Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
0 < \alpha < \dfrac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\sin \alpha > 0\\
cos\alpha > 0
\end{array} \right.\\
{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\cos \alpha > 0 \Rightarrow \cos \alpha = \sqrt {1 - {{\sin }^2}\alpha } = \dfrac{3}{5}\\
\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{4}{3}\\
b,\\
\sin 2\alpha = 2\sin \alpha .\cos \alpha = 2.\dfrac{4}{5}.\dfrac{3}{5} = \dfrac{{24}}{{25}}\\
\cos 2\alpha = 2{\cos ^2}\alpha - 1 = 2.{\left( {\dfrac{3}{5}} \right)^2} - 1 = - \dfrac{7}{{25}}\\
0 < \alpha < \dfrac{\pi }{2} \Rightarrow 0 < \dfrac{\alpha }{2} < \dfrac{\pi }{4} \Rightarrow \left\{ \begin{array}{l}
\sin \dfrac{\alpha }{2} > 0\\
\cos \dfrac{\alpha }{2} > 0
\end{array} \right.\\
\cos \alpha = 2{\cos ^2}\dfrac{\alpha }{2} - 1 = 1 - 2{\sin ^2}\dfrac{\alpha }{2} \Rightarrow \left\{ \begin{array}{l}
\sin \dfrac{\alpha }{2} = \dfrac{1}{{\sqrt 5 }}\\
\cos \dfrac{\alpha }{2} = \dfrac{2}{{\sqrt 5 }}
\end{array} \right.\\
\sin \dfrac{{5\alpha }}{2} = \sin \left( {2\alpha + \dfrac{\alpha }{2}} \right) = \sin 2\alpha .\cos \dfrac{\alpha }{2} + \cos 2\alpha .\sin \dfrac{\alpha }{2}\\
= \dfrac{{24}}{{25}}.\dfrac{2}{{\sqrt 5 }} + \dfrac{{ - 7}}{{25}}.\dfrac{1}{{\sqrt 5 }} = \dfrac{{41}}{{25\sqrt 5 }}
\end{array}\)
