Question Number 196242 by pete last updated on 20/Aug/23
$$\mathrm{Simplify}\:\left(\frac{\mathrm{1}+\mathrm{cos2}\theta\:+\mathrm{isin2}\theta}{\mathrm{1}+\mathrm{cos2}\theta\:−\mathrm{isin2}\theta}\right)^{\mathrm{30}} \\ $$
Answered by MM42 last updated on 20/Aug/23
$$\left(\frac{\mathrm{2}{cos}^{\mathrm{2}} \theta+\mathrm{2}{isin}\theta{cos}\theta}{\mathrm{2}{cos}^{\mathrm{2}} \theta−\mathrm{2}{isin}\theta{cos}\theta}\right)^{\mathrm{30}} =\left(\frac{{cos}\theta+{isin}\theta}{{cos}\theta−{isin}\theta}\right)^{\mathrm{30}} \\ $$$$={cos}\left(\mathrm{60}\theta\right)+{isin}\left(\mathrm{60}\theta\right)\:\checkmark \\ $$
Answered by Frix last updated on 20/Aug/23
$$\frac{\mathrm{1}+{a}+{b}\mathrm{i}}{\mathrm{1}+{a}−{b}\mathrm{i}}=\frac{\left({a}+\mathrm{1}+{b}\mathrm{i}\right)^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{We}\:\mathrm{have}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1}\:\Leftrightarrow\:{b}^{\mathrm{2}} =\mathrm{1}−{a}^{\mathrm{2}} \\ $$$$\frac{\left({a}+\mathrm{1}+{b}\mathrm{i}\right)^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{2}\left({a}+\mathrm{1}\right)\left({a}+{b}\mathrm{i}\right)}{\mathrm{2}\left({a}+\mathrm{1}\right)}={a}+{b}\mathrm{i} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta\:+\mathrm{i}\:\mathrm{sin}\:\mathrm{2}\theta\:=\mathrm{e}^{\mathrm{2i}\theta} \\ $$$$\left(\mathrm{e}^{\mathrm{2i}\theta} \right)^{\mathrm{30}} =\mathrm{e}^{\mathrm{60i}\theta} =\mathrm{cos}\:\mathrm{60}\theta\:+\mathrm{i}\:\mathrm{sin}\:\mathrm{60}\theta \\ $$