1-sin-1-8-pi-i-cos-1-8-pi-1-sin-1-8-pi-i-cos-1-8-pi-

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

\frac{1 + s i n (\frac{1}{8}) π + i c o s (\frac{1}{8}) π}{1 + s i n (\frac{1}{8}) π - i c o s (\frac{1}{8}) π} = ?

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

l e t s i n \frac{π}{8} + i c o s \frac{π}{8} = z, ∣ z ∣= 1

\frac{1 + z}{1 + \bar{z}}

= \frac{1 + z}{1 + \frac{1}{z}}

= \frac{1 + z}{\frac{1 + z}{z}}

= z

= s i n \frac{π}{8} + i c o s \frac{π}{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

t h a n k s f o r a l l

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

t h a n k y o u

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

\frac{1 + s i n a + i c o s a}{1 + s i n a - i c o s a}

= \frac{{(1 + s i n a + i c o s a)}^{2}}{{(1 + s i n a)}^{2} + {c o s}^{2} a} = \frac{1 + {s i n}^{2} a - {c o s}^{2} a + 2 s i n a + 2 i s i n a c o s a + 2 i c o s a}{1 + 2 s i n a + {s i n}^{2} a + {c o s}^{2} a}

= \frac{2 {s i n}^{2} a + 2 s i n a + 2 i s i n a c o s a + 2 i c o s a}{2 (1 + s i n a)}

= \frac{2 s i n a (1 + s i n a) + 2 i c o s a (1 + s i n a)}{2 (1 + s i n a)}

= \frac{2 (1 + s i n a) (s i n a + i c o s a)}{2 (1 + s i n a)} = s i n a + i c o s a

= s i n (\frac{π}{8}) + i c o s (\frac{π}{8})