A) \[-3/16\]
B) \[5/16\]
C) \[3/16\]
D) \[-5/16\]
Correct Answer: C
Solution :
\[\sin 20{}^\circ \sin {{40}^{o}}\sin 60{}^\circ \sin 80{}^\circ \]\[=\frac{1}{2}\sin 20{}^\circ \sin 60{}^\circ \,(2\sin {{40}^{o}}\sin 80{}^\circ )\] \[=\frac{1}{2}\sin 20{}^\circ \sin 60{}^\circ (\cos 40{}^\circ -\cos 120{}^\circ )\] \[=\frac{1}{2}.\frac{\sqrt{3}}{2}\sin 20{}^\circ \left( 1-2{{\sin }^{2}}20{}^\circ +\frac{1}{2} \right)\] \[=\frac{\sqrt{3}}{4}\sin 20{}^\circ \left( \frac{3}{2}-2{{\sin }^{2}}20{}^\circ \right)\] \[=\frac{\sqrt{3}}{8}(3\sin 20{}^\circ -4{{\sin }^{3}}20{}^\circ )\]\[=\frac{\sqrt{3}}{8}\sin 60{}^\circ =\frac{\sqrt{3}}{8}.\frac{\sqrt{3}}{2}=\frac{3}{16}\].
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