Giải thích các bước giải:
Dựa vào đường tròn lượng giác ta có:
\(\begin{array}{l}
\frac{{5\pi }}{2} < \alpha < 3\pi \Rightarrow \left\{ \begin{array}{l}
\sin \alpha > 0\\
\cos \alpha < 0
\end{array} \right.\\
\tan \alpha = \frac{{ - 4}}{3} \Rightarrow \left\{ \begin{array}{l}
\cot \alpha = \frac{1}{{\tan \alpha }} = - \frac{3}{4}\\
\frac{{\sin \alpha }}{{\cos \alpha }} = - \frac{4}{3} \Leftrightarrow \sin \alpha = - \frac{4}{3}\cos \alpha
\end{array} \right.\\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\Leftrightarrow {\left( { - \frac{4}{3}\cos \alpha } \right)^2} + {\cos ^2}\alpha = 1\\
\Leftrightarrow \frac{{25}}{9}{\cos ^2}\alpha = 1\\
\Leftrightarrow {\cos ^2}\alpha = \frac{9}{{25}} \Rightarrow \cos \alpha = - \frac{3}{5}\,\,\,\,\,\,\left( {\cos \alpha < 0} \right)\\
\sin \alpha = - \frac{4}{3}.\cos \alpha = \frac{4}{5}\\
\sin \left( {\alpha - \pi } \right) = \sin \left[ {\left( \pi \right) - \left( {\alpha - \pi } \right)} \right] = \sin \left( {2\pi - \alpha } \right) = \sin \left( { - \alpha } \right) = - \sin \alpha = - \frac{4}{5}\\
\sin \left( {\frac{\pi }{2} - \alpha } \right) = \cos \alpha = - \frac{3}{5}\\
\cos \left( {\alpha + \frac{\pi }{2}} \right) = \sin \left[ {\frac{\pi }{2} - \left( {\alpha + \frac{\pi }{2}} \right)} \right] = \sin \left( { - \alpha } \right) = - \sin \alpha = - \frac{4}{5}\\
\cos \left( {\pi + \alpha } \right) = - \cos \left[ {\pi - \left( {\pi + \alpha } \right)} \right] = - \cos \left( { - \alpha } \right) = - \cos \alpha = \frac{3}{5}
\end{array}\)
