Question

Prove that ${(\sin \theta +i\cos \theta )}^n=\cos n(\frac{\pi }{2}-\theta )+i\sin n(\frac{\pi }{2}-\theta )$

Answer 1

${(\sin \theta +i\cos \theta )}^n={(\cos (\frac{\pi }{2}-\theta )+i\sin (\frac{\pi }{2}-\theta ))}^n$

${(\sin \theta +i\cos \theta )}^n={\left(e^{i(\frac{\pi }{2}-\theta )}\right)}^n=e^{n(\frac{\pi }{2}-\theta )}$

${(\sin \theta +i\cos \theta )}^n=\cos \left(n(\frac{\pi }{2}-\theta )\right)+i\sin \left(n(\frac{\pi }{2}-\theta )\right)$