Proving $(\cos x +i\sin x)^n= \cos nx+i\sin nx$ by mathematical induction

Question

Prove $(\cos x +i\sin x)^n= \cos nx+i\sin nx$ by mathematical induction.

Hence, express $1-i$ in the form of $r(\cos x+i\sin x )$.

Answer 1

Hint: The case $n=1$ is trivial. Then show that

$$(\cos x+i\sin x)^2 = \cos 2x +i \sin 2x$$

by using

$$ \cos(a+b)=\cos a \cos b - \sin a \sin b \qquad (1)$$ $$\sin(a+b)= \sin a \cos b + \sin b \cos a. \qquad (2)$$

Then assume $$(\cos x + i \sin x )^n = \cos nx + i \sin nx$$ for $n\leq N$ then use the induction step $n \to n+1$

$$(\cos x + i \sin x )^{n+1} = (\cos x + i \sin x )^n(\cos x +i\sin x)=(\cos nx + i \sin nx)(\cos x +i\sin x)$$

and use $(1)$ and $(2)$.

Can you complete it from here?