4. Let $x$ and $y$ be real numbers such that $\sin x + \sin y = \frac{\sqrt{2}}{2}$ and $\cos x + \cos y = \frac{\sqrt{6}}{2}$. What is $\sin (x+y) ?$
III/4. If we square and add the equations, we get $\sin ^{2} x+2 \sin x \sin y+\sin ^{2} y+\cos ^{2} x+2 \cos x \cos y+\cos ^{2} y=\frac{1}{2}+\frac{3}{2}$ or $2(\sin x \sin y+\cos x \cos y)=0$ and from this $\cos (x-y)=0$. If we multiply the equations by each other, we get $\sin x \cos x+\sin x \cos y+\sin y \cos x+\sin y \cos y=\frac{\sqrt{3}}{2}$, which we can rewrite as $(\sin x \cos y+\sin y \cos x)+(\sin x \cos x+\sin y \cos y)=\frac{\sqrt{3}}{2}$ and further as $\sin (x+y)+\sin (x+y) \cos (x-y)=\frac{\sqrt{3}}{2}$. If we consider that $\cos (x-y)=0$, we have $\sin (x+y)=\frac{\sqrt{3}}{2}$, which is what we wanted to calculate.

Finding $\cos (x-y)=0$ 3 points

Finding $(\sin x \cos y+\sin y \cos x)+(\sin x \cos x+\sin y \cos y)=\frac{\sqrt{3}}{2} \ldots \ldots \ldots \ldots \ldots \ldots .2$ points

![](https://cdn.mathpix.com/cropped/2024_06_07_5a4348e8f3c4edef190ag-10.jpg?height=52&width=1694&top_left_y=451&top_left_x=181)

![](https://cdn.mathpix.com/cropped/2024_06_07_5a4348e8f3c4edef190ag-10.jpg?height=66&width=1690&top_left_y=498&top_left_x=186)