Answer
$sin~(x+y) = \frac{117}{125}$
$cos~(x+y) = -\frac{44}{125}$
$tan~(x+y) = -\frac{117}{44}$
$x+y$ is in quadrant II.
Work Step by Step
$sin~x = -\frac{3}{5}$
Since $x$ is in quadrant III, $cos~x$ is also negative.
We can find $cos~x$:
$sin^2~x+cos^2~x = 1$
$cos~x = -\sqrt{1-sin^2~x}$
$cos~x = -\sqrt{1-(-\frac{3}{5})^2}$
$cos~x = -\sqrt{1-\frac{9}{25}}$
$cos~x = -\frac{4}{5}$
$cos~y = -\frac{7}{25}$
Since $y$ is in quadrant III, $sin~y$ is also negative.
We can find $sin~y$:
$sin^2~y+cos^2~y = 1$
$sin~y = -\sqrt{1-cos^2~y}$
$sin~y = -\sqrt{1-(-\frac{7}{25})^2}$
$sin~y = -\sqrt{1-\frac{49}{625}}$
$sin~y = -\frac{24}{25}$
We can find $sin(x+y)$:
$sin~(x+y) = sin~x~cos~y+cos~x~sin~y$
$sin~(x+y) = (-\frac{3}{5})(-\frac{7}{25})+(-\frac{4}{5})(-\frac{24}{25})$
$sin~(x+y) = \frac{21}{125}+\frac{96}{125}$
$sin~(x+y) = \frac{117}{125}$
We can find $cos(x+y)$:
$cos~(x+y) = cos~x~cos~y-sin~x~sin~y$
$cos~(x+y) = (-\frac{4}{5})(-\frac{7}{25})-(-\frac{3}{5})(-\frac{24}{25})$
$cos~(x+y) = \frac{28}{125}-\frac{72}{125}$
$cos~(x+y) = -\frac{44}{125}$
We can find $tan~(x+y)$:
$tan~(x+y) = \frac{sin~(x+y)}{cos~(x+y)}$
$tan~(x+y) = \frac{ \frac{117}{125}}{-\frac{44}{125}}$
$tan~(x+y) = -\frac{117}{44}$
Since $sin~(x+y)$ is positive, while $cos~(x+y)$ and $tan~(x+y)$ are both negative, then $x+y$ must be in quadrant II.
