Four steps from determine the trigonometric values of 18\degree to 36\degree , and then using difference identities to get the value of cos 6\degree

1. Determine the value of sin 18°

Using the triple angle identity

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

Since 54 = 3\times 18

\sin 54\degree =\sin (90\degree -36\degree ) =\cos 36\degree

Using the double angle identity

(1) and (2) yield the equation

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}

2. Determine the value of cos 36°

Since 54 is the triple of 18,

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}

3. Determine the value of sin 36°

Using Pythagorean Theorem,

\sin 36° = \sqrt{1-\cos^2 36\degree }

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

=\dfrac{1}{4}\cdotp \sqrt{16-(6-2\sqrt{5} )}

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

4. Substituting the trig values of 36° to the following equations

\cos 36°= \dfrac{1+\sqrt{5} }{4},

\sin 36° =\dfrac{1}{4}\cdotp \sqrt{10-2\sqrt{5} }

\cos 6° = \cos(36° -30° ) = \cos 36°\cos 30° +\sin 36° \sin 30°

= \dfrac{1+\sqrt{5} }{4}\cdot \dfrac{\sqrt{3} }{2}+\dfrac{1}{2}\cdot \dfrac{1}{4}\cdotp \sqrt{10-2\sqrt{5} }

=\dfrac{1}{8}[\sqrt{3}(1+\sqrt{5} ) + \sqrt{10-2\sqrt{5} } ]

So the exact value of cos 6° is

\dfrac{1}{8}[\sqrt{3}(1+\sqrt{5} ) + \sqrt{10-2\sqrt{5} } ]