Estonian Math Competitions

As $\sin 10^{\circ }=\cos 80^{\circ }$, $\sin 30^{\circ }=\cos 60^{\circ }$, $\sin 50^{\circ }=\cos 40^{\circ }$, and $\sin 70^{\circ }=\cos 20^{\circ }$, the desired equality is equivalent to $$(\cos 20^{\circ }\cdot \cos 40^{\circ }\cdot \cos 60^{\circ }\cdot \cos 80^{\circ }{)}^2=\frac{1}{256}.$$ It is known that $\cos 60^{\circ }=\frac{1}{2}$. Concerning the other factors, we obtain $$\begin{array}{rl}\cos 20^{\circ }\cdot \cos 40^{\circ }\cdot \cos 80^{\circ }& =\frac{8\sin 20^{\circ }\cos 20^{\circ }\cdot \cos 40^{\circ }\cdot \cos 80^{\circ }}{8\sin 20^{\circ }}\\ & =\frac{4\sin 40^{\circ }\cos 40^{\circ }\cdot \cos 80^{\circ }}{8\sin 20^{\circ }}\\ & =\frac{2\sin 80^{\circ }\cos 80^{\circ }}{8\sin 20^{\circ }}\\ & =\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}=\frac{\sin 20^{\circ }}{8\sin 20^{\circ }}=\frac{1}{8}.\end{array}$$ Consequently, $$(\cos 20^{\circ }\cdot \cos 40^{\circ }\cdot \cos 60^{\circ }\cdot \cos 80^{\circ }{)}^2={\left(\frac{1}{2}\cdot \frac{1}{8}\right)}^2={\left(\frac{1}{16}\right)}^2=\frac{1}{256}.$$

---

Alternative solution.

As $\sin 10^{\circ }=\cos 80^{\circ }$, $\sin 30^{\circ }=\cos 60^{\circ }$, $\sin 50^{\circ }=\cos 40^{\circ }$, and $\sin 70^{\circ }=\cos 20^{\circ }$, the l.h.s. of the desired equality can be rewritten as $(\cos 20^{\circ }\cos 40^{\circ }\cos 60^{\circ }\cos 80^{\circ }{)}^2$. Hence the desired equality is equivalent to $$\cos 20^{\circ }\cos 40^{\circ }\cos 60^{\circ }\cos 80^{\circ }=\frac{1}{16}.(2)$$ As $\cos 60^{\circ }=\frac{1}{2}$ and $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$, we have $$\begin{array}{rl}\cos 40^{\circ }\cos 80^{\circ }& =\cos (60^{\circ }-20^{\circ })\cos (60^{\circ }+20^{\circ })\\ & =(\cos 60^{\circ }\cos 20^{\circ }+\sin 60^{\circ }\sin 20^{\circ })(\cos 60^{\circ }\cos 20^{\circ }-\sin 60^{\circ }\sin 20^{\circ })\\ & ={\cos }^{2}60^{\circ }{\cos }^{2}20^{\circ }-{\sin }^{2}60^{\circ }{\sin }^{2}20^{\circ }\\ & =\frac{1}{4}{\cos }^{2}20^{\circ }-\frac{3}{4}{\sin }^{2}20^{\circ }=\frac{1}{4}({\cos }^{2}20^{\circ }-3{\sin }^{2}20^{\circ }).\end{array}$$ Thus $$\cos 20^{\circ }\cos 40^{\circ }\cos 60^{\circ }\cos 80^{\circ }=\frac{1}{8}({\cos }^{3}20^{\circ }-3{\sin }^{2}20^{\circ }\cos 20^{\circ }).(3)$$ By applying the formula $\cos 3x={\cos }^{3}x-3{\sin }^{2}x\cos x$ to $x=20^{\circ }$, we get ${\cos }^{3}20^{\circ }-3{\sin }^{2}20^{\circ }\cos 20^{\circ }=\cos 60^{\circ }=\frac{1}{2}$. Consequently, (3) reduces to (2), completing the solution of the problem.