A) \[\frac{1}{2}\]
B) \[\frac{1}{4}\]
C) \[\frac{1}{6}\]
D) \[\frac{1}{8}\]
Correct Answer: D
Solution :
\[\sin 12{}^\circ \sin 48{}^\circ \sin 54{}^\circ \] \[=\sin 12{}^\circ \sin \,(60{}^\circ -12{}^\circ )\sin \,(90{}^\circ -36{}^\circ )\] \[=\frac{\sin 12{}^\circ \sin \,(60{}^\circ -12{}^\circ )\sin \,72{}^\circ \cos 36{}^\circ }{\sin 72{}^\circ }\] \[=\frac{\left[ \sin 12{}^\circ \sin \,(60{}^\circ -12{}^\circ )\sin \,(60{}^\circ +12{}^\circ )\cos 36{}^\circ \right]}{\sin 72{}^\circ }\] \[=\frac{\sin 12{}^\circ ({{\sin }^{2}}60{}^\circ -{{\sin }^{2}}12{}^\circ )\cos 36{}^\circ }{\sin 72{}^\circ }\] \[=\frac{\sin 12{}^\circ \left( \frac{3}{4}-{{\sin }^{2}}12{}^\circ \right)\cos 36{}^\circ }{\sin 72{}^\circ }\] \[=\frac{(3\sin 12{}^\circ -4{{\sin }^{3}}12{}^\circ )}{4}\cdot \frac{\cos 36{}^\circ }{\sin 72{}^\circ }\] \[=\frac{\sin 36{}^\circ \cos 36{}^\circ }{4\sin 72{}^\circ }\] \[=\frac{1}{2}\frac{\sin 72{}^\circ }{4\sin 72{}^\circ }=\frac{1}{8}\]
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