Question

$sin{36}^{\circ }sin{72}^{o}sin{108}^{\circ }sin{144}^{\circ }$=

• A. 1/4
• B. 1/16
• C. 3/4
• D. 5/16

Answer 1

The correct option is D 5/16

$\sin {144}^{\circ }=\sin ({180}^{\circ }-{36}^{\circ })=\sin {36}^{\circ }$

$\sin {108}^{\circ }\sin {72}^{\circ }=\sin ({90}^{\circ }+{18}^{\circ })\sin ({90}^{\circ }-{18}^{\circ })={\sin }^2{90}^{\circ }-{\sin }^2{18}^{\circ }=1-\frac{(\sqrt{5}-1{)}^2}{16}=\frac{16-5-1+2\sqrt{5}}{16}=\frac{5+\sqrt{5}}{8}$

$\sin {36}^{\circ }\sin {72}^{\circ }\sin {108}^{\circ }\sin {144}^{\circ }=\sin {36}^{\circ }\times \left(\sin {108}^{\circ }\sin {72}^{\circ }\right)\times \sin {36}^{\circ }$

$={\sin }^2{36}^{\circ }\times \frac{5+\sqrt{5}}{8}=\frac{10-2\sqrt{5}}{16}\times \frac{5+\sqrt{5}}{8}=\frac{5^2-5}{8^2}=\frac{5}{16}$