Question

Prove that:
$\sin 36°\sin 72°\sin 108°\sin 144°=\frac{5}{16}$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sin {36}^{\circ }\sin {72}^{\circ }\sin {108}^{\circ }\sin {144}^{\circ } \\ =\sin {36}^{\circ }\sin {72}^{\circ }\sin {72}^{\circ }\sin {36}^{\circ } \\ \left[\because \sin {144}^{\circ }=\sin \left({180}^{\circ }-{36}^{\circ }\right)=\sin {36}^{\circ } \\ \mathrm{and}\sin {108}^{\circ }=\sin \left({180}^{\circ }-{72}^{\circ }\right)=\sin {72}^{\circ }\right] \\ ={\left(\sin {36}^{\circ }\sin {72}^{\circ }\right)}^2 \\ ={\left(\frac{\sqrt{10-2\sqrt{5}}}{4}\times \frac{\sqrt{10+2\sqrt{5}}}{4}\right)}^2 \\ ={\left(\frac{\sqrt{100-20}}{4}\right)}^2 \\ ={\left(\frac{4\sqrt{5}}{16}\right)}^2 \\ =\frac{5}{16} \\ =\mathrm{RHS} \end{gathered}$$

Hence proved.