Answer
a. $ -\frac{4+6\sqrt 2}{15}$
b. $ \frac{3-8\sqrt 2}{15}$
c. $ \frac{54-25\sqrt 2}{28}$
Work Step by Step
Given $tan\alpha=-\frac{3}{4}$ with $\alpha$ in quadrant II, build a right triangle with angle $\alpha$ and sides $3,4,5$; we have $sin\alpha=\frac{3}{5}$ and $cos\alpha=-\frac{4}{5}$.
Similarly, given $cos\beta=\frac{1}{3}$ with $\beta$ in quadrant I, build a right triangle with angle $\beta$ and sides $1,2\sqrt 2,3$; we have $sin\beta=\frac{2\sqrt 2}{3}$ and $tan\beta=2\sqrt 2$.
Using the Addition Formulas, we have:
a. $cos(\alpha+\beta)=cos(\alpha) cos(\beta)-sin(\alpha) sin(\beta)=(-\frac{4}{5})(\frac{1}{3})-(\frac{3}{5})(\frac{2\sqrt 2}{3})=-\frac{4+6\sqrt 2}{15}$
b. $sin(\alpha+\beta)=sin(\alpha) cos(\beta)+cos(\alpha) sin(\beta)=(\frac{3}{5})(\frac{1}{3})+(-\frac{4}{5})(\frac{2\sqrt 2}{3})=\frac{3-8\sqrt 2}{15}$
c. $tan( \alpha+\beta)=\frac{tan(\alpha)+tan(\beta)}{1-tan(\alpha) tan(\beta)}=\frac{(-\frac{3}{4})+(2\sqrt 2)}{1-(-\frac{3}{4})(2\sqrt 2)}=\frac{54-25\sqrt 2}{28}$
