Given that cos 30 = 3 / 2 \cos 30 = \sqrt{3}/2 cos 30 = 3 /2 cos 36 = ( 1 + 5 ) / 4 \cos 36 = (1 + \sqrt{5})/4 cos 36 = ( 1 + 5 ) /4 cos 3 \cos 3 cos 3
Solution We will need sin 30 \sin 30 sin 30 sin 36 \sin 36 sin 36 ( cos θ ) 2 + ( sin θ ) 2 = 1 (\cos \theta)^2 + (\sin \theta)^2 = 1 ( cos θ ) 2 + ( sin θ ) 2 = 1
sin 30 = 1 − ( cos 30 ) 2 = 1 − 3 4 = 1 2 sin 36 = 1 − ( cos 36 ) 2 = 1 − 1 + 5 + 2 5 16 = 16 − 6 − 2 5 16 = 10 − 2 5 4 .
\begin{aligned}
\sin 30 &=& \sqrt{1 - (\cos 30)^2} = \sqrt{1 - \frac{3}{4}} = \frac{1}{2} \\
\sin 36 &=& \sqrt{1 - (\cos 36)^2} = \sqrt{1 - \frac{1 + 5 + 2 \sqrt{5}}{16}} \\
&=& \sqrt{\frac{16 - 6 - 2 \sqrt{5}}{16}} = \frac{\sqrt{10 - 2 \sqrt{5}}}{4}.
\end{aligned}
sin 30 sin 36 = = = 1 − ( cos 30 ) 2 = 1 − 4 3 = 2 1 1 − ( cos 36 ) 2 = 1 − 16 1 + 5 + 2 5 16 16 − 6 − 2 5 = 4 10 − 2 5 . After some thought and a bit of scratch work, we find that
cos 3 = cos 6 2 = cos ( 36 − 30 2 ) cos ( 36 − 30 ) = cos 36 cos 30 − sin 36 sin 30 = 1 + 5 4 ⋅ 3 2 − 10 − 2 5 4 ⋅ 1 2 = 3 + 5 + 10 − 2 5 8 cos 3 = cos ( 6 2 ) = 1 + cos 6 2 = 8 + 3 + 15 + 10 − 2 5 16 = 8 + 3 + 15 + 10 − 2 5 4
\begin{aligned}
\cos 3 &=& \cos \frac{6}{2} = \cos \left( \frac{36 - 30}{2} \right) \\
\cos (36 - 30) &=& \cos 36 \cos 30 - \sin 36 \sin 30 \\
&=& \frac{1+\sqrt{5}}{4} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{10 - 2 \sqrt{5}}}{4} \cdot \frac{1}{2} \\
&=& \frac{\sqrt{3} + \sqrt{5} + \sqrt{10 - 2 \sqrt{5}}}{8} \\
\cos 3 &=& \cos \left( \frac{6}{2} \right) = \sqrt{ \frac{1 + \cos{6}}{2}} \\
&=& \sqrt{ \frac{8 + \sqrt{3} + \sqrt{15} + \sqrt{10 - 2 \sqrt{5}}}{16}} \\
&=& \frac{ \sqrt{8 + \sqrt{3} + \sqrt{15} + \sqrt{10 - 2 \sqrt{5}}}}{4}
\end{aligned}
cos 3 cos ( 36 − 30 ) cos 3 = = = = = = = cos 2 6 = cos ( 2 36 − 30 ) cos 36 cos 30 − sin 36 sin 30 4 1 + 5 ⋅ 2 3 − 4 10 − 2 5 ⋅ 2 1 8 3 + 5 + 10 − 2 5 cos ( 2 6 ) = 2 1 + cos 6 16 8 + 3 + 15 + 10 − 2 5 4 8 + 3 + 15 + 10 − 2 5 Wow. Complicated!