Let $a$ and $b$ be real numbers such that $\sin{a} + \sin{b} = \frac{\sqrt{2}}{2}$ and $\cos {a} + \cos {b} = \frac{\sqrt{6}}{2}.$ Find $\sin{(a+b)}.$
$\text{(A) }\frac{1}{2} \qquad \text{(B) }\frac{\sqrt{2}}{2} \qquad \text{(C) }\frac{\sqrt{3}}{2} \qquad \text{(D) }\frac{\sqrt{6}}{2} \qquad \text{(E) }1$
Given $\begin{cases}\sin{a} + \sin{b} = \frac{\sqrt{2}}{2} \dots \textcircled{1}\\ \cos {a} + \cos {b} = \frac{\sqrt{6}}{2} \dots \textcircled{2}\end{cases}$
We multiply both sides of the syetem, $\textcircled{1} \times \textcircled{2}$, then  we get $(\sin{a}\cos{a}  + \sin{b} \cos{b}  )+( \sin{a}\cos{b}  + \sin{b} \cos{a} )= \frac{\sqrt{3}}{2}$. i.e. 
$(\sin{a}\cos{a}  + \sin{b} \cos{b}  )+\sin{(a+b)}= \frac{\sqrt{3}}{2}$.
We must get the sum of the first part of the equation, then we calculate $\textcircled{1}^2+\textcircled{2}^2$, we will 
get $\sin{a}\cos{a}  + \sin{b} \cos{b} = 0$ as $\sin^{2}{a}+\cos^{2}{a} = 1$ and $\sin^{2}{b}+\cos^{2}{b} = 1$. 
So $\sin{(a+b)} = \frac{\sqrt{3}}{2} \Longrightarrow \boxed{\mathrm{C}}$