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Question Number 87656 by M±th+et£s last updated on 05/Apr/20

((1+sin((1/8))π+i cos((1/8))π)/(1+sin((1/8))π−i cos((1/8))π))=?

$$\frac{\mathrm{1}+{sin}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\pi+{i}\:{cos}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\pi}{\mathrm{1}+{sin}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\pi−{i}\:{cos}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\pi}=? \\ $$

Commented by Tony Lin last updated on 05/Apr/20

let sin(π/8)+icos(π/8)=z ,∣z∣=1 ((1+z)/(1+z^� )) =((1+z)/(1+(1/z))) =((1+z)/((1+z)/z)) =z =sin(π/8)+icos(π/8) =((√(2−(√2)))/2)+((√(2+(√2)))/2)i

$${let}\:{sin}\frac{\pi}{\mathrm{8}}+{icos}\frac{\pi}{\mathrm{8}}={z}\:,\mid{z}\mid=\mathrm{1} \\ $$$$\frac{\mathrm{1}+{z}}{\mathrm{1}+\bar {{z}}} \\ $$$$=\frac{\mathrm{1}+{z}}{\mathrm{1}+\frac{\mathrm{1}}{{z}}} \\ $$$$=\frac{\mathrm{1}+{z}}{\frac{\mathrm{1}+{z}}{{z}}} \\ $$$$={z} \\ $$$$={sin}\frac{\pi}{\mathrm{8}}+{icos}\frac{\pi}{\mathrm{8}} \\ $$$$=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}{i} \\ $$

Commented by M±th+et£s last updated on 05/Apr/20

thanks for all

$${thanks}\:{for}\:{all} \\ $$$$ \\ $$

Commented by peter frank last updated on 05/Apr/20

thank you

$${thank}\:{you} \\ $$

Answered by TANMAY PANACEA. last updated on 05/Apr/20

((1+sina+icosa)/(1+sina−icosa)) =(((1+sina+icosa)^2 )/((1+sina)^2 +cos^2 a))=((1+sin^2 a−cos^2 a+2sina+2isinacosa+2icosa)/(1+2sina+sin^2 a+cos^2 a)) =((2sin^2 a+2sina+2isinacosa+2icosa)/(2(1+sina))) =((2sina(1+sina)+2icosa(1+sina))/(2(1+sina))) =((2(1+sina)(sina+icosa))/(2(1+sina)))=sina+icosa =sin((π/8))+icos((π/8))

$$\frac{\mathrm{1}+{sina}+{icosa}}{\mathrm{1}+{sina}−{icosa}} \\ $$$$=\frac{\left(\mathrm{1}+{sina}+{icosa}\right)^{\mathrm{2}} }{\left(\mathrm{1}+{sina}\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} {a}}=\frac{\mathrm{1}+{sin}^{\mathrm{2}} {a}−{cos}^{\mathrm{2}} {a}+\mathrm{2}{sina}+\mathrm{2}{isinacosa}+\mathrm{2}{icosa}}{\mathrm{1}+\mathrm{2}{sina}+{sin}^{\mathrm{2}} {a}+{cos}^{\mathrm{2}} {a}} \\ $$$$=\frac{\mathrm{2}{sin}^{\mathrm{2}} {a}+\mathrm{2}{sina}+\mathrm{2}{isinacosa}+\mathrm{2}{icosa}}{\mathrm{2}\left(\mathrm{1}+{sina}\right)} \\ $$$$=\frac{\mathrm{2}{sina}\left(\mathrm{1}+{sina}\right)+\mathrm{2}{icosa}\left(\mathrm{1}+{sina}\right)}{\mathrm{2}\left(\mathrm{1}+{sina}\right)} \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}+{sina}\right)\left({sina}+{icosa}\right)}{\mathrm{2}\left(\mathrm{1}+{sina}\right)}={sina}+{icosa} \\ $$$$={sin}\left(\frac{\pi}{\mathrm{8}}\right)+{icos}\left(\frac{\pi}{\mathrm{8}}\right) \\ $$$$ \\ $$

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