Answer
$sin~(x+y) = \frac{\sqrt{231}-2}{18}$
$cos~(x-y) = \frac{\sqrt{77}-2\sqrt{3}}{18}$
$tan~(x+y) = \frac{2-\sqrt{231}}{\sqrt{77}+2\sqrt{3}}$
$x+y~~$ is in quadrant II.
Work Step by Step
$sin~y = -\frac{1}{2}$
Since $y$ is in quadrant III, $cos~y$ is also negative.
We can find $cos~y$:
$sin^2~y+cos^2~y = 1$
$cos~y = -\sqrt{1-sin^2~y}$
$cos~y = -\sqrt{1-(\frac{-1}{2})^2}$
$cos~y = -\sqrt{1-\frac{1}{4}}$
$cos~y = -\frac{\sqrt{3}}{2}$
$cos~x = \frac{2}{9}$
Since $x$ is in quadrant IV, $sin~x$ is negative.
We can find $sin~x$:
$sin^2~x+cos^2~x = 1$
$sin~x = -\sqrt{1-cos^2~x}$
$sin~x = -\sqrt{1-(\frac{2}{9})^2}$
$sin~x = -\sqrt{1-\frac{4}{81}}$
$sin~x = -\frac{\sqrt{77}}{9}$
We can find $sin(x+y)$:
$sin~(x+y) = sin~x~cos~y+cos~x~sin~y$
$sin~(x+y) = (-\frac{\sqrt{77}}{9})(-\frac{\sqrt{3}}{2})+(\frac{2}{9})(-\frac{1}{2})$
$sin~(x+y) = \frac{\sqrt{231}}{18}-\frac{1}{9}$
$sin~(x+y) = \frac{\sqrt{231}-2}{18}$
We can find $cos(x-y)$:
$cos~(x-y) = cos~x~cos~(-y)-sin~x~sin~(-y)$
$cos~(x-y) = cos~x~cos~y-sin~x~(-sin~y)$
$cos~(x-y) = (\frac{2}{9})(-\frac{\sqrt{3}}{2})-(-\frac{\sqrt{77}}{9})(-\frac{-1}{2})$
$cos~(x-y) = -\frac{\sqrt{3}}{9}+\frac{\sqrt{77}}{18}$
$cos~(x-y) = \frac{\sqrt{77}-2\sqrt{3}}{18}$
We can find $cos(x+y)$:
$cos~(x+y) = cos~x~cos~y-sin~x~sin~y$
$cos~(x+y) = (\frac{2}{9})(-\frac{\sqrt{3}}{2})-(-\frac{\sqrt{77}}{9})(-\frac{1}{2})$
$cos~(x+y) = -\frac{\sqrt{3}}{9}-\frac{\sqrt{77}}{18}$
$cos~(x+y) = -(\frac{\sqrt{77}+2\sqrt{3}}{18})$
We can find $tan~(x+y)$:
$tan~(x+y) = \frac{sin~(x+y)}{cos~(x+y)}$
$tan~(x+y) = \frac{\frac{\sqrt{231}-2}{18}}{-(\frac{\sqrt{77}+2\sqrt{3}}{18})}$
$tan~(x+y) = \frac{2-\sqrt{231}}{\sqrt{77}+2\sqrt{3}}$
Since $sin~(x+y)$ is positive and $tan~(x+y)$ is negative, then $x+y$ must be in quadrant II.