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$
Solution
Squaring and adding: $(\sin a+\sin b)^2+(\cos a+\cos b)^2=2+2\cos(a-b)=\frac12+\frac64=2$, so $\cos(a-b)=0$.
By sum-to-product, $\sin a+\sin b=2\sin\frac{a+b}2\cos\frac{a-b}2$ and $\cos a+\cos b=2\cos\frac{a+b}2\cos\frac{a-b}2$. Both sums are nonzero, so dividing gives $t=\tan\frac{a+b}2=\dfrac{\sqrt2/2}{\sqrt6/2}=\dfrac1{\sqrt3}$.
This determines $\frac{a+b}2$ only up to multiples of $180^\circ$, but $\sin(a+b)=\dfrac{2t}{1+t^2}$ depends only on $t$: $\sin(a+b)=\dfrac{2/\sqrt3}{4/3}=\boxed{\dfrac{\sqrt3}2}$, answer (C).
