A) \[\cos 20{}^\circ \cos 40{}^\circ \cos 60{}^\circ \cos 80{}^\circ \]
B) \[\sin 20{}^\circ \sin 40{}^\circ \sin 60{}^\circ \sin 80{}^\circ \]
C) \[\frac{3}{15}\]
D) None of these
Correct Answer: A
Solution :
\[\sin \,\,{{12}^{o}}\,\,\sin \,\,{{24}^{o}}\,\,\sin \,\,{{48}^{o}}\,\,\sin \,\,{{84}^{o}}\] \[=\frac{1}{4}\,(2\,\,\sin \,\,{{12}^{o}}\,\sin \,\,{{48}^{o}})\,\,(2\,\,\sin \,\,{{24}^{o}}\,\,\sin \,\,{{84}^{o}})\] \[=\frac{1}{2}(\cos \,\,{{36}^{o}}-\cos \,\,{{60}^{o}})\,\,(\cos \,\,{{60}^{o}}-\cos \,\,{{108}^{o}})\] \[=\frac{1}{4}\,\left( \cos \,\,{{36}^{o}}-\frac{1}{2} \right)\,\,\left( \frac{1}{2}+\sin \,\,{{18}^{o}} \right)\] \[=\frac{1}{4}\left\{ \frac{1}{4}(\sqrt{5}+1)-\frac{1}{2} \right\}\,\left\{ \frac{1}{2}+\frac{1}{4}(\sqrt{5}-1) \right\}=\frac{1}{16}\] and \[\cos \,\,{{20}^{o}}\,\cos \,\,{{40}^{o}}\,\,\cos \,\,60\,\,\cos \,\,{{80}^{o}}\] \[=\frac{1}{2}[\cos \,({{60}^{o}}-{{20}^{o}})\,\cos \,\,{{20}^{o}}\,\cos \,({{60}^{o}}+{{20}^{o}})]\] \[=\frac{1}{2}\,\left[ \frac{1}{4}\cos \,\,3\,\,({{20}^{o}}) \right]=\frac{1}{8}\cos \,\,{{60}^{o}}=\frac{1}{2}\times \frac{1}{8}=\frac{1}{16}\].
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