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

$$\cos2\theta+\cos2\phi=2\cos(\theta+\phi)\cos(\theta-\phi),$$
$$a^2+b^2=2+2\cos(\theta-\phi)$$ and
$$b^2-a^2=\cos2\theta+\cos2\phi+2\cos(\theta+\phi).$$
Thus, $$\cos2\theta+\cos2\phi=2\cdot\frac{b^2-a^2-(\cos2\theta+\cos2\phi)}{2}\cdot\frac{a^2+b^2-2}{2},$$
which gives
$$\cos2\theta+\cos2\phi=\frac{(a^2+b^2-2)(b^2-a^2)}{a^2+b^2}.$$
