## Trigonometry (11th Edition) Clone

$sin~(x+y) = \frac{117}{125}$ $cos~(x+y) = -\frac{44}{125}$ $tan~(x+y) = -\frac{117}{44}$ $x+y$ is in quadrant II.
$sin~x = -\frac{3}{5}$ 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.