Question

Вычислите: cos(arcsin(-12/13)+arcsin4/5)

Answer 1

$cos(arcsin(- \frac{12}{13} )+arcsin \frac{4}{5} )=cos(arcsin \frac{4}{5}-arcsin \frac{12}{13} )$
Обозначим
$arcsin \frac{4}{5}= \alpha \\ arcsin\frac{12}{13} = \beta$
формула
$cos( \alpha - \beta )=cos \alpha \cdot cos \beta +sin \alpha \cdot sin \beta$
Так как
$arcsin \frac{4}{5}= \alpha \Rightarrow sin \alpha = \frac{4}{5}, \alpha \in(0; \frac{ \pi }{2})$
$cos \alpha = \sqrt{1-sin ^{2} \alpha }= \sqrt{1-( \frac{4}{5}) ^{2} }= \sqrt{1- \frac{16}{25} }= \sqrt{ \frac{9}{25} }= \frac{3}{5}$
Так как
$arcsin \frac{12}{13}= \beta \Rightarrow sin \beta = \frac{12}{13}, \beta \in(0; \frac{ \pi }{2})$
$cos\beta = \sqrt{1-sin ^{2}\beta }= \sqrt{1-( \frac{12}{13}) ^{2} }= \sqrt{1- \frac{144}{169} }= \sqrt{ \frac{25}{169} }= \frac{5}{13}$

$cos(arcsin(- \frac{12}{13} )+arcsin \frac{4}{5} )=cos(arcsin \frac{4}{5}-arcsin \frac{12}{13} )= \\ =cos( \alpha - \beta )=cos \alpha \cdot cos \beta +sin \alpha \cdot sin \beta= \frac{3}{5}\cdot \frac{5}{13}+ \frac{4}{5} \cdot \frac{12}{13}= \\ = \frac{15}{65}+ \frac{48}{65}= \frac{15+48}{65}= \frac{63}{65}$