﻿ Determine the value of \sin 18\degree

#### Question

Using the triple angle identity

\sin 3\alpha =3\sin \alpha -4\sin^3\alpha

Since 54 = 3\times 18

\sin 54\degree =3\sin 18\degree -4\sin^3 18\degree
(1)
\sin 54\degree =\sin (90\degree -36\degree ) =\cos 36\degree

Using the double angle identity

\cos 36\degree = 1-2\sin^2 18\degree
(2)

(1) and (2) yield the equation

3\sin 18\degree -4\sin^3 18\degree =1-2\sin^2 18\degree
(3)

Let x=\sin 18\degree , we get more clear view of the equation

4x^3-2x^2-3x+1=0

(2x^3-2x^2)+(2x^3-2x)-(x-1)=0

2x^2(x-1)+2x(x+1)(x-1)-(x-1)=0

2x^2(x-1)(4x^2+2x-1)=0

Cancel x=1 as \sin 18\degree \ne 1

Solve the equation

4x^2+2x-1=0

x=\dfrac{-2\pm \sqrt{4+16} }{8}

Cancel negative result

x=\dfrac{-1+\sqrt{5} }{4}

Therefore, \sin 18\degree = \dfrac{-1+\sqrt{5} }{4}

Steven Zheng posted 1 month ago

Using golden triangle to determine the value of \sin 18\degree

A golden triangle is an isosceles triangle with vertex angle of 36\degree and two congruent base angles of 72\degree

Draw a circle with point B as center and radius in the same measure as the length of BC. The circle intersects line AC at point D.

Since \triangle ABC and \triangle BCD are similar , we get

\dfrac{AC}{BC}=\dfrac{BC}{DC}

BC^2 = AC\cdotp DC

And

Let BC =1 , AC =x

1=x(x-1)

x^2-x-1=0

AC=x = \dfrac{1+\sqrt{5} }{2}
(1)

Drop an altitude from point A to BC

\sin \angle CAF = 18\degree =\dfrac{CF}{AC} = \dfrac{\dfrac{1}{2} }{ \dfrac{1+\sqrt{5} }{2} }

=\dfrac{1}{1+\sqrt{5} }

=\dfrac{-1+\sqrt{5} }{4}

Now we have determined the value of \sin 18° using geometric method.

Steven Zheng posted 4 weeks ago

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