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Question Number 144697 by ZiYangLee last updated on 28/Jun/21

$$\mathrm{Evaluate}{\left(\frac{1+\cos \frac{\pi }{10}-i\sin \frac{\pi }{10}}{1+\cos \frac{\pi }{10}+i\sin \frac{\pi }{10}}\right)}^{15}.$$

Answered by mathmax by abdo last updated on 28/Jun/21

$$1+\cos \left(\frac{\pi }{10}\right)+\mathrm{isin}\left(\frac{\pi }{10}\right)={2\cos }^2\left(\frac{\pi }{20}\right)+2\mathrm{isin}\left(\frac{\pi }{20}\right)\cos \left(\frac{\pi }{20}\right)$$$$=2\cos \left(\frac{\pi }{20}\right){\mathrm{e}}^{\frac{\mathrm{i}\pi }{20}}\mathrm{and}1+\cos \left(\frac{\pi }{10}\right)-\mathrm{isin}\left(\frac{\pi }{10}\right)=2\cos \left(\frac{\pi }{20}\right){\mathrm{e}}^{-\frac{\mathrm{i}\pi }{20}}\Rightarrow$$$$\mathrm{A}=\frac{1+\cos \left(\frac{\pi }{10}\right)-\mathrm{isin}\left(\frac{\pi }{10}\right)}{1+\cos \left(\frac{\pi }{10}\right)+\mathrm{isin}\left(\frac{\pi }{10}\right)}=\frac{{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{20}}}{{\mathrm{e}}^{\frac{\mathrm{i}\pi }{20}}}={\mathrm{e}}^{-\frac{\mathrm{i}\pi }{10}}\Rightarrow$$$${\mathrm{A}}^{15}={\mathrm{e}}^{-\frac{\mathrm{i}15\pi }{10}}={\mathrm{e}}^{\mathrm{i}\left(\frac{20-5)\pi }{10}\right)}={\mathrm{e}}^{2\mathrm{i}\pi }.{\mathrm{e}}^{-\frac{\mathrm{i}\pi }{2}}=-\mathrm{i}$$

Answered by som(math1967) last updated on 28/Jun/21

$$\frac{1+cos\theta -isin\theta }{1+cos\theta +isin\theta }[\theta =\frac{\pi }{10}]$$$$=\frac{{(1+cos\theta -isin\theta )}^2}{{(1+cos\theta )}^2-i^2{sin}^2\theta }$$$$=\frac{{(1+cos\theta )}^2-2isin\theta (1+cos\theta )+i^2{sin}^2\theta }{1+2cos\theta +{cos}^2\theta +{sin}^2\theta }$$$$=\frac{{(1+cos\theta )}^2-2isin\theta (1+cos\theta )-(1-{cos}^2\theta )}{2(1+cos\theta )}$$$$=\frac{\cancel{(1+cos\theta )}(1+cos\theta -2isin\theta -1+cos\theta )}{2\cancel{(1+cos\theta )}}$$$$=\frac{2(cos\theta -isin\theta )}{2}=(cos\theta -isin\theta )$$$$\therefore \left(\frac{1+cos\frac{\pi }{10}-isin\frac{\pi }{10}}{1+cos\frac{\pi }{10}+isin\frac{\pi }{10}}\right)=(cos\frac{\pi }{10}-isin\frac{\pi }{10})$$$${\left(\frac{1+cos\frac{\pi }{10}-isin\frac{\pi }{10}}{1+cos\frac{\pi }{10}+isin\frac{\pi }{10}}\right)}^{15}$$$$={(cos\frac{\pi }{10}-isin\frac{\pi }{10})}^{15}=cos\frac{3\pi }{2}-isin\frac{3\pi }{2}$$$$=0-i(-1)=ians$$

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