A) \[\sin \theta \]
B) \[\frac{\pi }{2}\]
C) \[0\]
D) \[1\]
Correct Answer: D
Solution :
\[{{2}^{n}}.\cos \theta .\,\,\cos \,{{2}^{1}}\theta .\,{{\cos }^{{{2}^{2}}}}\theta ......\cos \,{{2}^{n-1}}\theta \] \[={{2}^{n}}\frac{\sin {{2}^{n}}\theta }{{{2}^{n}}.\sin \theta }=\frac{\sin {{2}^{n}}\theta }{\sin \theta }\] \[\left\{ \because \,\,\,\theta =\frac{\pi }{{{2}^{n}}+1} \right\}\] \[(\because \,\,\,{{2}^{n}}.\theta =\pi -\theta )\] \[=\frac{\sin \,(\pi -\theta )}{\sin \theta }=\frac{\sin \theta }{\sin \theta }=1\]
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