Question

Solve : $\sin {20}^o.\sin {40}^o.\sin {60}^o.\sin {80}^o$

• A. $-\frac{3}{16}$
• B. $\frac{3}{16}$
• C. $\frac{3}{14}$
• D. $\frac{1}{16}$

Answer 1

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

$\sin 20\sin 40\sin 60\sin 80$

$\sin 60=\frac{\sqrt{3}}{2}$

$=\frac{\sqrt{3}}{2}\left(\sin 20sin40sin80\right)$

$\frac{\sqrt{3}}{4}\left(2\right(sin20sin40\left)sin80\right)$

On applying $\left[\cos \right(A-B\left)\cos \right(A+B)=2\sin A\sin B)$

$=\frac{\sqrt{3}}{4}\left(\right(\cos (40-20)-\cos (40+20)\left)\sin 80\right)$

$=\frac{\sqrt{3}}{4}\left(\right(\cos 20-\cos 60\left)\sin 80\right)$

$=\frac{\sqrt{3}}{4}(\cos 20\sin 80-\cos 60\sin 80)$

$=\frac{\sqrt{3}}{8}(2cos20\sin 80-\sin 80)$

$=\frac{\sqrt{3}}{8}\left(\sin \right(100)+\sin 60-\sin 80)$

$=\frac{\sqrt{3}}{8}(\frac{\sqrt{3}}{2}+\sin 100-\sin 80)$

$=\frac{\sqrt{3}}{8}(\frac{\sqrt{3}}{2}+\sin (180-80)-\sin 80)$

$=\frac{\sqrt{3}}{8}(\frac{\sqrt{3}}{2}+\sin 80-\sin 80)$

$=\frac{3}{16}$