Decimating a discrete time cosine function Let $x[n] = \cos(wn)\ \ $ be our signal. Let the other signal be $x_2[n] = \cos(2wn)\ \ $ . The spectrum of $x[n]$ and $x_2[n]$ are well defined and consist of impulses at the relevant frequencies with weight $\pi$. However, if we interpret $x_2[n]$ as the decimated version of $x[n]$, the spectrum of $x_2[n]$ must have impulses with weight $\frac{\pi}{2}$. I have the explanation but I would like to hear the alternatives, what is the best way to resolve  these seemingly contradicting facts?
The following correction of Jason R's answer completely resolves the problem: The frequency spectrum of $X_2(ω)$  consists of two components, which are scaled versions of $X(ω)$. Those are $4\pi$-periodic instead of being $2\pi$-periodic. When those $4\pi$-periodic versions are added up, the sum becomes $2π$-periodic again. THERE IS NO ALIASING or IMPULSES SUPERIMPOSED ON EACH OTHER. The derivation from
$$
X_2(\omega) = \frac{1}{2} \pi\left(\delta\left(\frac{\omega}{2}-\omega_0\right) + \delta\left(\frac{\omega}{2}+\omega_0\right) + \delta\left(\frac{\omega}{2}-\omega_0-\pi\right) + \delta\left(\frac{\omega}{2}+\omega_0-\pi\right)\right)
$$
to
$$
X_2(\omega) = \frac{1}{2} \pi\left(\delta\left(\omega-2\omega_0\right) + \delta\left(\omega+2\omega_0\right) + \delta\left(\omega-2\omega_0-2\pi\right) + \delta\left(\omega+2\omega_0-2\pi\right)\right)
$$
is not correct. Instead it must read 
$$
X_2(ω)=π(δ(ω−2ω_0 )+δ(ω+2ω_0)+δ(ω−2ω_0−2π)+δ(ω+2ω_0−2π)) \qquad (∗) 
$$ 
This is due to the scaling property of impulses 
$$
δ(aω)=\frac{δ(ω)}{|a|}   
$$ 
Then, observing that (*) is $2π$-periodic, two of the impulses are redundant and hence the DTFT of $x_2[n]=x[2n]$  becomes
$$
X_2(ω)=π(δ(ω−2ω_0)+δ(ω+2ω_0)) 
$$