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