Mathematical induction on $ [R(\cos(t) + i\sin(t) )]^n = R^n(\cos(nt) + i\sin(nt)) $

Question

Using Mathematical Induction, where $n$ is a positive integer, prove:

$$ [R( \cos(t) + i\sin(t) )]^n = R^n(\cos(nt) + i\sin(nt)) $$

I am not entirely aware of what $i\sin(x)$ would be, or what value $R$ is suppose to hold.

Answer 1

The base case is obvious. For the inductive step :

$$ \left(R(\cos t+ i\sin t) \right)^{n+1}=\left(R(\cos t+ i\sin t) \right)^{n}\left(R(\cos t+ i\sin t) \right)=$$ $$=R^{n+1}\left((\cos nt+ i\sin nt)(\cos t+ i\sin t) \right)= $$ $$ R^{n+1}\left(\cos nt\cos t+i\cos nt \sin t +i \sin nt \cos t -\sin nt \sin t \right)=R^{n+1}\left(\cos (nt+t)+i\sin (nt+t) \right) $$

Answer 2

As for more information about $i\sin(x)$, you are referred to the nature of a complex number. However, you need somehow to know a bit about polar coordinates in the plane( Argand Plane). Let $n=1$, so the equality holds. If $n=k\in\mathbb{Z}^{+}$ and the statments holds for $k$ then $$(\cos(x)+i\sin(x))^{k+1}=[\cos(kx)+i\sin(kx)][\cos(x)+i\sin(x)]=\cos(kx+x)+i\sin(kx+x)$$ which shows that your claim is true for $k+1$. If $n=k\in\mathbb{Z}^{-}$ then assume you have $k=-n$ which is a positive integer. Of course we have it for $n=0$. Just check this case mutually.

Answer 3

\begin{align}[R( \cos(t) + i\sin(t) )]^{n+1} &= [R( \cos(t) + i\sin(t) )]^{n}[R( \cos(t) + i\sin(t) )] \\ &= [R^n(\cos(nt) + i\sin(nt))] [R( \cos(t) + i\sin(t) )] \\&= R^{n+1}\left[ \cos(nt)\cos(t)-\sin(nt)\sin(t)+i(\cos(nt)\sin(t)+\sin(nt)\cos(t) \right]\end{align}

Using trigonometric addition formula to complete the rest.

Answer 4

The statement is trivially true for $n=1$. The term $\iota$ or $i$ is just treated like a constant. Now, let us assume hypothetically that $[R(\cos t+\iota \sin t)^k=R^k[\cos(kt)+\iota \sin(kt)]$. Now, the inductive step $[R(\cos t +\iota \sin t)]^{k+1}=R^{k+1}[\cos(k+1)t+\iota \sin(k+1)t]$ can be proved using basic addition formulae for sine and cosine. Hence, the theorem is proved using induction. This is the standard De-Moivre's Theorem.