sin-9-sin-21-sin-39-2-golden-ratio-m-n-

Question Number 153553 by mnjuly1970 last updated on 08/Sep/21

s i n (9) + s i n (21) + s i n (39) \overset{?}{=} \frac{φ}{\sqrt{2}}

φ := g o l d e n r a t i o

m . n

Answered by bramlexs22 last updated on 08/Sep/21

\sin (9 °) + \sin (21 °) + \sin (39 °) =

2 \sin 15 ° \cos 6 ° + \sin (45 ° - 6 °) =

2 \sin (45 ° - 30 °) \cos 6 ° + \frac{\sqrt{2}}{2} \cos 6 ° - \frac{\sqrt{2}}{2} \sin 6 ° =

2 \cos 6 ° (\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}) + \frac{\sqrt{2}}{2} \cos 6 ° - \frac{\sqrt{2}}{2} \sin 6 ° =

\frac{\sqrt{6}}{2} \cos 6 ° - \frac{\sqrt{2}}{2} \sin 6 ° = \sqrt{\frac{6 + 2}{4}} \cos (6 ° - 330 °)

= \sqrt{2} \cos 324 ° = \sqrt{2} \cos (270 ° + 54 °)

= \sqrt{2} \sin 54 ° = \sqrt{2} \cos 36 °

= \sqrt{2} (1 - 2 \sin^{2} 18 °)

= \sqrt{2} (1 - 2 {(\frac{\sqrt{5} - 1}{4})}^{2}) = \sqrt{2} (1 - (\frac{6 - 2 \sqrt{5}}{8}))

= \sqrt{2} (\frac{2 + 2 \sqrt{5}}{8}) = \sqrt{2} (\frac{\sqrt{5} + 1}{4}) = \frac{\sqrt{2}}{2} (\frac{\sqrt{5} + 1}{2})

= \frac{1}{\sqrt{2}} φ = \frac{φ}{\sqrt{2}}

Commented by Tawa11 last updated on 08/Sep/21

great sir

Facebook Tweet Pin