Question

$sin{20}^{\circ }sin{40}^{\circ }sin{60}^{\circ }sin{80}^{\circ }=$

• A. $-\frac{3}{16}$
  
• B. $\frac{5}{16}$
• C. $\frac{3}{16}$
• D. $-\frac{5}{16}$

Answer 1

The correct option is C $\frac{3}{16}$

$sin{20}^{\circ }sin{40}^{\circ }sin{60}^{\circ }sin{80}^{\circ }=\frac{1}{2}sin{20}^{\circ }sin{60}^{\circ }\left(2sin{40}^{\circ }sin{80}^{\circ }\right)$
$=\frac{1}{2}sin{20}^{\circ }sin{60}^{\circ }(cos{40}^{\circ }-cos{120}^{\circ })$
$=\frac{1}{2}.\frac{\sqrt{3}}{2}sin{20}^{\circ }(1-2si{n}^2{20}^{\circ }+\frac{1}{2})$
$=\frac{\sqrt{3}}{4}sin{20}^{\circ }(\frac{3}{2}-2si{n}^2{20}^{\circ })$
$=\frac{\sqrt{3}}{8}(3sin{20}^{\circ }-4si{n}^3{20}^{\circ })=\frac{\sqrt{3}}{8}sin{60}^{\circ }=\frac{\sqrt{3}}{8}.\frac{\sqrt{3}}{2}=\frac{3}{16}.$