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Question Number 200646 by Calculusboy last updated on 21/Nov/23

Answered by Frix last updated on 21/Nov/23

$${\left(\frac{1+\sin \theta +\mathrm{i}\cos \theta }{1+\sin \theta -\mathrm{i}\cos \theta }\right)}^n={(\sin \theta +\mathrm{i}\cos \theta )}^n=$$$$={(\cos (\frac{\pi }{2}-\theta )+\mathrm{i}\sin (\frac{\pi }{2}-\theta ))}^n={\mathrm{e}}^{\mathrm{i}(\frac{\pi }{2}-\theta )n}=$$$$=\cos \left((\frac{\pi }{2}-\theta )n\right)+\mathrm{i}\sin \left((\frac{\pi }{2}-\theta )n\right)$$

Commented by Calculusboy last updated on 22/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

Answered by Spillover last updated on 21/Nov/23

Commented by Calculusboy last updated on 22/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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