Question Number 87656 by M±th+et£s last updated on 05/Apr/20
$$\frac{\mathrm{1}+{sin}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\pi+{i}\:{cos}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\pi}{\mathrm{1}+{sin}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\pi−{i}\:{cos}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\pi}=? \\ $$
Commented by Tony Lin last updated on 05/Apr/20
$${let}\:{sin}\frac{\pi}{\mathrm{8}}+{icos}\frac{\pi}{\mathrm{8}}={z}\:,\mid{z}\mid=\mathrm{1} \\ $$$$\frac{\mathrm{1}+{z}}{\mathrm{1}+\bar {{z}}} \\ $$$$=\frac{\mathrm{1}+{z}}{\mathrm{1}+\frac{\mathrm{1}}{{z}}} \\ $$$$=\frac{\mathrm{1}+{z}}{\frac{\mathrm{1}+{z}}{{z}}} \\ $$$$={z} \\ $$$$={sin}\frac{\pi}{\mathrm{8}}+{icos}\frac{\pi}{\mathrm{8}} \\ $$$$=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}{i} \\ $$
Commented by M±th+et£s last updated on 05/Apr/20
$${thanks}\:{for}\:{all} \\ $$$$ \\ $$
Commented by peter frank last updated on 05/Apr/20
$${thank}\:{you} \\ $$
Answered by TANMAY PANACEA. last updated on 05/Apr/20
$$\frac{\mathrm{1}+{sina}+{icosa}}{\mathrm{1}+{sina}−{icosa}} \\ $$$$=\frac{\left(\mathrm{1}+{sina}+{icosa}\right)^{\mathrm{2}} }{\left(\mathrm{1}+{sina}\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} {a}}=\frac{\mathrm{1}+{sin}^{\mathrm{2}} {a}−{cos}^{\mathrm{2}} {a}+\mathrm{2}{sina}+\mathrm{2}{isinacosa}+\mathrm{2}{icosa}}{\mathrm{1}+\mathrm{2}{sina}+{sin}^{\mathrm{2}} {a}+{cos}^{\mathrm{2}} {a}} \\ $$$$=\frac{\mathrm{2}{sin}^{\mathrm{2}} {a}+\mathrm{2}{sina}+\mathrm{2}{isinacosa}+\mathrm{2}{icosa}}{\mathrm{2}\left(\mathrm{1}+{sina}\right)} \\ $$$$=\frac{\mathrm{2}{sina}\left(\mathrm{1}+{sina}\right)+\mathrm{2}{icosa}\left(\mathrm{1}+{sina}\right)}{\mathrm{2}\left(\mathrm{1}+{sina}\right)} \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}+{sina}\right)\left({sina}+{icosa}\right)}{\mathrm{2}\left(\mathrm{1}+{sina}\right)}={sina}+{icosa} \\ $$$$={sin}\left(\frac{\pi}{\mathrm{8}}\right)+{icos}\left(\frac{\pi}{\mathrm{8}}\right) \\ $$$$ \\ $$