Question

If $ \theta =\frac{\pi }{{{2}^{n}}+1}, $ then the value of $ {{2}^{n}}\cos \theta \,\cos \,2\theta \,\cos \,{{2}^{2}}\theta .....\cos {{2}^{n-1}}\theta $ is

• A. $ \sin \theta $
• B. $ \frac{\pi }{2} $
• C. $ 0 $
• D. $ 1 $

Answer 1

Answer: (D)

$ {{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 $