# 4100.22 – Cosine of Three

Given that $\cos 30 = \sqrt{3}/2$ and that $\cos 36 = (1 + \sqrt{5})/4$, find (using formulas for the cosines of the sums and differences of angles) an exact expression for $\cos 3$. Your expression may use radicals as they are exact.

Solution

We will need $\sin 30$ and $\sin 36$ as well as the given cosines. These follow from the Pythagorean formula: $(\cos \theta)^2 + (\sin \theta)^2 = 1$. Using this we find that

\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}

After some thought and a bit of scratch work, we find that

\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}

Wow. Complicated!