Answer
a. $-\frac{63}{65}$
b. $-\frac{56}{65}$
c. $\frac{63}{16}$
d. $\frac{24}{25}$
e. $\frac{5\sqrt {26}}{26}$
Work Step by Step
Given
$tan\alpha=\frac{4}{3}, \pi\lt \alpha \lt \frac{3\pi}{2}$
and
$tan\beta=\frac{5}{12}, 0\lt \beta \lt\frac{\pi}{2} $
For angle $\alpha$, form a triangle of sides $4,3,5$; we have:
$sin\alpha=-\frac{4}{5}, cos\alpha=-\frac{3}{5}$;
For angle $\beta$, form a triangle of sides $5,12,13$; we have
$sin\beta=\frac{5}{13}, cos\beta=\frac{12}{13}$.
a. Use the Addition Formulas,
$sin(\alpha+\beta)=sin\alpha cos\beta+ cos\alpha sin\beta =(-\frac{4}{5})(\frac{12}{13})+(-\frac{3}{5})(\frac{5}{13})=-\frac{63}{65}$
b. Use the Subtraction Formulas,
$cos(\alpha-\beta)=cos\alpha cos\beta+ sin\alpha sin\beta =(-\frac{3}{5})(\frac{12}{13})+(-\frac{4}{5})(\frac{5}{13})=-\frac{56}{65}$
c. Use the Addition Formulas,
$tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}=\frac{(\frac{4}{3})+(\frac{5}{12})}{1-(\frac{4}{3})(-\frac{5}{12})}=\frac{63}{16}$
d. Use the Double-Angle Formula,
$sin2\alpha = 2 sin\alpha cos\alpha = 2(-\frac{4}{5})(-\frac{3}{5})=\frac{24}{25}$
e. Use the Half-Angle Formula,
$0\lt \frac{\beta}{2} \lt\frac{\pi}{4} $ and $cos\frac{\beta}{2}=\sqrt {\frac{1+cos\beta}{2}}=\sqrt {\frac{1+(\frac{12}{13})}{2}}=\frac{5\sqrt {26}}{26}$
