Question

The value of $\left( {\cos \dfrac{{\pi}}{2} + i\sin \dfrac{{\pi}}{2}} \right)\left( {\cos \dfrac{{\pi}}{4} + i\sin \dfrac{{\pi}}{4}} \right)\left( {\cos \dfrac{{\pi}}{8} + i\sin \dfrac{{\pi}}{8}} \right).....\infty$ is
A. $1$
B. $0$
C. $ - 1$
D. None of these

Answer 1

Hint: First, we shall analyze the given information so that we are able to solve this problem. Here, we are given $\left( {\cos \dfrac{{\pi}}{2} + i\sin \dfrac{{\pi}}{2}} \right)\left( {\cos \dfrac{{\pi}}{4} + i\sin \dfrac{{\pi}}{4}} \right)\left( {\cos \dfrac{{\pi}}{8} + i\sin \dfrac{{\pi}}{8}} \right).....\infty$ and we are asked to calculate the value of $\left( {\cos \dfrac{{\pi}}{2} + i\sin \dfrac{{\pi}}{2}} \right)\left( {\cos \dfrac{{\pi}}{4} + i\sin \dfrac{{\pi}}{4}} \right)\left( {\cos \dfrac{{\pi}}{8} + i\sin \dfrac{{\pi}}{8}} \right).....\infty$.
Since we all know ${e^{i\theta }} = \cos \theta + i\sin \theta $ , we need to substitute it in the given expression. We also need to apply the formula of the sum of the GP in this problem.
 Formula to be used:
The formula to calculate the sum of the geometric series is as follows.
$a + ax + a{x^2} + .... + \infty = \dfrac{a}{{1 - x}}$

Complete step by step answer:
The given expression is $\left( {\cos \dfrac{{\pi}}{2} + i\sin \dfrac{{\pi}}{2}} \right)\left( {\cos \dfrac{{\pi}}{4} + i\sin \dfrac{{\pi}}{4}} \right)\left( {\cos \dfrac{{\pi}}{8} + i\sin \dfrac{{\pi}}{8}} \right).....\infty$
Since we all know ${e^{i\theta }} = \cos \theta + i\sin \theta $ , we need to substitute it in the given expression.
We have $\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2} = {e^{i\dfrac{\pi }{2}}}$ ,$\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4} = {e^{i\dfrac{\pi }{4}}}$ ,$\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8} = {e^{i\dfrac{\pi }{8}}}$
$ \Rightarrow \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right).....\infty = {e^{i\dfrac{\pi }{2}}}.{e^{i\dfrac{\pi }{4}}}.{e^{i\dfrac{\pi }{8}}}....\infty $
$ \Rightarrow \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right).....\infty = {e^{i\left( {\dfrac{\pi }{2} + \dfrac{\pi }{4} + \dfrac{\pi }{8} + ... + \infty } \right)}}$
$ \Rightarrow \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right).....\infty = {e^{i\dfrac{\pi }{2}\left( {1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ... + \infty } \right)}}$…………..$\left( 1 \right)$
Here $\left( {1 + \dfrac{1}{2} + \dfrac{1}{4} + ... + \infty } \right)$ is an infinite geometric series.
The formula to calculate the sum of the geometric series is as follows.
$a + ax + a{x^2} + .... + \infty = \dfrac{a}{{1 - x}}$
Thus, $\left( {1 + \dfrac{1}{2} + \dfrac{1}{4} + ... + \infty } \right) = \dfrac{1}{{1 - \dfrac{1}{2}}}$
$ \Rightarrow \left( {1 + \dfrac{1}{2} + \dfrac{1}{4} + ... + \infty } \right) = 2$
Now, we shall substitute the above result in the equation $\left( 1 \right)$.
$ \Rightarrow \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right).....\infty = {e^{i\dfrac{\pi }{2} \times 2}}$
$ \Rightarrow \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right).....\infty = {e^{i\pi }}$
Since we all know ${e^{i\theta }} = \cos \theta + i\sin \theta $ , we need to substitute it in the above equation.
$ \Rightarrow \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right).....\infty = \cos \pi + i\sin \pi $
We know that $\cos \pi = - 1$ and $\sin \pi = 0$
Thus, we get
$ \Rightarrow \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right).....\infty = - 1 + i \times 0$
$ \Rightarrow \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right).....\infty = - 1$
Therefore, we calculated$\left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right).....\infty = - 1$

So, the correct answer is “Option C”.

Note: We know that ${e^{i\theta }} = \cos \theta + i\sin \theta $ . Generally, a complex number consists of a real part and an imaginary part. For example, let us consider a complex number $z = 5 + 7i$. In this example, the real part of $z$ is $5$ and the imaginary part of $z$ is $7$.
Similarly, ${e^{i\theta }} = \cos \theta + i\sin \theta $, the real part of ${e^{i\theta }}$ is $\cos \theta $ and the imaginary part of ${e^{i\theta }}$ is $\sin \theta $