Answer
a. $\frac{33}{65}$
b. $\frac{16}{65}$
c. $-\frac{33}{56}$
d. $\frac{24}{25}$
e. $\frac{2\sqrt {13}}{13}$
Work Step by Step
Given
$sin\alpha=\frac{3}{5}, 0\lt \alpha \lt \frac{\pi}{2}$
and
$sin\beta=\frac{12}{13}, \frac{\pi}{2}\lt \beta \lt\pi $
For angle $\alpha$, form a triangle with sides $3,4,5$. We have
$cos\alpha=\frac{4}{5}, tan\alpha=\frac{3}{4}$.
For angle $\beta$, form a triangle of sides $12,5,13$. We have
$cos\beta=-\frac{5}{13}, tan\beta=-\frac{12}{5}$.
a. Use the Addition Formulas
$sin(\alpha+\beta)=sin\alpha cos\beta+ cos\alpha sin\beta =(\frac{3}{5})(-\frac{5}{13})+(\frac{4}{5})(\frac{12}{13})=\frac{33}{65}$
b. Use the Subtraction Formulas
$cos(\alpha-\beta)=cos\alpha cos\beta+ sin\alpha sin\beta =(\frac{4}{5})(-\frac{5}{13})+(\frac{3}{5})(\frac{12}{13})=\frac{16}{65}$
c. Use the Addition Formulas
$tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}=\frac{(\frac{3}{4})+(-\frac{12}{5})}{1-(\frac{3}{4})(-\frac{12}{5})}=-\frac{33}{56}$
d. Use the Double-Angle Formula
$sin2\alpha = 2 sin\alpha cos\alpha = 2(\frac{3}{5})(\frac{4}{5})=\frac{24}{25}$
e. Use the Half-Angle Formula
$\frac{\pi}{4}\lt \frac{\beta}{2} \lt\frac{\pi}{2} $
and
$cos\frac{\beta}{2}=\sqrt {\frac{1+cos\beta}{2}}=\sqrt {\frac{1+(-\frac{5}{13})}{2}}=\frac{2\sqrt {13}}{13}$
