Conversion from exponential to cosine
Andrew Henderson 
I'm trying to understand the following expansion. The question was
Show that if $Y(t) = X(t+a) - X(t-a)$ and $X(t)$ is WSS, then $$S_Y(\omega) = 4S_X(\omega)sin^2a\omega$$
The solution is
\begin{equation} \begin{aligned} S_Y(\tau) &= 2S_X(\omega) - e^{j2\omega a}S_X(\omega) - e^{-j2\omega a}S_X(\omega) \\ &= [2 - e^{-j2\omega a} - e^{j2\omega a})S_X(\omega) \\ &= [2 - 2cos(2a\omega)]S_X(\omega) \\ &=4S_X(\omega)sin^2a\omega, \end{aligned} \end{equation}
I don't understand how the exponential expression is converted to the cosine expression. Is it because of any identity that I'm unaware of?
$\endgroup$2 Answers
$\begingroup$I assume you are using $j$ for the imaginary unit, which is more commonly written as $i$ by English-speaking mathematicians (although electronics engineers like to use $j$, because they use $i$ for current).
The identity you seem to be unaware of is known as Euler's formula:
$$e^{ix} = \cos x + i \sin x$$
FWIW, that formula is valid for complex $x$ as well as real $x$.
Substitution of $-x$ gives:
$$\begin{align} e^{-ix} & = \cos (-x) + i \sin (-x)\\ & = \cos x - i \sin x\\ \end{align}$$
Thus,
$$e^{ix} + e^{-ix} = 2\cos x$$
and $$e^{ix} - e^{-ix} = 2i\sin x$$
$\endgroup$ $\begingroup$It relies on the fact that (using your notation with "$j$" for what is usually written as "$i$")$$e^{jt}=\cos t + j\sin t$$ From this it follows that $$e^{-jt}=\cos t - j\sin t$$ so that $$\cos t = \frac12(e^{jt}+e^{-jt})$$
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