Answer
a. $1$
b. $-\frac{3}{5}$
c. undefined
d. $-\frac{3}{5}$
e. $-\frac{\sqrt {10+3\sqrt {10}}}{2\sqrt 5}$
Work Step by Step
Given
$tan\alpha=-3,\frac{\pi}{2}\lt \alpha \lt\pi$
and $cot\beta=-3, \frac{3\pi}{2}\lt \beta \lt 2\pi $,
For angle $\alpha$, form a right triangle of sides $3,1,\sqrt {10}$; we have
$sin\alpha=\frac{3\sqrt {10}}{10}, cos\alpha=-\frac{\sqrt {10}}{10}$;
For angle $\beta$, form a right triangle of sides $1,3,\sqrt {10}$; we have
$sin\beta=-\frac{\sqrt {10}}{10}, cos\beta=\frac{3\sqrt {10}}{10}, tan\beta=-\frac{1}{3}$.
a. Use the Addition Formulas,
$sin(\alpha+\beta)=sin\alpha cos\beta+ cos\alpha sin\beta =(\frac{3\sqrt {10}}{10})(\frac{3\sqrt {10}}{10})+(-\frac{\sqrt {10}}{10})(-\frac{\sqrt {10}}{10})=1$
b. Use the Subtraction Formulas,
$cos(\alpha-\beta)=cos\alpha cos\beta+ sin\alpha sin\beta =(-\frac{\sqrt {10}}{10})(\frac{3\sqrt {10}}{10})+(\frac{3\sqrt {10}}{10})(-\frac{\sqrt {10}}{10})=-\frac{3}{5}$
c. Use the Addition Formulas,
$tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}=\frac{(-3)+(-\frac{1}{3})}{1-(-3)(-\frac{1}{3})}=-\infty$ undefined
d. Use the Double-Angle Formula,
$sin2\alpha = 2 sin\alpha cos\alpha = 2(\frac{3\sqrt {10}}{10})(-\frac{\sqrt {10}}{10})=-\frac{3}{5}$
e. Use the Half-Angle Formula,
$ \frac{3\pi}{4}\lt \frac{\beta}{2} \lt \pi $
and
$cos\frac{\beta}{2}=-\sqrt {\frac{1+cos\beta}{2}}=-\sqrt {\frac{1+\frac{3\sqrt {10}}{10}}{2}}$=$-\frac{\sqrt {10+3\sqrt {10}}}{2\sqrt 5}$
