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Question

Mathematics Question on Trigonometric Functions

If θ=π2n+1,\theta =\frac{\pi }{{{2}^{n}}+1}, then the value of 2ncos⁡θ cos⁡ 2θ cos⁡ 22θ.....cos⁡2n−1θ{{2}^{n}}\cos \theta \,\cos \,2\theta \,\cos \,{{2}^{2}}\theta .....\cos {{2}^{n-1}}\theta is

A

sin⁡θ\sin \theta

B

π2\frac{\pi }{2}

C

00

D

11

Answer

11

Explanation

Solution

2n.cos⁡θ.  cos⁡ 21θ. cos⁡22θ......cos⁡ 2n−1θ{{2}^{n}}.\cos \theta .\,\,\cos \,{{2}^{1}}\theta .\,{{\cos }^{{{2}^{2}}}}\theta ......\cos \,{{2}^{n-1}}\theta
=2nsin⁡2nθ2n.sin⁡θ=sin⁡2nθsin⁡θ={{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\\}
(∵   2n.θ=π−θ)(\because \,\,\,{{2}^{n}}.\theta =\pi -\theta )
=sin⁡ (π−θ)sin⁡θ=sin⁡θsin⁡θ=1=\frac{\sin \,(\pi -\theta )}{\sin \theta }=\frac{\sin \theta }{\sin \theta }=1