Answer
$sin~(x+y) = \frac{2+3\sqrt{7}}{10}$
$cos~(x-y) = \frac{2\sqrt{3}+\sqrt{21}}{10}$
$tan~(x+y) = \frac{2+3\sqrt{7}}{2\sqrt{3}-\sqrt{21}}$
$x+y~~$ is in quadrant II.
Work Step by Step
$sin~x = -\frac{1}{2}$
Since $x$ is in quadrant III, $cos~x$ is also negative.
We can find $cos~x$:
$sin^2~x+cos^2~x = 1$
$cos~x = -\sqrt{1-sin^2~x}$
$cos~x = -\sqrt{1-(\frac{-1}{2})^2}$
$cos~x = -\sqrt{1-\frac{1}{4}}$
$cos~x = -\frac{\sqrt{3}}{2}$
$cos~y = -\frac{2}{5}$
Since $y$ is in quadrant III, $sin~y$ is also negative.
We can find $sin~y$:
$sin^2~y+cos^2~y = 1$
$sin~y = -\sqrt{1-cos^2~y}$
$sin~y = -\sqrt{1-(\frac{-2}{5})^2}$
$sin~y = -\sqrt{1-\frac{4}{25}}$
$sin~y = -\frac{\sqrt{21}}{5}$
We can find $sin(x+y)$:
$sin~(x+y) = sin~x~cos~y+cos~x~sin~y$
$sin~(x+y) = (-\frac{1}{2})(-\frac{2}{5})+(-\frac{\sqrt{3}}{2})(-\frac{\sqrt{21}}{5})$
$sin~(x+y) = \frac{1}{5}+\frac{\sqrt{63}}{10}$
$sin~(x+y) = \frac{2+3\sqrt{7}}{10}$
We can find $cos(x-y)$:
$cos~(x-y) = cos~x~cos~(-y)-sin~x~sin~(-y)$
$cos~(x-y) = cos~x~cos~y-sin~x~(-sin~y)$
$cos~(x-y) = (-\frac{\sqrt{3}}{2})(-\frac{2}{5})-(\frac{-1}{2})(-\frac{-\sqrt{21}}{5})$
$cos~(x-y) = \frac{\sqrt{3}}{5}+\frac{\sqrt{21}}{10}$
$cos~(x-y) = \frac{2\sqrt{3}+\sqrt{21}}{10}$
We can find $cos(x+y)$:
$cos~(x+y) = cos~x~cos~y-sin~x~sin~y$
$cos~(x+y) = (-\frac{\sqrt{3}}{2})(-\frac{2}{5})-(\frac{-1}{2})(-\frac{\sqrt{21}}{5})$
$cos~(x+y) = \frac{\sqrt{3}}{5}-\frac{\sqrt{21}}{10}$
$cos~(x+y) = \frac{2\sqrt{3}-\sqrt{21}}{10}$
We can find $tan~(x+y)$:
$tan~(x+y) = \frac{sin~(x+y)}{cos~(x+y)}$
$tan~(x+y) = \frac{\frac{2+3\sqrt{7}}{10}}{\frac{2\sqrt{3}-\sqrt{21}}{10}}$
$tan~(x+y) = \frac{2+3\sqrt{7}}{2\sqrt{3}-\sqrt{21}}$
Since $sin~(x+y)$ is positive while $tan~(x+y)$ is negative, then $x+y$ must be in quadrant II.
