Question Number 144697 by ZiYangLee last updated on 28/Jun/21
$$\mathrm{Evaluate}\:\left(\frac{\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{10}}−{i}\mathrm{sin}\:\frac{\pi}{\mathrm{10}}}{\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{10}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{10}}}\right)^{\mathrm{15}} . \\ $$
Answered by mathmax by abdo last updated on 28/Jun/21
$$\mathrm{1}+\mathrm{cos}\left(\frac{\pi}{\mathrm{10}}\right)+\mathrm{isin}\left(\frac{\pi}{\mathrm{10}}\right)=\mathrm{2cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{20}}\right)+\mathrm{2isin}\left(\frac{\pi}{\mathrm{20}}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{20}}\right) \\ $$$$=\mathrm{2cos}\left(\frac{\pi}{\mathrm{20}}\right)\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{20}}} \:\:\mathrm{and}\:\mathrm{1}+\mathrm{cos}\left(\frac{\pi}{\mathrm{10}}\right)−\mathrm{isin}\left(\frac{\pi}{\mathrm{10}}\right)=\mathrm{2cos}\left(\frac{\pi}{\mathrm{20}}\right)\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{20}}} \:\Rightarrow \\ $$$$\mathrm{A}=\frac{\mathrm{1}+\mathrm{cos}\left(\frac{\pi}{\mathrm{10}}\right)−\mathrm{isin}\left(\frac{\pi}{\mathrm{10}}\right)}{\mathrm{1}+\mathrm{cos}\left(\frac{\pi}{\mathrm{10}}\right)+\mathrm{isin}\left(\frac{\pi}{\mathrm{10}}\right)}=\frac{\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{20}}} }{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{20}}} }=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{10}}} \:\Rightarrow \\ $$$$\mathrm{A}^{\mathrm{15}} \:=\mathrm{e}^{−\frac{\mathrm{i15}\pi}{\mathrm{10}}} \:=\mathrm{e}^{\mathrm{i}\left(\frac{\left.\mathrm{20}−\mathrm{5}\right)\pi}{\mathrm{10}}\right)} \:=\mathrm{e}^{\mathrm{2i}\pi} .\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{2}}} \:=−\mathrm{i} \\ $$
Answered by som(math1967) last updated on 28/Jun/21
![((1+cosθ−isinθ)/(1+cosθ+isinθ)) [θ=(π/(10))] =(((1+cosθ−isinθ)^2 )/((1+cosθ)^2 −i^2 sin^2 θ)) =(((1+cosθ)^2 −2isinθ(1+cosθ)+i^2 sin^2 θ)/(1+2cosθ+cos^2 θ+sin^2 θ)) =(((1+cosθ)^2 −2isinθ(1+cosθ)−(1−cos^2 θ))/(2(1+cosθ))) =(((1+cosθ)(1+cosθ−2isinθ−1+cosθ))/(2(1+cosθ))) =((2(cosθ−isinθ))/2)=(cosθ−isinθ) ∴ (((1+cos(π/(10))−isin(π/(10)))/(1+cos(π/(10))+isin(π/(10)))))=(cos(π/(10))−isin(π/(10))) (((1+cos(π/(10))−isin(π/(10)))/(1+cos(π/(10))+isin(π/(10)))))^(15) =(cos(π/(10))−isin(π/(10)))^(15) =cos((3π)/2)−isin((3π)/2) =0−i(−1)=i ans](Q144703.png)
$$\frac{\mathrm{1}+{cos}\theta−{isin}\theta}{\mathrm{1}+{cos}\theta+{isin}\theta}\:\:\:\left[\theta=\frac{\pi}{\mathrm{10}}\right] \\ $$$$=\frac{\left(\mathrm{1}+{cos}\theta−{isin}\theta\right)^{\mathrm{2}} }{\left(\mathrm{1}+{cos}\theta\right)^{\mathrm{2}} −{i}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta} \\ $$$$=\frac{\left(\mathrm{1}+{cos}\theta\right)^{\mathrm{2}} −\mathrm{2}{isin}\theta\left(\mathrm{1}+{cos}\theta\right)+{i}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}{\mathrm{1}+\mathrm{2}{cos}\theta+{cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta} \\ $$$$=\frac{\left(\mathrm{1}+{cos}\theta\right)^{\mathrm{2}} −\mathrm{2}{isin}\theta\left(\mathrm{1}+{cos}\theta\right)−\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right)}{\mathrm{2}\left(\mathrm{1}+{cos}\theta\right)} \\ $$$$=\frac{\cancel{\left(\mathrm{1}+{cos}\theta\right)}\left(\mathrm{1}+{cos}\theta−\mathrm{2}{isin}\theta−\mathrm{1}+{cos}\theta\right)}{\mathrm{2}\cancel{\left(\mathrm{1}+{cos}\theta\right)}} \\ $$$$=\frac{\mathrm{2}\left({cos}\theta−{isin}\theta\right)}{\mathrm{2}}=\left({cos}\theta−{isin}\theta\right) \\ $$$$\therefore\:\left(\frac{\mathrm{1}+{cos}\frac{\pi}{\mathrm{10}}−{isin}\frac{\pi}{\mathrm{10}}}{\mathrm{1}+{cos}\frac{\pi}{\mathrm{10}}+{isin}\frac{\pi}{\mathrm{10}}}\right)=\left({cos}\frac{\pi}{\mathrm{10}}−{isin}\frac{\pi}{\mathrm{10}}\right) \\ $$$$\left(\frac{\mathrm{1}+{cos}\frac{\pi}{\mathrm{10}}−{isin}\frac{\pi}{\mathrm{10}}}{\mathrm{1}+{cos}\frac{\pi}{\mathrm{10}}+{isin}\frac{\pi}{\mathrm{10}}}\right)^{\mathrm{15}} \\ $$$$=\left({cos}\frac{\pi}{\mathrm{10}}−{isin}\frac{\pi}{\mathrm{10}}\right)^{\mathrm{15}} ={cos}\frac{\mathrm{3}\pi}{\mathrm{2}}−{isin}\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$$$=\mathrm{0}−{i}\left(−\mathrm{1}\right)={i}\:{ans} \\ $$