## Precalculus (6th Edition) Blitzer

a. $1$ b. $\frac{4\sqrt 2}{9}$ c. undefined d. $\frac{4\sqrt 2}{9}$ e. $-\frac{\sqrt {3}}{3}$
Given $sin\alpha=-\frac{1}{3}, \pi\lt \alpha \lt \frac{3\pi}{2}$ and $cos\beta=-\frac{1}{3}, \pi\lt \beta \lt\frac{3\pi}{2}$ For angle $\alpha$, form a right triangle of sides $1,2\sqrt 2,3$; we have $cos\alpha=-\frac{2\sqrt 2}{3}, tan\alpha=\frac{\sqrt 2}{4}$; For angle $\beta$, form a right triangle of sides $2\sqrt 2,1,3$; we have $sin\beta=-\frac{2\sqrt 2}{3}, tan\beta=2\sqrt 2$. a. Use the Addition Formulas, $sin(\alpha+\beta)=sin\alpha cos\beta+ cos\alpha sin\beta =(-\frac{1}{3})(-\frac{1}{3})+(-\frac{2\sqrt 2}{3})(-\frac{2\sqrt 2}{3})=1$ b. Use the Subtraction Formulas, $cos(\alpha-\beta)=cos\alpha cos\beta+ sin\alpha sin\beta =(-\frac{2\sqrt 2}{3})(-\frac{1}{3})+(-\frac{1}{3})(-\frac{2\sqrt 2}{3})=\frac{4\sqrt 2}{9}$ c. Use the Addition Formulas, $tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}=\frac{(\frac{\sqrt 2}{4})+(2\sqrt 2)}{1-(\frac{\sqrt 2}{4})(2\sqrt 2)}=\infty$, undefined d. Use the Double-Angle Formula, $sin2\alpha = 2 sin\alpha cos\alpha = 2(-\frac{1}{3})(-\frac{2\sqrt 2}{3})=\frac{4\sqrt 2}{9}$ e. Use the Half-Angle Formula, $\frac{\pi}{2}\lt \frac{\beta}{2} \lt\frac{3\pi}{4}$ and $cos\frac{\beta}{2}=-\sqrt {\frac{1+cos\beta}{2}}=-\sqrt {\frac{1+(-\frac{1}{3})}{2}}=-\frac{\sqrt {3}}{3}$