=\sum_{k=0}^{n+1} (-1)^k \dbinom{2(n+1)}{2k}\cos^{2((n+1)-k)}(x)\cdot \sin^{2k}(x)$$ Using the induction hypothesis, we get: $$=(-1)^{n+1}\sin^{2(n+1)}(x)+\cos(2nx)$$

This is, however, not true (does not hold for $n=1$ at least). Could anyone point out where I made a mistake?

Answer 1

Not by induction:

$$\begin{align*} \cos 2nx &= \Re[\cos 2nx + i\sin 2nx]\\ &= \Re[(\cos x + i\sin x)^{2n}]\\ &= \Re\left[\sum_{j=0}^{2n}\binom{2n}{j}\cos^{2n-j} x\cdot i^j\sin^j x\right]\\ &= \sum_{k=0}^{n}\binom{2n}{2k}\cos^{2n-2k} x\cdot i^{2k}\sin^{2k} x &&(j = 2k)\\ &= \sum_{k=0}^{n} (-1)^{k}\binom{2n}{2k}\cos^{2(n-k)} x\cdot \sin^{2k} x \end{align*}$$

Answer 2

Hint $$\cos(2nx)=\frac{e^{i2nx}+e^{-i2nx}}{2}$$ now use induction