Question
Determine the value of \sin 18\degree
Answer 1
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)
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}
Answer 2
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
BC = BD = AD
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.