# Induction problem 2

### Solution

Proof by induction on n.

Base: $$n = 0$$: $$(\cos(x) + i \sin(x))^0 = 1$$ and $$\cos(0x) + i \sin(0x) = 1 + i0 = 1$$. The two are equal and so the claim is true for the base case.

Inductive hypothesis: Suppose that $$(\cos(x) + i \sin(x))^n = \cos(nx) + i \sin(nx)$$ holds for $$n = 0,1, \ldots, k$$.

Rest of the inductive step: By the inductive hypothesis, we know that $$(\cos(x) + i \sin(x))^k = \cos(kx) + i \sin(kx)$$.

Also recall from high school trignometry that $$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$$ $$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$$

Using these equations, we can compute

$$\begin{eqnarray*} (\cos(x) + i \sin(x))^{k+1} &=& (\cos(x) + i \sin(x))^{k}(\cos(x) + i\sin(x)) \\ &=& (\cos(x) + i \sin(x))\cdot(\cos(kx) + i \sin(kx)) \\ &=& \cos(x)\cos(kx) + i\cos(x)\sin(kx) + i\sin(x)\cos(kx) - \sin(x)\sin(kx) \\ &=& (\cos(x)\cos(kx) - \sin(x)\sin(kx)) + i(\sin(kx)\cos(x) + \sin(x)\cos(kx)) \\ &=& \cos(x + kx) + i\sin(x + kx) \\ &=& \cos((k+1)x) + i\sin((k+1)x) \end{eqnarray*}$$

So $$(\cos(x) + i \sin(x))^{k+1} = \cos((k+1)x) + i\sin((k+1)x)$$ which is what we needed to show.