what-is-value-sin-36-plese-give-me-answer-

Question Number 8289 by rhm last updated on 07/Oct/16

w h a t i s v a l u e

s i n 36 °

p l e s e g i v e m e a n s w e r

Commented by rhm last updated on 07/Oct/16

s i r a n s w e r

Commented by Rasheed Soomro last updated on 07/Oct/16

s i n 36 ° = \frac{\sqrt{2 \sqrt{5} - 10}}{4}

Answered by ridwan balatif last updated on 15/Oct/16

let : {\sin 18}^{o} = \sin θ

{\sin 36}^{o} = {\cos 54}^{o}

\sin 2 θ = \cos 3 θ

2 \sin θ \cos θ = \cos 2 θ \cos θ - \sin θ \sin 2 θ

2 \sin θ \cos θ = \cos 2 θ \cos θ - \sin θ . (2 \sin θ \cos θ)

2 \sin θ = \cos 2 θ - {2 \sin}^{2} θ

2 \sin θ = 1 - {2 \sin}^{2} θ - {2 \sin}^{2} θ

{4 \sin}^{2} θ + 2 \sin θ - 1 = 0

\sin θ_{1, 2} = \frac{- b \pm \sqrt{D}}{2 a}

\sin θ_{1, 2} = \frac{- 2 \pm \sqrt{20}}{2.4}

\sin θ_{1, 2} = \frac{- 1 \pm \sqrt{5}}{4}

we take positive, because {\sin 18}^{o} has positive value

\sin θ = \frac{- 1 + \sqrt{5}}{4}

so, {\sin 18}^{o} = \frac{- 1 + \sqrt{5}}{4} and {\cos 18}^{o} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}

{\sin 36}^{o} = {2 \sin 18}^{o} {\cos 18}^{o}

{\sin 36}^{o} = (\frac{- 1 + \sqrt{5}}{8}) (\sqrt{10 + 2 \sqrt{5}})