Question

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

Answer 1

Hint: Here we have to find the value of a given complex exponential function. Firstly as we know that the complex exponential function can be written in the form of sine and cosine using Euler’s Formula which states that for any real number $x$ the value ${{e}^{ix}}=\cos x+i\sin x$ . So we will replace this value in the fraction given and simplify it to get our desired answer.

Complete step by step answer:
We have to find the value of below fraction:
$\dfrac{{{e}^{ix}}+{{e}^{ix}}}{2i}$….$\left( 1 \right)$
By Euler’s formula we know that for any real number $x$ the value of ${{e}^{ix}}$is given as follows:
${{e}^{ix}}=\cos x+i\sin x$
On substituting the above value in equation (1) we get,
$\Rightarrow \dfrac{\left( \cos x+i\sin x \right)+\left( \cos x+i\sin x \right)}{2i}$
$\Rightarrow \dfrac{\cos x+i\sin x+\cos x+i\sin x}{2i}$
On simplifying further we get,
$\Rightarrow \dfrac{2\cos x+2i\sin x}{2i}$
Taking $2$ common from the numerator values we get,
$\Rightarrow \dfrac{2\left( \cos x+i\sin x \right)}{2i}$
$\Rightarrow \dfrac{\cos x+i\sin x}{i}$
Rationalize the denominator by multiplying and dividing by $i$ in above value,
$\Rightarrow \dfrac{\cos x+i\sin x}{i}\times \dfrac{i}{i}$
$\Rightarrow \dfrac{i\cos x+{{i}^{2}}\sin x}{{{i}^{2}}}$
We know ${{i}^{2}}=-1$ use it above,
$\Rightarrow \dfrac{i\cos x-\sin x}{-1}$
$\Rightarrow \sin x-i\cos x$
Hence we got the answer as $\dfrac{{{e}^{ix}}+{{e}^{ix}}}{2i}=\sin x-i\cos x$ .

Note:
Numbers that are expressed in the form $a+ib$ are known as complex numbers where $i=$ Imaginary number and $a,b$ are the real numbers. We should know that the value of the square root of a negative term can’t be determined in a real number line so the concept of complex number came into existence and we know that ${{i}^{2}}=-1$ is most important for simplifying complex problems.