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: We are given an equation \[\left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty \]
Now we write it in terms of exponential, then equate it to some variable and find the product, then we solve the series using G.P. In simplification, we will get our result.

Formula used:
We have been using the following formula:
1. \[{e^{ix}} = \cos x + i\sin x\,\,\,\]
2. \[{S_n} = \dfrac{a}{{1 - r}}\]
3. \[\cos \pi = - 1\]
4. \[\sin \pi = 0\]
5. \[{a^m}{a^n}{a^p} = {a^{m + n + p}}\]

Complete step-by-step solution:
We are given that \[\left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty - - - (1)\]
Now we rewrite the given equation as:
\[
  \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty \\
  \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{{{{\left( 2 \right)}^2}}} + i\sin \dfrac{\pi }{{{{\left( 2 \right)}^2}}}} \right)\left( {\cos \dfrac{\pi }{{{{\left( 2 \right)}^3}}} + i\sin \dfrac{\pi }{{{{\left( 2 \right)}^3}}}} \right)...\infty - - - \left( 2 \right) \\ \]
Now we know the exponential formula \[{e^{ix}} = \cos x + i\sin x\,\,\,\]
Therefore, we can write equation (2) as:
 \[\left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty = {e^{i\,\dfrac{\pi }{2}}} \cdot {e^{\,i\,\dfrac{\pi }{{{{\left( 2 \right)}^2}}}}} \cdot {e^{\,i\,\dfrac{\pi }{{{{\left( 2 \right)}^3}}}}}...\infty \]
Now, we will simplify the exponent powers using \[{a^m}{a^n}{a^p} = {a^{m + n + p}}\]
\[{e^{i\,\dfrac{\pi }{2}}} \cdot {e^{\,i\,\dfrac{\pi }{{{{\left( 2 \right)}^2}}}}} \cdot {e^{\,i\,\dfrac{\pi }{{{{\left( 2 \right)}^3}}}}}...\infty = {e^{\left( {i\,\dfrac{\pi }{2} + i\,\dfrac{\pi }{{{{\left( 2 \right)}^2}}} + i\,\dfrac{\pi }{{{{\left( 2 \right)}^3}}}...\infty } \right)}}\]
Now we know that a given series is an infinite G.P series because each succeeding term is produced by multiplying each preceding term by a fixed number in which the first term is \[{e^{i\,\,\dfrac{\pi }{2}}}\] and the common difference is \[{e^{i\,\,\dfrac{\pi }{2}}}\]
Now, we apply the formula of the sum of series which is \[{S_n} = \dfrac{a}{{1 - r}}\]
\[
  \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty = {e^{\left( {i\,\dfrac{\pi }{2} + i\,\dfrac{\pi }{{{{\left( 2 \right)}^2}}} + i\,\dfrac{\pi }{{{{\left( 2 \right)}^3}}}...\infty } \right)}} \\
  \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty = {e^{i\,\dfrac{\pi }{2}\left( {\dfrac{1}{{1 - \dfrac{1}{2}}}} \right)}} \\
  \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty = {e^{i\,\dfrac{\pi }{2}\left( {\dfrac{1}{{\dfrac{{2 - 1}}{2}}}} \right)}} \\
  \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty = {e^{i\,\dfrac{\pi }{2}\left( {\dfrac{1}{{\dfrac{1}{2}}}} \right)}} \\
 \]
Further Solving,
\[\left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty = {e^{i\,\pi }}\]
Further solving we get,
\[
  \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty = \,\cos \pi + i\sin \pi \\
  \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty = - 1 + 0i \\
  \left( {\cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}} \right)\left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\left( {\cos \dfrac{\pi }{8} + i\sin \dfrac{\pi }{8}} \right)...\infty = - 1 \\
 \]
Therefore, 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 \[ - 1\].
Hence, option (C) is correct answer

Note: Students become confused when solving equations and wonder which equation to enter. Converting equations having exponential values to equations including sine and cosine values can be problematic.