4100.22 – Cosine of Three

Given that cos⁡(30∘)=32\cos(30^\circ) = \dfrac{\sqrt{3}}{2} and that cos⁡(36∘)=(1+5)4\cos(36^\circ) = \dfrac{(1 + \sqrt{5})}{4}, find (using formulas for the cosines of the sums and differences of angles) an exact expression for cos⁡(3∘)\cos(3^\circ). Your expression may use radicals as they are exact.

Solution

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

sin⁡(30∘)=1−cos⁡2(30∘)=1−34=12sin⁡(36∘)=1−cos⁡2(36∘)=1−1+5+2516=16−6−2516=10−254.\begin{aligned}\sin(30^\circ) &= \sqrt{1 - \cos^2(30^\circ)} = \sqrt{1 - \dfrac{3}{4}} = \dfrac{1}{2} \\[1em]\sin(36^\circ) &= \sqrt{1 - \cos^2(36^\circ)} = \sqrt{1 - \dfrac{1 + 5 + 2 \sqrt{5}}{16}} \\&= \sqrt{\dfrac{16 - 6 - 2 \sqrt{5}}{16}} = \dfrac{\sqrt{10 - 2 \sqrt{5}}}{4}.\end{aligned}

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+54⋅32−10−254⋅12=3+5+10−258cos⁡(3∘)=cos⁡(6∘2)=1+cos⁡(6∘)2=8+3+15+10−2516=8+3+15+10−254\begin{aligned}\cos(3^\circ) &= \cos\left(\dfrac{6^\circ}{2}\right) = \cos \left( \dfrac{36^\circ - 30^\circ}{2} \right) \\[1em]\cos(36^\circ - 30^\circ) &= \cos(36^\circ)\cos(30^\circ) - \sin(36^\circ) \sin(30^\circ) \\[0.5em]&= \dfrac{1+\sqrt{5}}{4} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{10 - 2 \sqrt{5}}}{4} \cdot \dfrac{1}{2} \\[1em]&= \dfrac{\sqrt{3} + \sqrt{5} + \sqrt{10 - 2 \sqrt{5}}}{8} \\[1em]\cos(3^\circ) &= \cos\left(\dfrac{6^\circ}{2}\right) = \sqrt{ \dfrac{1 + \cos(6^\circ)}{2}} \\[1em]&= \sqrt{ \dfrac{8 + \sqrt{3} + \sqrt{15} + \sqrt{10 - 2 \sqrt{5}}}{16}} \\[1em]&= \dfrac{ \sqrt{8 + \sqrt{3} + \sqrt{15} + \sqrt{10 - 2 \sqrt{5}}}}{4}\end{aligned}

Wow. Complicated!