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

$$\frac{1+sin\left(\frac{1}{8}\right)\pi +icos\left(\frac{1}{8}\right)\pi }{1+sin\left(\frac{1}{8}\right)\pi -icos\left(\frac{1}{8}\right)\pi }=?$$

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

$$letsin\frac{\pi }{8}+icos\frac{\pi }{8}=z,\mid z\mid =1$$$$\frac{1+z}{1+\overline{z}}$$$$=\frac{1+z}{1+\frac{1}{z}}$$$$=\frac{1+z}{\frac{1+z}{z}}$$$$=z$$$$=sin\frac{\pi }{8}+icos\frac{\pi }{8}$$$$=\frac{\sqrt{2-\sqrt{2}}}{2}+\frac{\sqrt{2+\sqrt{2}}}{2}i$$

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

$$thanksforall$$$$$$

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

$$thankyou$$

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

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

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