## Trigonometry (11th Edition) Clone

$sin~(x+y) = \frac{2-2\sqrt{30}}{15}$ $cos~(x-y) = \frac{\sqrt{5}-4\sqrt{6}}{15}$ $tan~(x+y) = \frac{2-2\sqrt{30}}{\sqrt{5}+4\sqrt{6}}$ $x+y~~$ is in quadrant IV.
$sin~y = -\frac{2}{3}$ 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.