Prove-that-sin-36-10-2-5-4-

Question Number 163171 by ZiYangLee last updated on 04/Jan/22

Prove that \sin 36 ° = \frac{\sqrt{10 - 2 \sqrt{5^{}}}}{4}

Answered by mr W last updated on 04/Jan/22

\sin (3 \times 18^{°}) = \sin (54 °) = \cos (36 °) = \cos (2 \times 18 °)

3 \sin 18 ° - 4 \sin^{3} 18 ° = 1 - 2 \sin^{2} 18 °

l e t t = \sin 18 °

3 t - 4 t^{3} = 1 - 2 t^{2}

4 t^{3} - 2 t^{2} - 3 t + 1 = 0

4 t^{3} - 4 t^{2} + 2 t^{2} - 2 t - t + 1 = 0

(t - 1) (4 t^{2} + 2 t - 1) = 0

t \neq 1

\Rightarrow 4 t^{2} + 2 t - 1 = 0

\Rightarrow t = \frac{- 1 + \sqrt{5}}{4} = \sin 18 °

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

\sin 36 ° = 2 \sin 18 ° \cos 18 °

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

= \frac{\sqrt{(10 + 2 \sqrt{5}) (6 - 2 \sqrt{5})}}{8}

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

Commented by peter frank last updated on 05/Jan/22

thank you

Commented by henderson last updated on 07/Jan/22

please, sir W!

how you get the second line \dots ?

Commented by mr W last updated on 07/Jan/22

t h e f o r m u l a s f o r \sin 3 α a n d \cos 3 α

a r e v e r y u s e f u l . y o u s h o u l d m e m o r i s e

t h e m . b u t i f y o u {c a n}^{'} t, y o u c a n a l s o

g e t t h e m e a s i l y l i k e i n f o l l o w i n g w a y :

\sin 3 α

= \sin (2 α + α)

= \sin 2 α \cos α + \cos 2 α \sin α

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

= 2 \sin α (1 - \sin^{2} α) + (1 - 2 \sin^{2} α) \sin α

= 2 \sin α - 2 \sin^{3} α + \sin α - 2 \sin^{3} α

= 3 \sin α - 4 \sin^{3} α

= (3 - 4 \sin^{2} α) \sin α

Commented by greg_ed last updated on 07/Jan/22

very good

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