How does the golden ratio come in the play?

Geometry Level 3

Let θ \theta denote the value of the golden ratio , 1 + 5 2 \dfrac{1+\sqrt5}2 , find the value of sin ⁡ − 1 ( 1 2 θ ) \sin^{-1} \left( \dfrac1{2\theta} \right) in degrees.


The answer is 18.

Joel Yip by Joel Yip , 21, Singapore

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1 solution

Rishabh Jain
Feb 7, 2016

Substituting value of θ ( = 5 + 1 2 ) \theta(=\dfrac{\sqrt5+1}{2}) the given expression simplifies to: sin ⁡ − 1 ( 1 2 ( ( 5 + 1 ) 2 ) ) = sin ⁡ − 1 1 ( 5 + 1 ) \sin^{ -1 }(\dfrac{1 }{\small{2(\frac{(\sqrt5+1)}{2})}}) =\sin^{-1}\dfrac{1}{(\sqrt5+1)} = sin ⁡ − 1 ( 5 − 1 4 ) = 18 ° =\sin^{-1}(\dfrac{\sqrt5-1}{4})=\huge\boxed{\color{#007fff}{18°}} ________________________________ \Large\color{#D61F06}{\text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}} L e t 18 = x ⇒ 5 x = 90 ⇒ 3 x = 90 − 2 x ⇒ sin ⁡ 3 x = c o s 2 x ( sin ⁡ 3 x = 3 sin ⁡ x − 4 sin ⁡ 3 x a n d cos ⁡ 2 x = 1 − 2 sin ⁡ 2 x ) ⇒ 4 sin ⁡ 3 − 2 sin ⁡ 2 x − 3 sin ⁡ x + 1 = 0 ( sin ⁡ x − 1 ) ( 4 sin ⁡ 2 x + 2 sin ⁡ x − 1 ) = 0 Solving the quadratic, we get sin ⁡ x = sin ⁡ 18 = 5 − 1 4 ( ∵ sin ⁡ 18 > 0 a n d sin ⁡ 18 ≠ 1 ) \small{\color{forestgreen}{\boxed{\color{#302B94}{Let~18=x\\ \Rightarrow 5x=90\\ \Rightarrow 3x=90-2x\\ \Rightarrow \sin 3x=cos 2x\\ \color{#EC7300}{(\sin 3x=3\sin x-4\sin^3 x~and~\cos 2x=1-2\sin^2 x)}\\ \Rightarrow 4\sin^3-2\sin^2 x-3\sin x+1=0\\ (\sin x-1)(4\sin^2x+2\sin x-1)=0\\ \text{Solving the quadratic, we get}\\ \sin x=\sin 18=\mathcal{\large{\dfrac{\sqrt5-1}{4}}}\\~~(\because\sin 18>0~ and~ \sin 18\neq1)}}}}

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Joel Yip - 5 years, 4 months ago

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You sure have patience ;P

PS I have edited something ;)

Nihar Mahajan - 5 years, 4 months ago

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Why is the centre appearing distorted?? \color{#D61F06}{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\color{#20A900}{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\color{#CEBB00}{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\color{#302B94}{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\color{plum}{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\color{#007fff}{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\color{magenta}{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\color{#0C6AC7}{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\color{#624F41}{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{~~}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}

Rishabh Jain - 5 years, 4 months ago

used Microsoft Word 2016

Joel Yip - 5 years, 4 months ago

how can you proof the last part?

Joel Yip - 5 years, 4 months ago

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Check the updated solution... : }

Rishabh Jain - 5 years, 4 months ago

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