Evaluate-sin-9-

Question Number 17497 by tawa tawa last updated on 06/Jul/17

Evaluate : \sin (9) °

Answered by mrW1 last updated on 06/Jul/17

18 ° + 2 \times 18 ° = 90 ° - 2 \times 18 °

\cos (18 ° + 2 \times 18 °) = \cos (90 ° - 2 \times 18 °) = \sin (2 \times 18 °)

\cos 18 \cos (2 \times 18) - \sin 18 \sin (2 \times 18) = \sin (2 \times 18)

\cos 18 \cos (2 \times 18) - (\sin 18 + 1) \sin (2 \times 18) = 0

\cos 18 (2 \cos^{2} 18 - 1) - 2 (\sin 18 + 1) \sin 18 \cos 18 = 0

2 \cos^{2} 18 - 1 - 2 \sin^{2} 18 - 2 \sin 18 = 0

1 - 4 \sin^{2} 18 - 2 \sin 18 = 0

4 \sin^{2} 18 + 2 \sin 18 - 1 = 0

\Rightarrow \sin 18 = \frac{- 2 + \sqrt{20}}{8} = \frac{\sqrt{5} - 1}{4}

\Rightarrow \cos 18 = \sqrt{1 - {(\frac{\sqrt{5} - 1}{4})}^{2}} = \sqrt{(1 + \frac{\sqrt{5} - 1}{4}) (1 - \frac{\sqrt{5} - 1}{4})} = \frac{1}{4} \sqrt{(3 + \sqrt{5}) (5 - \sqrt{5})} = \frac{1}{4} \sqrt{10 + 2 \sqrt{5}}

\Rightarrow 1 - {2 \sin}^{2} 9 = \frac{1}{4} \sqrt{10 + 2 \sqrt{5}}

\Rightarrow \sin 9 = \sqrt{\frac{1}{2} - \frac{1}{8} \sqrt{10 + 2 \sqrt{5}}}

\Rightarrow \sin 9 ° = \frac{1}{4} \sqrt{8 - 2 \sqrt{10 + 2 \sqrt{5}}}

Commented by tawa tawa last updated on 06/Jul/17

Wow, God bless you sir .