de Moivres theorem
Task number: 2797
Prove de Moivre's theorem by induction: \( (\cos \alpha + i \sin\alpha)^n = \cos(n\alpha)+ i \sin(n\alpha) \)
Hint
Use the additive identities:
\(\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\),
\(\sin(\alpha+\beta)=\cos(\alpha)\sin(\beta)+\sin(\alpha)\cos(\beta)\).Resolution
For \(n=1\) both sides are identical.
Assuming that the statement is true for \(n\), we show that it holds for \(n+1\).
\( (\cos \alpha + i \sin\alpha)^{n+1}= (\cos \alpha + i \sin\alpha)^n(\cos \alpha + i \sin\alpha)= (\cos(n\alpha)+ i \sin(n\alpha))(\cos \alpha + i \sin\alpha)=\\ \cos(n\alpha)\cos \alpha - \sin(n\alpha)\sin\alpha + i(\cos(n\alpha)\sin \alpha - \sin(n\alpha)\cos\alpha)= \cos((n+1)\alpha)+ i \sin((n+1)\alpha) \)
