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Question: If \(x_{r} = \cos\left( \frac{\pi}{2^{r}} \right)\)+\(i\sin\left( \frac{\pi}{2^{r}} \right)\), then ...

If xr=cos⁡(π2r)x_{r} = \cos\left( \frac{\pi}{2^{r}} \right)+isin⁡(π2r)i\sin\left( \frac{\pi}{2^{r}} \right), then x1.x2.x3...........∞x_{1}.x_{2}.x_{3}...........\infty is

A

–3

B

–2

C

–1

D

0

Answer

–1

Explanation

Solution

x1.x2.x3.....x_{1}.x_{2}.x_{3}..... upto ∞=(cos⁡π2+isin⁡π2)\infty = \left( \cos \frac { \pi } { 2 } + i \sin \frac { \pi } { 2 } \right) (cos⁡π22+isin⁡π22)\left( \cos\frac{\pi}{2^{2}} + i\sin\frac{\pi}{2^{2}} \right)………………..

= cos⁡(π2+π22+...)+isin⁡(π2+π22+....)\mathbf{\cos}\left( \frac{\mathbf{\pi}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{\pi}}{\mathbf{2}^{\mathbf{2}}}\mathbf{+ ...} \right)\mathbf{+ i}\mathbf{\sin}\left( \frac{\mathbf{\pi}}{\mathbf{2}}\mathbf{+}\frac{\mathbf{\pi}}{\mathbf{2}^{\mathbf{2}}}\mathbf{+ ....} \right)

= cos⁡(π21−12)+isin⁡(π21−12)\cos \left( \frac { \frac { \pi } { 2 } } { 1 - \frac { 1 } { 2 } } \right) + i \sin \left( \frac { \frac { \pi } { 2 } } { 1 - \frac { 1 } { 2 } } \right)= cos⁡π+isin⁡π=−1\mathbf{\cos}\mathbf{\pi}\mathbf{+ i}\mathbf{\sin}\mathbf{\pi}\mathbf{=}\mathbf{-}\mathbf{1}