Giải thích các bước giải:
Ta có:
$\begin{array}{l}
+ )\sin A = \dfrac{3}{5};\widehat A < {90^0}\\
\Rightarrow \cos A = \sqrt {1 - {{\sin }^2}A} = \dfrac{4}{5}\\
\Rightarrow \tan A = \dfrac{{\sin A}}{{\cos A}} = \dfrac{3}{4}\\
+ )\sin B = \dfrac{8}{{17}};\widehat B > {90^0}\\
\Rightarrow \cos B = - \sqrt {1 - {{\sin }^2}B} = - \dfrac{{15}}{{17}}\\
\Rightarrow \tan B = - \dfrac{8}{{15}}
\end{array}$
Khi đó:
$\begin{array}{l}
+ )\sin \left( {A - B} \right) = \sin A\cos B - \sin B\cos A\\
= \dfrac{3}{5}.\left( {\dfrac{{ - 15}}{{17}}} \right) - \dfrac{8}{{17}}.\dfrac{4}{5}\\
= \dfrac{{ - 77}}{{85}}\\
+ )\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\\
= \dfrac{4}{5}.\left( {\dfrac{{ - 15}}{{17}}} \right) + \dfrac{3}{5}.\dfrac{8}{{17}}\\
= \dfrac{{ - 36}}{{85}}\\
+ )\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A.\tan B}}\\
= \dfrac{{\dfrac{3}{4} - \left( {\dfrac{{ - 8}}{{15}}} \right)}}{{1 + \dfrac{3}{4}.\left( {\dfrac{{ - 8}}{{15}}} \right)}}\\
= \dfrac{{77}}{{35}}
\end{array}$
Vậy $\sin \left( {A - B} \right)$;$\cos \left( {A + B} \right) = \dfrac{{ - 36}}{{85}}$;$\tan \left( {A - B} \right) = \dfrac{{77}}{{35}}$
