A) \[\frac{\sin {{2}^{n}}A}{{{2}^{n}}\sin A}\]
B) \[\frac{{{2}^{n}}\sin {{2}^{n}}A}{\sin A}\]
C) \[\frac{{{2}^{n}}\sin A}{\sin {{2}^{n}}A}\]
D) \[\frac{\sin A}{{{2}^{n}}\sin {{2}^{n}}A}\]
Correct Answer: A
Solution :
It is a standard result. \[\cos A\cos 2A\cos {{2}^{2}}A...\cos {{2}^{n-1}}A\] \[=\frac{\sin {{2}^{n}}A}{{{2}^{n}}\sin A}\]
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