Find $\cos(x+y)$ and $\sin(x+y)$ given that $\cos x + \cos y = a$ and $\sin x + \sin y = b$ If $\cos x + \cos y = a$ and $\sin x + \sin y = b$.
Find $\cos(x+y)$  and $\sin(x+y)$. 
I only need some hints to start as I am not able to get any way to go forward to.
 A: HINT:
Let $u = \cos x + i \sin x$, $v = \cos y + i \sin y$. Then 
\begin{eqnarray}
u + v = a+ i b \\
\frac{1}{u} + \frac{1}{v} = a -i b
\end{eqnarray}
so 
$$u v = (a+ib)/(a-ib)$$
But $uv = \cos(x+y) + i \sin(x+y)$. Now take real and imaginary parts.
A: Here is a geometric way to find $\{(\cos x, \sin x), (\cos y, \sin y)\}$ and $(\cos(x+y), \sin(x+y))$.
The vector $(a,b)$ is a sum of two  vectors $(\cos x, \sin x)$, $(\cos y, \sin y)$ starting at the origin and ending on the unit circle therefore the points $(0,0)$, $(\cos x, \sin x)$,  $(\cos y, \sin y)$, $(a,b)$ are the vertices of a rhombus. Take the segment bisector of the diagonal $(0,0) (a,b)$ and find its intersections with the unit circle $(\cos x, \sin x)$ and $(\cos y, \sin y)$. To find the point $(\cos(x+y), \sin(x+y))$ draw the chord  through from $(1,0)$ perpendicular to the line $(0,0) (a,b)$ and take its other end. 
A neat picture made by @Semiclassical $\ $--$\ $thanks!! $\ \ \ $  (using Chrome's Geogebra ) 
Reflect $(1,0)$ with respect to $(0,0) (a,b)$

blue: $(0,0)$, $(1,0)$, $(a,b)$
A: Using Prosthaphaeresis Formulas
$$2\sin\frac{x+y}2\cos\frac{x-y}2=a$$
and $$2\cos\frac{x+y}2\cos\frac{x-y}2=b$$
Divide to find $\tan\dfrac{x+y}2$ assuming $ab\cos\dfrac{x-y}2\ne0$ 
Now apply Weierstrass substitution

Alternatively find $a^2+b^2,a^2-b^2,ab$ and use Prosthaphaeresis Formulas
