Question

The value of $\sin i = $
$
  \left( a \right)\dfrac{{{e^2} - 1}}{2} \\
  \left( b \right)\dfrac{{{e^2} - 1}}{{2e}} \\
  \left( c \right)i\left( {\dfrac{{{e^2} - 1}}{{2e}}} \right) \\
  \left( d \right)i\left( {\dfrac{{{e^2} - 1}}{2}} \right) \\
 $

Answer 1

Hint-In this question, we use the concept of Euler's form. Euler is a mathematical formula in complex analysis that establishes the fundamental relationship between the trigonometric functions and the complex exponential function. Euler's form, ${e^{i\theta }} = \cos \theta + i\sin \theta $.

Complete step-by-step solution -
Now, we use Euler's form to find the value of $\sin i$.
Euler's form, ${e^{i\theta }} = \cos \theta + i\sin \theta ...........\left( 1 \right)$
Now, we substitute $ - \theta {\text{ in place of }}\theta $ in (1) equation.
${e^{i\left( { - \theta } \right)}} = \cos \left( { - \theta } \right) + i\sin \left( { - \theta } \right)$
We know according to trigonometric function, $\cos \left( { - \theta } \right) = \cos \theta {\text{ and }}\sin \left( { - \theta } \right) = - \sin \theta $
${e^{ - i\theta }} = \cos \theta - i\sin \theta ...........\left( 2 \right)$
Now, subtract (2) equation from (1) equation.
$
  {e^{i\theta }} - {e^{ - i\theta }} = \left( {\cos \theta + i\sin \theta } \right) - \left( {\cos \theta - i\sin \theta } \right) \\
   \Rightarrow {e^{i\theta }} - {e^{ - i\theta }} = 2i\sin \theta \\
   \Rightarrow i\sin \theta = \dfrac{{{e^{i\theta }} - {e^{ - i\theta }}}}{2} \\
$
Now, Multiply by $i$ on both sides of the equation.
\[ \Rightarrow {i^2}\sin \theta = i\left( {\dfrac{{{e^{i\theta }} - {e^{ - i\theta }}}}{2}} \right)\]
As we know, ${i^2} = - 1$
\[
   \Rightarrow - \sin \theta = i\left( {\dfrac{{{e^{i\theta }} - {e^{ - i\theta }}}}{2}} \right) \\
   \Rightarrow \sin \theta = i\left( {\dfrac{{{e^{ - i\theta }} - {e^{i\theta }}}}{2}} \right) \\
   \Rightarrow \sin \theta = i\left( {\dfrac{{\dfrac{1}{{{e^{i\theta }}}} - {e^{i\theta }}}}{2}} \right) \\
   \Rightarrow \sin \theta = i\left( {\dfrac{{1 - {e^{i2\theta }}}}{{2{e^{i\theta }}}}} \right) \\
 \]
Now, we have to find $\sin i$ so put the value of $\theta = i$ .
\[ \Rightarrow \sin i = i\left( {\dfrac{{1 - {e^{2{i^2}}}}}{{2{e^{{i^2}}}}}} \right)\]
As we know, ${i^2} = - 1$
\[
   \Rightarrow \sin i = i\left( {\dfrac{{1 - {e^{ - 2}}}}{{2{e^{ - 1}}}}} \right) \\
   \Rightarrow \sin i = i\left( {\dfrac{{1 - \dfrac{1}{{{e^2}}}}}{{\dfrac{2}{e}}}} \right) \\
   \Rightarrow \sin i = i\left( {\dfrac{{{e^2} - 1}}{{2e}}} \right) \\
 \]
So, the correct option is (c).

Note-In such types of problems we use some important points to solve questions in an easy way. Like first we use the value of Euler’s form ${e^{i\theta }} = \cos \theta + i\sin \theta $ and find the value of its conjugate ${e^{ - i\theta }} = \cos \theta - i\sin \theta $ then if we find the value of $\sin i$ so we subtract both equations and put the value of $\theta = i$ .