Argument of the function $2 \frac{\sin(x)}{x} + \frac{\sin(x/2)}{x/2} e^{-i x/2} $ 
Plot the argument (phase ) of the Complex function
  $$2 \frac{\sin(x)}{x} + \frac{\sin(x/2)}{x/2} e^{-i x/2}$$

This can be written as
$$\frac{1}{ix} (e^{ix} + 1 - 2 e^{-ix})$$
Wolfram shows the following answer
But I dont see how to write a expression the argument or something like that. I know how to plot the phase for both functions, but not for the sum. I know I should sum the complex and real parts, but I also cant see a good form using it.
Thanks in advance!
 A: $$
2 \frac{\sin(x)}{x} + \frac{\sin(x/2)}{x/2} e^{-i x/2}
$$
since we know 
$$
\sin x = \frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i}
$$
thus
$$
\frac{2}{x}\frac{\mathrm{e}^{ix}-\mathrm{e}^{-ix}}{2i} + \frac{2}{x}\frac{\mathrm{e}^{i\frac{x}{2}}-\mathrm{e}^{-i\frac{x}{2}}}{2i}e^{-i x/2}
$$
which is equal to 
$$
\frac{1}{ix}\left(\mathrm{e}^{ix}-\mathrm{e}^{-ix}\right)+ \frac{1}{ix}\left(\mathrm{e}^{i\frac{x}{2}}e^{-i x/2}-\mathrm{e}^{-i\frac{x}{2}}e^{-i x/2}\right)
$$
now
$$
\left(\mathrm{e}^{i\frac{x}{2}}e^{-i x/2}-\mathrm{e}^{-i\frac{x}{2}}e^{-i x/2}\right) = 1-\mathrm{e}^{-ix}
$$
now all together
$$
\frac{1}{ix}\left(\mathrm{e}^{ix}-\mathrm{e}^{-ix} + \left(1-\mathrm{e}^{-ix}\right)\right) = \frac{1}{ix}\left(\mathrm{e}^{ix} +1 -2\mathrm{e}^{-ix}\right)
$$
thus we can write it using
$$
\mathrm{e}^{\pm ix} = \cos x \pm i \sin x
$$
thus we have
$$
\begin{align}
\frac{1}{ix}\left(\cos x + i \sin x + 1 -2\cos x + 2i\sin x\right) &=& \frac{1}{x}\left(3\sin x\right) + \frac{1}{ix}\left(1-\cos x\right) \\
&=&\frac{3\sin x}{x} -i\frac{1-\cos x}{x}
\end{align}
$$
so you can see the real and imaginary part.
ps This can be done simply from the first equation, but i thought you did not know how to derive the second expression.
