Question

If \[{x_r} = \cos (\dfrac{\pi }{{{3^r}}}) - i\sin (\dfrac{\pi }{{{3^r}}})\],(where $i = \sqrt { - 1} $) then the value of ${x_1}.{x_2}.{x_3}...........\infty $ is
A.1
B.-1
C.-$i$
D.$i$

Answer 1

Hint: To solve this question first of all, apply Euler’s formula given as
$ \Rightarrow {e^{ - i\theta }} = \cos \theta - i\sin \theta $
After solving the equation, you will get the value of ${x_1}.{x_2}.{x_3}...........\infty $ i.e.
$ \Rightarrow {x_1}.{x_2}.{x_3}...........\infty = {e^{i(\dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + .......\infty )}}$
Now apply the formula for infinite G.P as you can see $\dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + .......\infty $series are in G.P and the formula for sum of G.P is given by
$ \Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}$, where a is first term and r is common multiple.

Complete step-by-step answer:
In this question, an equation is given i.e.
\[ \Rightarrow {x_r} = \cos (\dfrac{\pi }{{{3^r}}}) - i\sin (\dfrac{\pi }{{{3^r}}})\] ……..(1)
Let’s start with solving this equation by equating it with Euler’s formula we get,
$ \Rightarrow {e^{ - i\theta }} = \cos \theta - i\sin \theta $
$ \Rightarrow {e^{ - i(\dfrac{\pi }{{{3^r}}})}} = \cos (\dfrac{\pi }{{{3^r}}}) - i\sin (\dfrac{\pi }{{{3^r}}})$ ……..(2)
From (1) and (2)
 $ \Rightarrow {x_r} = {e^{ - i(\dfrac{\pi }{{{3^r}}})}}$ …….(3)
To find the value of ${x_1}.{x_2}.{x_3}...........\infty $, we will find ${x_1},{x_2},{x_3}...........\infty $by putting value of r= 1,2,3…… respectively in equation 3 we get,
 \[\]$ \Rightarrow {x_1} = {e^{ - i(\dfrac{\pi }{3})}},{x_2} = {e^{ - i(\dfrac{\pi }{{{3^2}}})}},{x_3} = {e^{ - i(\dfrac{\pi }{{{3^3}}})}}..........\infty $
 Therefore, the value of ${x_1}.{x_2}.{x_3}...........\infty $i.e.
 $
   \Rightarrow {x_1}.{x_2}.{x_3}...........\infty = {e^{ - i(\dfrac{\pi }{3})}}.{e^{ - i(\dfrac{\pi }{{{3^2}}})}}.{e^{ - i(\dfrac{\pi }{{{3^3}}})}}..........\infty \\
   \Rightarrow {x_1}.{x_2}.{x_3}...........\infty = {e^{ - i(\dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + \dfrac{\pi }{{{3^3}}} + ........\infty )}} \\
$
                                        …….(4)

As $\dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + \dfrac{\pi }{{{3^3}}} + ........\infty $makes an infinite G.P with a=$\dfrac{\pi }{3}$and r=$\dfrac{1}{3}$. So, apply the formula for sum of infinite G.P i.e.
$ \Rightarrow {S_\infty } = \dfrac{a}{{1 - r}}$
$
   \Rightarrow {S_\infty } = \dfrac{{\dfrac{\pi }{3}}}{{1 - \dfrac{1}{3}}} \\
   \Rightarrow {S_\infty } = \dfrac{\pi }{{3 - 1}} = \dfrac{\pi }{2} \\
$
 So, $\dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + \dfrac{\pi }{{{3^3}}} + ........\infty = \dfrac{\pi }{2}$, put this value in equation 4 we get,
 $ \Rightarrow {x_1}.{x_2}.{x_3}...........\infty = {e^{ - i(\dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + \dfrac{\pi }{{{3^3}}} + ........\infty )}} = {e^{ - i(\dfrac{\pi }{2})}}$
$ \Rightarrow {x_1}.{x_2}.{x_3}...........\infty = \cos \dfrac{\pi }{2} - i\sin \dfrac{\pi }{2} = - i$

Hence, the correct option is C.

Note: We can also do this question by directly putting the value of r but it gets complex by using Euler’s formula, it gets easier to solve. Common mistakes done by students while applying the formula is, they directly apply the formula by mistake. They can think of it as an A.P series but there is a multiple which is common not the difference. The Euler’s formula for $\cos \theta + i\sin \theta $ is
$ \Rightarrow {e^{i\theta }} = \cos \theta + i\sin \theta $