﻿ Determine the exact value of \sin 3\degree

#### Question

Determine the exact value of \sin 3\degree

Collected in the board: Trigonometry

Steven Zheng posted 4 months ago

Since 3 is the difference of 18 and 15,

\sin 3\degree can be derived from trig values of 18 \degree and 15 \degree using sum identity for sines function

\sin(\alpha -\beta ) = \sin \alpha\cos \beta -\cos \alpha \sin \beta

\sin 3\degree = \sin(18\degree-15\degree)

=\sin 18\degree\cos 15\degree-\cos18\degree\sin 15\degree
(1)

Now we are going to determine the values of \sin 18\degree,\cos 15\degree,\cos18\degree,\sin 15\degree

Using the sum identities, \sin 15\degree and \cos 15\degree can be transformed to product and sum of trig values of 45\degree and 30 \degree

\sin 15\degree=\sin(45\degree -\sin 30\degree )

=\sin 45\degree \cos 30\degree -\cos 45\sin 30\degree

=\dfrac{\sqrt{2} }{2} \cdotp \dfrac{\sqrt{3} }{2} - \dfrac{\sqrt{2} }{2} \cdotp\dfrac{1}{2}

=\dfrac{\sqrt{6}-\sqrt{2} }{4}
(2)

\cos 15\degree =\cos(45\degree -30\degree )

=\cos 45\degree\cos 30\degree+\sin 45\degree\sin 30\degree

=\dfrac{\sqrt{2} }{2}\cdotp \dfrac{\sqrt{3} }{2} +\dfrac{\sqrt{2} }{2}\cdotp \dfrac{1 }{2}

=\dfrac{\sqrt{6}+\sqrt{2} }{4}
(3)

The values of \sin 18\degree and \cos 18\degree can be determined with either trigonometric or geometric method. Since 18\degree is the half of the vertex angle of a golden triangle, its trigonometric values can be determined using the 36-72-72 isosceles triangle.

Since 54 is the triple of 18 and 36 is the double of 18, the triple and double identities are used to derive the trig values of 18\degree

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

\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

And then we have the following equation

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

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}
(4)

Now let's derive the value of \cos 18\degree Using the triple angle identity for cosine function

\cos 3\alpha =4\cos^3\alpha -3\cos \alpha

\cos 54° = 4\cos^3 18° -3\cos 18°

\cos 54°=\cos (90°-36°) = \sin 36°

Using the double angle identity leads to

2\sin 18°\cos 18°=4\cos^3 18° -3\cos 18°

Since \cos 18° \ne0, the equation is simplified as

4\cos^2 18° -2\sin 18° -3 =0

Using Pythagorean Identity

4(1-\sin^2 18°) -2\sin 18° -3 =0

4\sin^2 18°+2\sin 18°-1=0,

Solving the quadratic equation, we get

\sin 18° = \dfrac{-2+\sqrt{4+16} }{8} , (negative value is disregarded)

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

Now using Pythagorean Identity

\cos 18° = \sqrt{1-\sin^2 18° }

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

=\dfrac{\sqrt{16-(6-2\sqrt{5} )} }{4}

=\dfrac{\sqrt{10+2\sqrt{5} } }{4} Therefore,

\cos 18° =\dfrac{\sqrt{10+2\sqrt{5} } }{4}
(5)

Now we have ,

\sin 15\degree=\dfrac{\sqrt{6}-\sqrt{2} }{4}

\cos 15\degree=\dfrac{\sqrt{6}+\sqrt{2} }{4}

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

\cos 18° =\dfrac{\sqrt{10+2\sqrt{5} } }{4}

Substituting (2),(3),(4) and (5) into (1) gives exact value of

\sin 3\degree =\sin 18\degree\cos 15\degree-\cos18\degree\sin 15\degree

=\dfrac{\sqrt{5}-1 }{4} \cdotp \dfrac{\sqrt{6}+\sqrt{2} }{4} -\dfrac{\sqrt{10+2\sqrt{5} } }{4} \cdotp \dfrac{\sqrt{6}-\sqrt{2} }{4}

=\dfrac{(\sqrt{5}-1)(\sqrt{6}+\sqrt{2} )-\sqrt{10+2\sqrt{5} }(\sqrt{6}-\sqrt{2})}{16}

Steven Zheng posted 4 months ago

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