Question

Find the required value for the expression ${x_1}{x_2}{x_{_3}} \ldots {x_n}$(say, n is $\infty $) given that: ${x_n} = \cos \left( {\dfrac{\pi }{{{2^n}}}} \right) + i\sin \left( {\dfrac{\pi }{{{2^n}}}} \right)$.
(a) $0$
(b) $ - 1$
(c) $i$
(d) $ - i$

Answer 1

Hint: In the given question, we need to find the value of the expression provided to us. We will be going to use the Euler’s formula of the complex relations for the number of iterations, assumed and then substituting it into the formula. Then, we will simplify it using the rules of indices so that we can get to the required answer.

Complete step by step answer:
The condition is related to the complex number as there exists the parameter ‘i’ where the value of instance ‘i’ is $i = \sqrt { - 1} $ respectively.
Therefore, using the complex relation for trigonometric solution we can reach the desired value.
Since, we have given the condition:
${x_n} = \cos \left( {\dfrac{\pi }{{{2^n}}}} \right) + i\sin \left( {\dfrac{\pi }{{{2^n}}}} \right)$ which intricate with one of the formula of the complex number that is;
$\cos \theta + i\sin \theta = {e^{i\theta }}$respectively.
So, relating the given condition with above formula, as a result for a desired condition we get the following equation,
${x_n} = \cos \left( {\dfrac{\pi }{{{2^n}}}} \right) + i\sin \left( {\dfrac{\pi }{{{2^n}}}} \right)$
Here, we have $\theta = \dfrac{\pi }{{{2^n}}}$
When n $ = 1$,
(Substituting it to the given condition, we get)
\[{x_1} = \cos \left( {\dfrac{\pi }{2}} \right) + i\sin \left( {\dfrac{\pi }{2}} \right) = {e^{i\left( {\dfrac{\pi }{2}} \right)}}\]
Similarly, when n $ = 2$,
Simplifying it as likely to the above in equation, we get
${x_2} = \cos \left( {\dfrac{\pi }{{{2^2}}}} \right) + i\sin \left( {\dfrac{\pi }{{{2^2}}}} \right) = \cos \left( {\dfrac{\pi }{4}} \right) + i\sin \left( {\dfrac{\pi }{4}} \right) = {e^{i\left( {\dfrac{\pi }{4}} \right)}}$
When n $ = 3$, we get
${x_3} = \cos \left( {\dfrac{\pi }{{{2^3}}}} \right) + i\sin \left( {\dfrac{\pi }{{{2^3}}}} \right) = \cos \left( {\dfrac{\pi }{8}} \right) + i\sin \left( {\dfrac{\pi }{8}} \right) = {e^{i\left( {\dfrac{\pi }{8}} \right)}}$
When n $ = \infty $, we get
\[{x_n} = \cos \left( {\dfrac{\pi }{{{2^n}}}} \right) + i\sin \left( {\dfrac{\pi }{{{2^n}}}} \right) = {e^{i\left( {\dfrac{\pi }{{{2^n}}}} \right)}}\]
Now, hence we get all the required factors for a solution. Substituting it in the required expression, we get
${x_1}{x_2}{x_3} \ldots {x_n} = \left[ {{e^{i\left( {\dfrac{\pi }{2}} \right)}}} \right]\left[ {{e^{i\left( {\dfrac{\pi }{4}} \right)}}} \right]\left[ {{e^{i\left( {\dfrac{\pi }{8}} \right)}}} \right] \ldots $ nth terms
$ = {e^{i\pi \left( {\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} \ldots \infty } \right)}}$
$ = {e^{i\pi \left[ {\dfrac{1}{2}\left( {1 + \dfrac{1}{2} + \dfrac{1}{4} \ldots \infty } \right)} \right]}}$
=${e^{i\pi \left( {\dfrac{{\dfrac{1}{2}}}{{1 - \dfrac{1}{2}}}} \right)}}$
$ \Rightarrow $Where, $\dfrac{1}{2}\left( {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \infty } \right) = \left( {\dfrac{{\dfrac{1}{2}}}{{1 - \dfrac{1}{2}}}} \right)$
\[ = {e^{i\pi \left( {\dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{2}}}} \right)}}\]
$ = {e^{i\pi }}$
Where, $\theta = \pi $
As a result, we know the relation for ${e^{i\theta }}$, we get to know that
$ = {e^{i\pi }} = \cos \pi + i\sin \pi $
Now, since we know the basic trigonometric relation for the different angles particularly.
For $\cos \pi = - 1$ and $\sin \pi = 0$,
Hence, we get
$ = - 1 + i\left( 0 \right)$
${x_1}{x_2}{x_3} \ldots {x_n} = - 1$
Therefore, the value of ${x_1}{x_2}{x_{_3}} \ldots {x_n}$ is $ - 1$. The correct option is (b).

Note:
One must know the euler representation of a complex number to solve the question. Euler’s formula can be stated that $\cos \theta + i\sin \theta = {e^{i\theta }}$, when $x = \pi $, we get the Euler’s identity., i.e. ${e^{i\pi }} = - 1$. As a result, by putting the given values in the question the required answer can be obtained. One must take care of the calculations to be sure of the final answer.