A Quick Proof of $\displaystyle \cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{3\pi}{7}=\frac{1}{8}$

Let $ABC$ be an isosceles triangle, $AC=BC,$ with $\displaystyle \angle C=\frac{\pi}{7}.$ Mark point $D$ on $AC$ such that $AB=AD.$ Define, for convenience, $AB=1,$ $BC=x,$ and $BD=y.$ Then also $AD=1$ and $CD=y.$

In $\Delta BCD,$ $\displaystyle \cos C=\cos\frac{\pi}{7}=\frac{x/2}{y}=\frac{x}{2y}.$

In $\Delta ABD,$ $\displaystyle \cos D=\cos\frac{2\pi}{7}=\frac{y/2}{1}=\frac{y}{2}.$

In $\Delta ABC,$ $\displaystyle \cos A=\cos\frac{3\pi}{7}=\frac{1/2}{x}=\frac{1}{2x}.$

We thus have

$\displaystyle \cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{3\pi}{7}=\frac{x}{2y}\cdot\frac{y}{2x}\cdot\frac{1}{2x}=\frac{1}{8}.$

Let, for convenience, $\displaystyle \alpha=\frac{\pi}{7}.$ Then, using the double argument formulas for sine,

$\displaystyle\begin{align} \cos\alpha\cdot\cos 2\alpha\cdot\cos 3\alpha&=\frac{1}{2\sin\alpha}\sin 2\alpha\cos 2\alpha\cos 3\alpha\\ &=\frac{1}{4\sin\alpha}\sin 4\alpha\cos 3\alpha\\ &=\frac{1}{8\sin\alpha}\left(\sin\alpha+\sin 7\alpha\right)=\frac{1}{8}. \end{align}$

This is Quickie 846 proposed and proved (Solution 1) by John P. Hoyt (Mathematics Magazine, Vol. 69, No. 1 (Feb., 1996), p 73).