Find $\cos2\theta+\cos2\phi$, given $\sin\theta + \sin\phi = a$ and $\cos\theta+\cos\phi = b$ 
If 
  $$\sin\theta + \sin\phi = a \quad\text{and}\quad \cos\theta+\cos\phi = b$$
then find the value of   $$\cos2\theta+\cos2\phi$$

My attempt:
Squaring both sides of the second given equation:
$$\cos^2\theta+ \cos^2\phi + 2\cos\theta\cos\phi= b^2$$
Multiplying by 2 and subtracting 2 from both sides we obtain, 
$$\cos2\theta+ \cos2\phi = 2b^2-2 - 4\cos\theta\cos\phi$$
How do I continue from here? 
PS: I also found the value of $\sin(\theta+\phi)= \dfrac{2ab}{a^2+b^2}$
Edit: I had also tried to use $\cos2\theta + \cos2\phi= \cos(\theta+\phi)\cos(\theta-\phi)$ but that didn't seem to be of much use

Let $s= \cos^2(\theta)+\cos^2(\phi)$, we want $2(s-1)$. Square the first equation and use Pythagorus
\begin{eqnarray*}
a^2+s-2 =2 \sin( \theta) \sin (\phi) \\
(a^2+s-2)^2 = 4(1- \cos^2( \theta))(1- \cos^2 (\phi)) \\
(a^2-2)^2-4+2a^2 s+s^2 =4 \cos^2( \theta) \cos^2 (\phi)
\end{eqnarray*}
From the second equation
\begin{eqnarray*}
(b^2-s)^2=4 \cos^2( \theta) \cos^2 (\phi)
\end{eqnarray*}
Put these together & we have
\begin{eqnarray*}
s= \frac{b^4+4a^2-a^4}{2(a^2+b^2)} \\
\cos(2 \theta)+ \cos( 2\phi)=(b^2-a^2) \left(1- \frac{2}{a^2+b^2} \right).
\end{eqnarray*}
