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