Question

If ${X_n} = \cos \dfrac{\pi }{{{3^n}}} + i\sin \dfrac{\pi }{{{3^n}}}$, then ${X_1} \times {X_2} \times {X_3} \times ...$ equal to
A) 1
B) -1
C) i
D) -i

Answer 1

Hint: This question of a complex number in trigonometric form. In this question, ${X_n} = \cos \dfrac{\pi }{{{3^n}}} + i\sin \dfrac{\pi }{{{3^n}}}$ is given. And we want to find out the value of ${X_1} \times {X_2} \times {X_3} \times ...$. To find this value, let us substitute the value of n is 1 in the equation ${X_n}$, the value of n is 2 in the equation ${X_n}$, and so on. Now, substitute the above values in ${X_1} \times {X_2} \times {X_3} \times ...$.

Complete step-by-step solution:
In this question, given that
${X_n} = \cos \dfrac{\pi }{{{3^n}}} + i\sin \dfrac{\pi }{{{3^n}}}$.................(1)
Let us substitute the value of n is equal to 1 in the equation (1).
${X_n} = \cos \dfrac{\pi }{{{3^n}}} + i\sin \dfrac{\pi }{{{3^n}}}$
Here, $n = 1$.
$ \Rightarrow {X_1} = \cos \dfrac{\pi }{{{3^1}}} + i\sin \dfrac{\pi }{{{3^1}}}$
Substitute the value of ${3^1} = 3$.
 $ \Rightarrow {X_1} = \cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}$
Now, take$n = 2$.
$ \Rightarrow {X_2} = \cos \dfrac{\pi }{{{3^2}}} + i\sin \dfrac{\pi }{{{3^2}}}$
Now, take$n = 3$.
$ \Rightarrow {X_3} = \cos \dfrac{\pi }{{{3^3}}} + i\sin \dfrac{\pi }{{{3^3}}}$
Substitute the values of ${X_1},{X_2},{X_3}$in ${X_1} \times {X_2} \times {X_3} \times ...$
$ \Rightarrow {X_1} \times {X_2} \times {X_3} \times ...$
That is equal to
$ \Rightarrow \left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right) \times \left( {\cos \dfrac{\pi }{{{3^2}}} + i\sin \dfrac{\pi }{{{3^2}}}} \right) \times \left( {\cos \dfrac{\pi }{{{3^3}}} + i\sin \dfrac{\pi }{{{3^3}}}} \right) \times ...$
Let us write the real part and the imaginary part together.
$ \Rightarrow \cos \left( {\dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + \dfrac{\pi }{{{3^3}}} + ... + \infty } \right) + i\sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + \dfrac{\pi }{{{3^3}}} + ... + \infty } \right)$ …………….....(2)
Now, $\dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + \dfrac{\pi }{{{3^3}}} + ... + \infty = \dfrac{{\dfrac{\pi }{3}}}{{1 - \dfrac{1}{3}}}$
Let us take the least common factor to the denominator.
$ \Rightarrow \dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + \dfrac{\pi }{{{3^3}}} + ... + \infty = \dfrac{{\dfrac{\pi }{3}}}{{\dfrac{{3 - 1}}{3}}}$
So,
 $ \Rightarrow \dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + \dfrac{\pi }{{{3^3}}} + ... + \infty = \dfrac{{\dfrac{\pi }{3}}}{{\dfrac{2}{3}}}$
That is equal to
$ \Rightarrow \dfrac{\pi }{3} + \dfrac{\pi }{{{3^2}}} + \dfrac{\pi }{{{3^3}}} + ... + \infty = \dfrac{\pi }{2}$
Substitute the above value in equation (2).
$ \Rightarrow \cos \left( {\dfrac{\pi }{2}} \right) + i\sin \left( {\dfrac{\pi }{2}} \right)$
The value of $\cos \dfrac{\pi }{2} = 0$ and $\sin \dfrac{\pi }{2} = 1$.
So,
$ \Rightarrow 0 + i(1)$
That is equal to $i$
Hence, our answer is option (C), ‘$i$’.

Note: Here, we must remember the trigonometry ratios and the value of ratio at the angles 0, 30, 45, 60, and 90. And also know about trigonometry function in quadrant coordinate.
The value of $\sin 0^\circ = 0$
The value of $\sin 30^\circ = \dfrac{1}{2}$
The value of $\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}$
The value of $\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$
The value of $\sin 90^\circ = 1$
The value of $\cos 0^\circ = 1$
The value of $\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}$
The value of $\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$
The value of $\cos 60^\circ = \dfrac{1}{2}$
The value of $\cos 90^\circ = 0$
The value of $\tan 0^\circ = 0$
The value of $\tan 30^\circ = \dfrac{1}{{\sqrt 3 }}$
The value of $\tan 45^\circ = 1$
The value of $\tan 60^\circ = \sqrt 3 $
The value of $\tan 90^\circ $ is not defined.