If you have angle $\theta$ in quadrant $1$, you can find its "corresponding" angle in quadrant $2$ by $(\pi - \theta)$, in quadrant $3$ by $(\pi+\theta)$, and in quadrant $4$ by $(2\pi-\theta)$. For example, $\frac{\pi}{4}$ corresponds to $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, and $\frac{7\pi}{8}$ in quadrants $2$, $3$, and $4$, respectively. (That's how I always think of them at least.)
Also, recall sine functions correspond to the height of the right triangle ($y$-axis), so they are positive in quadrants $1$ and $2$. Cosine functions correspond to base of the right triangle ($x$-axis), so they are positive in quadrants $2$ and $4$. (Tangent functions can be found through sine and cosine functions.)
You can use the following identities (which are derived from the aforementioned facts).
$$\sin\bigg(\frac{\pi}{2}+\theta\bigg) = \cos\theta \quad \sin\bigg(\frac{\pi}{2}-\theta\bigg) = \cos\theta$$
$$\cos\bigg(\frac{\pi}{2}+\theta\bigg) = -\sin\theta \quad \cos\bigg(\frac{\pi}{2}-\theta\bigg) = \sin\theta$$
$$\tan\bigg(\frac{\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{\pi}{2}-\theta\bigg) = \cot\theta$$
$$\sin\bigg(\pi+\theta\bigg) = -\sin\theta \quad \sin\bigg(\pi-\theta\bigg) = \sin\theta$$
$$\cos\bigg(\pi+\theta\bigg) = -\cos\theta \quad \cos\bigg(\pi-\theta\bigg) = -\cos\theta$$
$$\tan\bigg(\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(\pi-\theta\bigg) = -\tan\theta$$
$$\sin\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cos\theta \quad \sin\bigg(\frac{3\pi}{2}-\theta\bigg) = -\cos\theta$$
$$\cos\bigg(\frac{3\pi}{2}+\theta\bigg) = \sin\theta \quad \cos\bigg(\frac{3\pi}{2}-\theta\bigg) = -\sin\theta$$
$$\tan\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{3\pi}{2}-\theta\bigg) = \cot\theta$$
$$\sin\bigg(2\pi+\theta\bigg) = \sin\theta \quad \sin\bigg(2\pi-\theta\bigg) = -\sin\theta$$
$$\cos\bigg(2\pi+\theta\bigg) = \cos\theta \quad \cos\bigg(2\pi-\theta\bigg) = \cos\theta$$
$$\tan\bigg(2\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(2\pi-\theta\bigg) = -\tan\theta$$
I certainly wouldn't recommend memorizing these though since knowing how the unit circle works basically means you know them already.
For example, in an equation you reach $$\cos \theta = -\frac{\sqrt{3}}{2}$$
You already know that $\cos {\frac{\pi}{6}} = \frac{\sqrt{3}}{2}$ and you also know cosine is negative in quadrants $2$ and $3$, so all you need to do is find the corresponding angle for ${\frac{\pi}{6}}$ in those quadrants.
$$\text{Quadrant II} \implies \theta = \pi-{\frac{\pi}{6}} = \frac{5\pi}{6}$$
$$\text{Quadrant III} \implies \theta = \pi+{\frac{\pi}{6}} = \frac{7\pi}{6}$$
This might take a bit of practice, but once you get this whole "corresponding" angle concept, it all becomes simple. Perhaps you can start by trying to visualize this by solving equations with a unit circle. You'll eventually get the hang of it.
Best Answer
Use sum to product formulas for sine and cosine functions stated as follows: $$\cos x+\cos y=2\cos\biggl(\dfrac{x+y}{2}\biggr)\sin \biggl(\dfrac{x-y}{2}\biggr) \tag1 $$ $$ \sin x+\sin y=2\sin \biggl(\dfrac{x+y}{2}\biggr)\cos\biggl(\dfrac{x-y}{2}\biggr) \tag2$$Notice that dividing $(2)$ by $(1)$ gets us rid of the expression of $\cos\bigl(\frac{x-y}{2}\bigr)$ and gives us the equation only in terms of $\tan\bigl(\frac{x+y}{2}\bigr)$.$$\dfrac{\sin x+\sin y}{\cos x+\cos y}=\frac{1}{2}=\tan\biggl(\dfrac{x+y}{2}\biggr)\implies \dfrac{1}{4}=\tan^2\biggl(\dfrac{x+y}{2}\biggr)$$Now using the half-angle formula for tangent gives us the expression in terms of $\cos(x+y)$.$$\dfrac{1}{4}=\dfrac{1-\cos(x+y)}{1+\cos(x+y)}\implies \dfrac{3}{5}=\cos(x+y)$$Now finding $\sin(x+y)$ is simply a matter of applying the Pythagorean identity. In fact doing so gives us $\sin(x+y)=4/5$.
Aliter: As suggested by @lab bhattacharjee, a quicker and better way would be to use the direct relation between $\sin 2\theta$ and $\tan\theta$. This makes sense because we're supposed to find $\sin (x+y)$ from $\tan\bigl(\frac{x+y}{2}\bigr)$ which we've extracted from the information provided. Also, in my opinion this is a better approach because it allows us to skip the step having to do with applying the Pythagorean identity. Also, I believe this is the thought process one should go through while attempting such questions because it minimizes the time taken to solve, which is key in exam situations and otherwise as well. $$\sin(x+y)=\dfrac{2\tan\bigl(\frac{x+y}{2}\bigr)}{1+\tan^2\bigl(\frac{x+y}{2}\bigr)}=\dfrac{1}{1+\frac{1}{4}}=\dfrac{4}{5}$$