Question


Answer 1

\cos 36\degree =\cos (2\times 18\degree)

=1-2\sin^2 18\degree


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

\sin^2 18\degree =\dfrac{5+1-2\sqrt{5} }{16} =\dfrac{3-\sqrt{5} }{8}

\cos 36\degree =1-\dfrac{3-\sqrt{5} }{4}

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

Steven Zheng posted 1 month ago


Answer 2

Using golden triangle to determine the value of \cos 36\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 altitude from point A to BC

Using Pythagorean Theorem

AG = \sqrt{AC^2-(\dfrac{1}{2}BC)^2}

=\sqrt{(\dfrac{1+\sqrt{5} }{2} )^2-\dfrac{1}{4} }

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

Drop altitude from point B to AC

Using the area formula for a triangle,

A_{ABC} = \dfrac{1}{2}AG\cdotp BC = \dfrac{1}{2}BF\cdotp AC

BF = \dfrac{AG\cdotp BC}{AC}

=\dfrac{\dfrac{1}{2}\sqrt{5+2\sqrt{5} } }{\dfrac{1+\sqrt{5} }{2}}

=\dfrac{\sqrt{5+2\sqrt{5} } }{1+\sqrt{5}}

\sin 36\degree =\dfrac{BF}{AC}

=\dfrac{\dfrac{\sqrt{5+2\sqrt{5} } }{1+\sqrt{5}} }{\dfrac{1+\sqrt{5} }{2}}

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

=\dfrac{\sqrt{5+2\sqrt{5} }}{3+\sqrt{5} }

Using Pythagorean identity

\cos 36\degree = \sqrt{1-\sin^2 36\degree }

=\sqrt{1-(\dfrac{\sqrt{5+2\sqrt{5} }}{3+\sqrt{5} } )^2}

=\sqrt{1-\dfrac{5+2\sqrt{5} }{14+6\sqrt{5} } }

=\sqrt{\dfrac{9+4\sqrt{5} }{14+6\sqrt{5} } }

=\sqrt{\dfrac{(2+\sqrt(5))^2}{(3+\sqrt(5))^2} }

=\dfrac{2+\sqrt{5} }{3+\sqrt{5} }

=\dfrac{2+\sqrt{5} }{3+\sqrt{5} } \cdotp \dfrac{3-\sqrt{5} }{3-\sqrt{5} }

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

Now we have determined the value of \cos 36\degree using geometric method


Steven Zheng posted 4 weeks ago

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