Answer
$sin~(x+y) = \frac{2-2\sqrt{30}}{15}$
$cos~(x-y) = \frac{\sqrt{5}-4\sqrt{6}}{15}$
$tan~(x+y) = \frac{2-2\sqrt{30}}{\sqrt{5}+4\sqrt{6}}$
$x+y~~$ is in quadrant IV.
Work Step by Step
$sin~y = -\frac{2}{3}$
Since $y$ is in quadrant III, $cos~y$ is also negative.
We can find $cos~y$:
$sin^2~y+cos^2~y = 1$
$cos~y = -\sqrt{1-sin^2~y}$
$cos~y = -\sqrt{1-(\frac{-2}{3})^2}$
$cos~y = -\sqrt{1-\frac{4}{9}}$
$cos~y = -\frac{\sqrt{5}}{3}$
$cos~x = -\frac{1}{5}$
Since $x$ is in quadrant II, $sin~x$ is positive.
We can find $sin~x$:
$sin^2~x+cos^2~x = 1$
$sin~x = \sqrt{1-cos^2~x}$
$sin~x = \sqrt{1-(\frac{-1}{5})^2}$
$sin~x = \sqrt{1-\frac{1}{25}}$
$sin~x = \frac{\sqrt{24}}{5}$
We can find $sin(x+y)$:
$sin~(x+y) = sin~x~cos~y+cos~x~sin~y$
$sin~(x+y) = (\frac{\sqrt{24}}{5})(-\frac{\sqrt{5}}{3})+(-\frac{1}{5})(-\frac{2}{3})$
$sin~(x+y) = -\frac{\sqrt{120}}{15}+\frac{2}{15}$
$sin~(x+y) = \frac{2-2\sqrt{30}}{15}$
We can find $cos(x-y)$:
$cos~(x-y) = cos~x~cos~(-y)-sin~x~sin~(-y)$
$cos~(x-y) = cos~x~cos~y-sin~x~(-sin~y)$
$cos~(x-y) = (-\frac{1}{5})(-\frac{\sqrt{5}}{3})-(\frac{\sqrt{24}}{5})(-\frac{-2}{3})$
$cos~(x-y) = \frac{\sqrt{5}}{15}-\frac{4\sqrt{6}}{15}$
$cos~(x-y) = \frac{\sqrt{5}-4\sqrt{6}}{15}$
We can find $cos(x+y)$:
$cos~(x+y) = cos~x~cos~y-sin~x~sin~y$
$cos~(x+y) = (-\frac{1}{5})(-\frac{\sqrt{5}}{3})-(\frac{\sqrt{24}}{5})(\frac{-2}{3})$
$cos~(x+y) = \frac{\sqrt{5}}{15}+\frac{4\sqrt{6}}{15}$
$cos~(x+y) = \frac{\sqrt{5}+4\sqrt{6}}{15}$
We can find $tan~(x+y)$:
$tan~(x+y) = \frac{sin~(x+y)}{cos~(x+y)}$
$tan~(x+y) = \frac{\frac{2-2\sqrt{30}}{15}}{\frac{\sqrt{5}+4\sqrt{6}}{15}}$
$tan~(x+y) = \frac{2-2\sqrt{30}}{\sqrt{5}+4\sqrt{6}}$
Since $sin~(x+y)$ and $tan~(x+y)$ are both negative, then $x+y$ must be in quadrant IV.
