The identity ∑k=0∞akcos⁡(kx)=1−acos⁡x1−2acos⁡x+a2,|a|<1 can be derived by using the fact that ∑k=0∞akcos⁡(kx)=Re∑k=0∞(aeix)k But can it...

Using the identity,
$\sum _{k=0}^{\mathrm{\infty }}{a}^k\cos (kx)=\frac{1-a\cos x}{1-2a\cos x+a^2},|a|<1$
the infinite series in question may be rewritten as a double infinite series over a triangle. Changing the order of summation (if you're like me and the transformation gymnastics with multiple indices makes you dizzy, here's a very handy cheat-sheet), we're left with fairly elementary summations:
$$\sum _{n=0}^{\mathrm{\infty }}{a}^n\cos (nx)=\sum _{n=0}^{\mathrm{\infty }}{a}^n\sum _{k=0}^{[\frac{n}{2}]}(-1{)}^k\left(\begin{array}{c}n\\ 2k\end{array}\right){\sin }^{2k}(x){\cos }^{n-2k}(x)$$
$$=\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{[\frac{n}{2}]}(-1{)}^k\left(\begin{array}{c}n\\ 2k\end{array}\right)a^n{\sin }^{2k}(x){\cos }^{n-2k}(x)$$
$$=\sum _{n=0}^{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k\left(\begin{array}{c}2k+n\\ 2k\end{array}\right)a^{2k+n}{\sin }^{2k}(x){\cos }^n(x)$$
$$=\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{a}^{2k}{\sin }^{2k}(x)\sum _{n=0}^{\mathrm{\infty }}\left(\begin{array}{c}2k+n\\ 2k\end{array}\right)[a\cos (x){]}^n$$
${\displaystyle =\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k{a}^{2k}{\sin }^{2k}\left(x\right)\frac{1}{{(1-a\cos \left(x\right))}^{2k+1}}}$
${\displaystyle =\frac{1}{1-a\cos \left(x\right)}\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k{\left[\frac{a\sin \left(x\right)}{1-a\cos \left(x\right)}\right]}^{2k}}$
${\displaystyle =\frac{1}{1-a\cos \left(x\right)}\cdot \frac{1}{1+{\left[\frac{a\sin \left(x\right)}{1-a\cos \left(x\right)}\right]}^2}}$
${\displaystyle =\frac{1}{1-a\cos \left(x\right)}\cdot \frac{{(1-a\cos \left(x\right))}^2}{{(1-a\cos \left(x\right))}^2+a^2{\sin }^2\left(x\right)}}$