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Question: The value of \(\sin i = \) \( \left( a \right)\dfrac{{{e^2} - 1}}{2} \\\ \left( b \right)...

The value of sin⁡i=\sin i =
(a)e2−12 (b)e2−12e (c)i(e2−12e) (d)i(e2−12)  \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) \\\

Explanation

Solution

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, eiθ=cos⁡θ+isin⁡θ{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\sin i.
Euler's form, eiθ=cos⁡θ+isin⁡θ...........(1){e^{i\theta }} = \cos \theta + i\sin \theta ...........\left( 1 \right)
Now, we substitute −θ in place of θ- \theta {\text{ in place of }}\theta in (1) equation.
ei(−θ)=cos⁡(−θ)+isin⁡(−θ){e^{i\left( { - \theta } \right)}} = \cos \left( { - \theta } \right) + i\sin \left( { - \theta } \right)
We know according to trigonometric function, cos⁡(−θ)=cos⁡θ and sin⁡(−θ)=−sin⁡θ\cos \left( { - \theta } \right) = \cos \theta {\text{ and }}\sin \left( { - \theta } \right) = - \sin \theta
e−iθ=cos⁡θ−isin⁡θ...........(2){e^{ - i\theta }} = \cos \theta - i\sin \theta ...........\left( 2 \right)
Now, subtract (2) equation from (1) equation.
eiθ−e−iθ=(cos⁡θ+isin⁡θ)−(cos⁡θ−isin⁡θ) ⇒eiθ−e−iθ=2isin⁡θ ⇒isin⁡θ=eiθ−e−iθ2  {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 ii on both sides of the equation.
⇒i2sin⁡θ=i(eiθ−e−iθ2)\Rightarrow {i^2}\sin \theta = i\left( {\dfrac{{{e^{i\theta }} - {e^{ - i\theta }}}}{2}} \right)
As we know, i2=−1{i^2} = - 1

⇒−sin⁡θ=i(eiθ−e−iθ2) ⇒sin⁡θ=i(e−iθ−eiθ2) ⇒sin⁡θ=i(1eiθ−eiθ2) ⇒sin⁡θ=i(1−ei2θ2eiθ)  \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\sin i so put the value of θ=i\theta = i .
⇒sin⁡i=i(1−e2i22ei2)\Rightarrow \sin i = i\left( {\dfrac{{1 - {e^{2{i^2}}}}}{{2{e^{{i^2}}}}}} \right)
As we know, i2=−1{i^2} = - 1

⇒sin⁡i=i(1−e−22e−1) ⇒sin⁡i=i(1−1e22e) ⇒sin⁡i=i(e2−12e)  \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 eiθ=cos⁡θ+isin⁡θ{e^{i\theta }} = \cos \theta + i\sin \theta and find the value of its conjugate e−iθ=cos⁡θ−isin⁡θ{e^{ - i\theta }} = \cos \theta - i\sin \theta then if we find the value of sin⁡i\sin i so we subtract both equations and put the value of θ=i\theta = i .