Trigonometry (11th Edition) Clone

$sin~(x+y) = \frac{2+3\sqrt{7}}{10}$ $cos~(x-y) = \frac{2\sqrt{3}+\sqrt{21}}{10}$ $tan~(x+y) = \frac{2+3\sqrt{7}}{2\sqrt{3}-\sqrt{21}}$ $x+y~~$ is in quadrant II.
$sin~x = -\frac{1}{2}$ 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.