Problem E-48 Arhimede Mathematical Journal

Proposed by José Luis Díaz-Barrero, BarcelonaTech, Barcelona, Spain. 

Without the aid of a computer, show that

$$\cot36^{\circ}\cot72^{\circ}=\frac{\sqrt{5}}{5}$$

Solution.

First we will find $\sin (18^{\circ})$. Calling $\alpha=18^{\circ}$, we have to

$5\alpha=90^{\circ} \to 3\alpha+2\alpha=90^{\circ} \to 2\alpha=90^{\circ}-3\alpha$. 

 After passing sine on both sides, we have:

$$\sin{2\alpha}=\sin{(90^{\circ}-3\alpha)}=\cos{3\alpha}=4\cos^3{\alpha}-3\cos{\alpha}$$

$$2\sin{\alpha}\cos{\alpha}=4\cos^3{\alpha}-3\cos{\alpha} \,\,\,\, (\cos\alpha\neq0)$$

$$2\sin{\alpha}=4\cos^2{\alpha}-3=4(1-\sin^2{\alpha})-3$$

$$4\sin^2{\alpha}+2\sin{\alpha}-1=0 \to 0<\sin{\alpha}=\sin{18^{\circ}}=\frac{-1+\sqrt{5}}{4}=\cos{72^{\circ}}$$

Note that: $$\cos^2{36^{\circ}}=\frac{1}{2}+\frac{\cos{72^{\circ}}}{2} \to \cos{(36^{\circ})}=\sqrt{\frac{3+\sqrt{5}}{8}}$$

e $$\sin^2{36^{\circ}}=\frac{1}{2}-\frac{\cos{72^{\circ}}}{2} \to \sin{(36^{\circ})}=\sqrt{\frac{5-\sqrt{5}}{8}}$$

Note que: $$\sin{72^{\circ}}=2\sin{(36^{\circ})}\cos{(36^{\circ})}=2\sqrt{\frac{3+\sqrt{5}}{8}}\sqrt{\frac{5-\sqrt{5}}{8}}$$

Soon: $$\cot36^{\circ}\cot72^{\circ}=\left(\frac{\frac{-1+\sqrt{5}}{4}}{2\sqrt{\frac{3+\sqrt{5}}{8}}\sqrt{\frac{5-\sqrt{5}}{8}}}\right)\left(\frac{\sqrt{\frac{3+\sqrt{5}}{8}}}{\sqrt{\frac{5-\sqrt{5}}{8}}}\right)=\left(\frac{\frac{-1+\sqrt{5}}{8}}{\frac{5-\sqrt{5}}{8}}\right)=\frac{-1+\sqrt{5}}{5-\sqrt{5}}=\frac{\sqrt{5}}{5}.$$