Exact-value-of-sin-9-

Question Number 10750 by Joel576 last updated on 24/Feb/17

Exact value of

\sin 9 ° = \dots

Answered by ridwan balatif last updated on 24/Feb/17

first we must know the value of \sin 18^{o}

let x = 18^{o}

5 x = 90^{o}

3 x = 90 - 2 x

\cos (3 x) = \cos (90 - 2 x)

\cos (3 x) = \sin (2 x)

\cos (2 x + x) = \sin (2 x)

\cos 2 x \times cosx - sinx \times \sin 2 x = 2 sinxcosx

(\cos^{2} x - \sin^{2} x) \times cosx - sinx \times (2 sinxcosx) = 2 sinxcosx

(\cos^{2} x - \sin^{2} x) - {2 \sin}^{2} x = 2 sinx

(1 - {2 \sin}^{2} x) - {2 \sin}^{2} x = 2 sinx

{4 \sin}^{2} x + 2 sinx - 1 = 0

D = b^{2} - 4 ac = 20

{sinx}_{1, 2} = \frac{- 2 \pm \sqrt{20}}{2.4} = \frac{- 1 \pm \sqrt{5}}{4}

\sin 18^{o} must be positive

{\sin 18}^{o} = \frac{\sqrt{5} - 1}{4}

{\cos 18}^{o} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}

{\sin 9}^{o} = \sqrt{\frac{1 - {\cos 18}^{o}}{2}}

= \sqrt{\frac{1 - (\frac{\sqrt{10 + 2 \sqrt{5}}}{4})}{2}}

= \sqrt{\frac{\frac{4 - (\sqrt{10 + 2 \sqrt{5}})}{4}}{2}}

= \sqrt{\frac{4 - \sqrt{10 + 2 \sqrt{5}}}{8}}

= \frac{1}{2} \sqrt{2 - \frac{1}{2} \sqrt{10 + 2 \sqrt{5}}}

Commented by Joel576 last updated on 24/Feb/17

t h a n k y o u v e r y m u c h

Facebook Tweet Pin