with-out-cientopic-calculeter-sin9-

Question Number 139724 by mathlove last updated on 30/Apr/21

w i t h o u t c i e n t o p i c c a l c u l e t e r

s i n 9 = ?

Commented by Dwaipayan Shikari last updated on 30/Apr/21

s i n (9 °) ?

Commented by malwan last updated on 30/Apr/21

s i n 9 = \frac{1}{8} \sqrt{2} (1 + \sqrt{5}) -

\frac{1}{8} (\sqrt{5} - 1) \sqrt{5 + \sqrt{5}}

= \frac{1}{2} \sqrt{2 - \sqrt{\frac{1}{2} (5 + \sqrt{5})}}

Answered by Dwaipayan Shikari last updated on 30/Apr/21

s i n (\frac{π}{10}) = \frac{\sqrt{5} - 1}{4}

2 s i n (\frac{π}{20}) \sqrt{1 - {s i n}^{2} (\frac{π}{20})} = \frac{\sqrt{5} - 1}{4}

\Rightarrow {s i n}^{2} (\frac{π}{20}) - {s i n}^{4} (\frac{π}{20}) = {(\frac{\sqrt{5} - 1}{8})}^{2}

\Rightarrow {s i n}^{4} (\frac{π}{20}) - {s i n}^{2} (\frac{π}{20}) + \frac{3 - \sqrt{5}}{32} = 0

\Rightarrow {s i n}^{2} (\frac{π}{20}) = \frac{1 \pm \sqrt{1 - \frac{3 - \sqrt{5}}{8}}}{2} = \frac{1 \pm \sqrt{\frac{5 + \sqrt{5}}{8}}}{2}

s i n (\frac{π}{20}) = \sqrt{\frac{1 - \frac{1}{2} \sqrt{\sqrt{5} ϕ}}{2}} = \frac{1}{2} \sqrt{2 - {(5)}^{1 / 4} \sqrt{ϕ}} ϕ = G o l d e n R a t i o

Commented by mathlove last updated on 01/May/21

w h a t i s \emptyset ?

Commented by Dwaipayan Shikari last updated on 01/May/21

G o l d e n R a t i o ϕ = \frac{1 + \sqrt{5}}{2} = 1.618 . .