Question

What does $\dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2i}$ equal?

Answer 1

Hint: Assume the given expression as E. Now, consider $i$ as the imaginary number $\sqrt{-1}$ and use the Euler’s formula given as ${{e}^{i\theta }}=\left( \cos \theta +i\sin \theta \right)$ to find the value of ${{e}^{ix}}$. Further to find the value of ${{e}^{-ix}}$ substitute $-x$ in place of x in the expression obtained for ${{e}^{ix}}$. Take the difference of the two expressions and simplify the expression E by cancelling the common factors to get the answer.

Complete step by step solution:
Here we have been provided with the expression $\dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2i}$ and we are asked to find its value. Let us assume the given expression as E, so we have,
$\Rightarrow E=\dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2i}$
Now, in the above expression we can see that we have the letter $i$ in the denominator as well as in the exponents of the Euler’s number e. In mathematics, this letter is the notation of the imaginary number $\sqrt{-1}$ which is the root of the quadratic equation ${{x}^{2}}+1=0$. Using Euler’s formula of complex number given as ${{e}^{i\theta }}=\left( \cos \theta +i\sin \theta \right)$ we get,
$\Rightarrow {{e}^{ix}}=\left( \cos x+i\sin x \right)$
Substituting $-x$ in place of x in the above relation we get,
$\Rightarrow {{e}^{i\left( -x \right)}}=\left( \cos \left( -x \right)+i\sin \left( -x \right) \right)$
Using the properties of the sine and cosine function given as $\cos \left( -x \right)=\cos x$ and $\sin \left( -x \right)=-\sin x$ we get,
$\Rightarrow {{e}^{-ix}}=\left( \cos x-i\sin x \right)$
Substituting these values in the expression E we get,
$\begin{align}
  & \Rightarrow E=\dfrac{\left( \cos x+i\sin x \right)-\left( \cos x-i\sin x \right)}{2i} \\
 & \Rightarrow E=\dfrac{2i\sin x}{2i} \\
 & \therefore E=\sin x \\
\end{align}$
Hence, the above relation is our answer.

Note: Note that there is one more theorem in complex number given as ${{\left( \cos x+i\sin x \right)}^{n}}=\left( \cos nx+i\sin nx \right)={{e}^{inx}}$. This formula is known as Demoivre’s formula. There is a hyperbolic function given as $\sinh \left( x \right)=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$ so if you will substitute $ix$ in place of x in this formula then the relation will become $\sinh \left( ix \right)=\dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2}$ and the answer you will get is $\dfrac{\sinh \left( ix \right)}{i}$. However, the above answer we have got is more simplified so it will be preferred to use the above approach of the solution.