三角関数

と
$$\begin{array}{rl}\sin x& =\frac{{\mathrm{e}}^{ix}–{\mathrm{e}}^{-ix}}{2i}\\ \cos x& =\frac{{\mathrm{e}}^{ix}+{\mathrm{e}}^{-ix}}{2}\end{array}$$
に注意して、問題の式を全て書き直すと

$$\begin{array}{rl}& \sin {20}^{\circ }\cdot \sin {40}^{\circ }\cdot \sin {60}^{\circ }\cdot \sin {80}^{\circ }\\ & =\frac{1}{(2i{)}^4}({\mathrm{e}}^{\frac{\pi i}{9}}–{\mathrm{e}}^{-\frac{\pi i}{9}})\cdot ({\mathrm{e}}^{\frac{2\pi i}{9}}–{\mathrm{e}}^{-\frac{2\pi i}{9}})\cdot ({\mathrm{e}}^{\frac{3\pi i}{9}}–{\mathrm{e}}^{-\frac{3\pi i}{9}})\cdot ({\mathrm{e}}^{\frac{4\pi i}{9}}–{\mathrm{e}}^{-\frac{4\pi i}{9}})\\ & =\frac{1}{16}({\mathrm{e}}^{\frac{3\pi i}{9}}–{\mathrm{e}}^{-\frac{\pi i}{9}}–{\mathrm{e}}^{\frac{\pi i}{9}}+{\mathrm{e}}^{-\frac{3\pi i}{9}})\cdot ({\mathrm{e}}^{\frac{7\pi i}{9}}–{\mathrm{e}}^{-\frac{\pi i}{9}}–{\mathrm{e}}^{\frac{\pi i}{9}}+{\mathrm{e}}^{-\frac{7\pi i}{9}})\\ & =\frac{1}{16}({\mathrm{e}}^{\frac{10\pi i}{9}}–{\mathrm{e}}^{\frac{2\pi i}{9}}–{\mathrm{e}}^{\frac{4\pi i}{9}}+{\mathrm{e}}^{-\frac{4\pi i}{9}}\\ & –{\mathrm{e}}^{\frac{6\pi i}{9}}+{\mathrm{e}}^{-\frac{2\pi i}{9}}+1–{\mathrm{e}}^{-\frac{8\pi i}{9}}\\ & –{\mathrm{e}}^{\frac{8\pi i}{9}}+1+{\mathrm{e}}^{\frac{2\pi i}{9}}–{\mathrm{e}}^{-\frac{6\pi i}{9}}\\ & +{\mathrm{e}}^{\frac{4\pi i}{9}}–{\mathrm{e}}^{-\frac{4\pi i}{9}}–{\mathrm{e}}^{-\frac{2\pi i}{9}}+{\mathrm{e}}^{-\frac{10\pi i}{9}})\\ & =\frac{1}{16}(2\cos (\frac{10\pi }{9})–2\cos (\frac{6\pi }{9})+2–2\cos (\frac{8\pi }{9}))\\ & =\frac{3}{16}\end{array}$$
と求まる。