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Question Number 196242 by pete last updated on 20/Aug/23

$$\mathrm{Simplify}{\left(\frac{1+\cos 2\theta +\mathrm{isin}2\theta }{1+\cos 2\theta -\mathrm{isin}2\theta }\right)}^{30}$$

Answered by MM42 last updated on 20/Aug/23

$${\left(\frac{2{cos}^2\theta +2isin\theta cos\theta }{2{cos}^2\theta -2isin\theta cos\theta }\right)}^{30}={\left(\frac{cos\theta +isin\theta }{cos\theta -isin\theta }\right)}^{30}$$$$=cos\left(60\theta \right)+isin\left(60\theta \right)✓$$

Answered by Frix last updated on 20/Aug/23

$$\frac{1+a+b\mathrm{i}}{1+a-b\mathrm{i}}=\frac{{(a+1+b\mathrm{i})}^2}{{(a+b)}^2+1}$$$$\mathrm{We}\mathrm{have}{a}^2+b^2=1\iff b^2=1-a^2$$$$\frac{{(a+1+b\mathrm{i})}^2}{{(a+b)}^2+1}=\frac{2(a+1)(a+b\mathrm{i})}{2(a+1)}=a+b\mathrm{i}$$$$\cos 2\theta +\mathrm{i}\sin 2\theta ={\mathrm{e}}^{2\mathrm{i}\theta }$$$${\left({\mathrm{e}}^{2\mathrm{i}\theta }\right)}^{30}={\mathrm{e}}^{60\mathrm{i}\theta }=\cos 60\theta +\mathrm{i}\sin 60\theta$$

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