A) \[\frac{\sin 2n\alpha }{2n\sin \alpha }\]
B) \[\frac{\sin {{2}^{n}}\alpha }{{{2}^{n}}\sin {{2}^{n-1}}\alpha }\]
C) \[\frac{\sin {{4}^{n-1}}\alpha }{{{4}^{n-1}}\sin \alpha }\]
D) \[\frac{\sin {{2}^{n}}\alpha }{{{2}^{n}}\sin \alpha }\]
E) none of these
Correct Answer: D
Solution :
\[\cos \alpha \,\cos 2\alpha \cos 4\alpha ....\cos {{2}^{n-1}}\alpha =\frac{\sin {{2}^{n}}\alpha }{{{2}^{n}}\sin \alpha }\]
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