Simplifying the expressions for the magnitude and phase of a Fourier transform $$h[n] = 2( \delta[n-2]-\delta[n-1]-\delta[n-3])$$
i computed my frequency response and i have this now: $$H[e^{j \omega}] = 2[ e^{-2 j \omega} - e^{-j \omega}-e^{-3 j \omega}]$$
$$H[e^{j \omega}] = 2[ \cos(2\omega) -i \sin(2\omega)-\cos(\omega)+i\sin(\omega)-\cos(3\omega)+i\sin(3\omega)]$$
And i need to compute magnitude and phase.
But how can i simplify that? there was a hint given to use trigonometric identities but i can't find some that suit my example.
 A: Expanding my comment above. From
$$H\left[ e^{-\mathrm{i}\omega }\right] =2\left[ \mathrm{e}^{-\mathrm{i}\left( 2\omega
\right) }-\mathrm{e}^{-\mathrm{i}\omega }-\mathrm{e}^{-\mathrm{i}\left(
3\omega \right) }\right] =2\mathrm{e}^{-\mathrm{i}\left( 2\omega \right) }%
\left[ 1-\mathrm{e}^{\mathrm{i}\omega }-\mathrm{e}^{-\mathrm{i}\omega }%
\right]$$
we get the following 
$$\begin{eqnarray*}
\left\vert H\left[ e^{-\mathrm{i}\omega }\right] \right\vert  &=&\left\vert 2\mathrm{e%
}^{-\mathrm{i}\left( 2\omega \right) }\left[ 1-\mathrm{e}^{\mathrm{i}\omega
}-\mathrm{e}^{-\mathrm{i}\omega }\right] \right\vert  \\
&=&\left\vert 2\right\vert \left\vert \mathrm{e}^{-\mathrm{i}\left( 2\omega
\right) }\right\vert \left\vert 1-\mathrm{e}^{\mathrm{i}\omega }-\mathrm{e}%
^{-\mathrm{i}\omega }\right\vert  \\
&=&2\cdot 1\cdot \left\vert 1-\cos \omega -\mathrm{i}\sin \omega -\cos
\omega +\mathrm{i}\sin \omega \right\vert  \\
&=&2\left\vert 1-2\cos \omega \right\vert  \\
\end{eqnarray*}$$
A: Expanding upon Américo and samanwita's answers:
$$H[e^{-j\omega }] =2e^{-j 2\omega}\left(1-2\cos(\omega)\right) \; ,$$
from which Américo deduced the magnitude to be
$$|H[e^{-j\omega }]| =2\left|1-2\cos(\omega)\right| \; .$$
The phase is then simply
$$e^{j\phi[\omega]}=e^{-j 2\omega}\frac{\left(1-2\cos(\omega)\right)}{\left|1-2\cos(\omega)\right| }$$
or
$$\phi[\omega] = \begin{cases} -2\omega \; , \text{ if } \omega \in \left]\frac{\pi}{3}+2k\pi,\frac{5\pi}{3}+2k\pi\right[\\ 
-2\omega + \pi \; , \text{ if } \omega \in \left]-\frac{\pi}{3}+2k\pi,\frac{\pi}{3}+2k\pi\right[\end{cases} \text{ for } k \in \mathbb{Z} \; .$$
A: It's a bit confusing here so to sum it up:
The frequency response is 
$ H[e^{j \omega}] = 2[ \cos(2\omega) -i \sin(2\omega)-\cos(\omega)+i\sin(\omega)-\cos(3\omega)+i\sin(3\omega)] $
$$ H[e^{j \omega}] = 2[1-2*cos(\omega)] $$ 
To get magnitude and phase i need real and imaginary parts:
$$ H_R[e^{j \omega}] = 2[ \cos(2\omega) -\cos(\omega)-\cos(3\omega)] $$
$$  H_I[e^{j \omega}] = 2[ -\sin(2\omega)+\sin(\omega)+\sin(3\omega)]  $$
$$ |H[e^{j \omega}]| = (H_R[e^{j \omega}]^2 + H_I[e^{j \omega}]^2)^{\frac{1}{2}} $$
And that is $$|H[e^{j \omega}]| = \sqrt{(2 \cos \omega - 1)^2}$$
PLOTS
