Question

The value of $\sin {20}^{\circ }\sin {40}^0\sin {60}^{\circ }\sin {100}^0$ is equal to

• A. $\frac{3}{4}$
• B. $\frac{1}{8}$
• C. $\frac{3}{2}$
• D. $\frac{3}{16}$

Answer 1

Answer: (D)

$\frac{3}{16}$

$(\frac{2sin{40}^{\circ }sin{20}^{\circ }}{2}\cdot sin{60}^{\circ }\cdot sin{100}^{\circ })=\left(\frac{1}{2}\right)\left[cos\right({40}^{\circ }-{20}^{\circ })-cos({40}^{\circ }+{20}^{\circ }\left)\right]\cdot \left(\frac{\sqrt{3}}{2}\right)sin{100}^{\circ }=\left(\frac{1}{2}\right)\left[cos\right({20}^{\circ }-\left(\frac{1}{2}\right)\left)\right(\frac{\sqrt{3}}{2}\left)si{n}^{\circ }\right({90}^{\circ }+{10}^{\circ })=(\frac{\sqrt{3}}{2}\left)si{n}^{\circ }\right({90}^{\circ }+{10}^{\circ })\cdot cos{10}^{\circ }(\frac{\sqrt{3}}{4}\left)\right[\frac{2sin{20}^{\circ }sin{10}^{\circ }}{2}-\left(\frac{1}{2}cos{10}^{\circ }\right)]=(\frac{\sqrt{3}}{4}\left)\right[\frac{cos({20}^{\circ }+{10}^{\circ })+cos({20}^{\circ }-{10}^{\circ }))}{2}-\left(\frac{1}{2}cos{10}^{\circ }\right)]=(\frac{\sqrt{3}}{4}\left)\right[(\frac{\sqrt{3}+cos{10}^{\circ }}{2}-(\frac{1}{2}cos{10}^{\circ }\left)\right]=\left(\frac{\sqrt{3}}{4}\right)\cdot \left(\frac{\sqrt{3}}{4}\right)=\left(\frac{3}{16}\right)$