Đáp án:
$\begin{array}{l}
a)sinx = \frac{5}{{13}} \Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x} = \frac{{12}}{{13}}\\
{\mathop{\rm cosy}\nolimits} = \frac{3}{5} \Rightarrow \sin y = \sqrt {1 - {{\cos }^2}y} = \frac{4}{5}\\
\Rightarrow \sin \left( {x + y} \right) = {\mathop{\rm s}\nolimits} {\rm{inx}}{\rm{.cosy + siny}}{\rm{.cosx}}\\
{\rm{ = }}\frac{5}{{13}}.\frac{3}{5} + \frac{{12}}{{13}}.\frac{4}{5}\\
= \frac{{63}}{{65}}\\
+ \cos \left( {x + y} \right) = \cos x.\cos y - \sin x.\sin y\\
= \frac{{12}}{{13}}.\frac{3}{5} - \frac{5}{{13}}.\frac{4}{5}\\
= \frac{{16}}{{65}}\\
b)\sin x = \frac{1}{{\sqrt 5 }} \Rightarrow \cos x = \frac{2}{{\sqrt 5 }}\\
\sin y = \frac{1}{{\sqrt {10} }} \Rightarrow \cos y = \frac{3}{{\sqrt {10} }}\\
\sin \left( {x + y} \right) = \sin x.\sin y + \cos x.\cos y\\
= \frac{1}{{\sqrt 5 }}.\frac{1}{{\sqrt {10} }} + \frac{2}{{\sqrt 5 }}.\frac{3}{{\sqrt {10} }}\\
= \frac{7}{{5\sqrt 2 }} = \frac{{7\sqrt 2 }}{{10}}
\end{array}$
