Đáp án:

\[\sin \frac{{5\alpha }}{2} = \frac{{41}}{{25\sqrt 5 }}\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
0 < \alpha  < \dfrac{\pi }{2} \Rightarrow \left\{ \begin{array}{l}
\sin \alpha  > 0\\
cos\alpha  > 0
\end{array} \right.\\
{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\
\cos \alpha  > 0 \Rightarrow \cos \alpha  = \sqrt {1 - {{\sin }^2}\alpha }  = \dfrac{3}{5}\\
\tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{4}{3}\\
\sin 2\alpha  = 2\sin \alpha .\cos \alpha  = 2.\dfrac{4}{5}.\dfrac{3}{5} = \dfrac{{24}}{{25}}\\
\cos 2\alpha  = 2{\cos ^2}\alpha  - 1 = 2.{\left( {\dfrac{3}{5}} \right)^2} - 1 =  - \dfrac{7}{{25}}\\
0 < \alpha  < \dfrac{\pi }{2} \Rightarrow 0 < \dfrac{\alpha }{2} < \dfrac{\pi }{4} \Rightarrow \left\{ \begin{array}{l}
\sin \dfrac{\alpha }{2} > 0\\
\cos \dfrac{\alpha }{2} > 0
\end{array} \right.\\
\cos \alpha  = 2{\cos ^2}\dfrac{\alpha }{2} - 1 = 1 - 2{\sin ^2}\dfrac{\alpha }{2} \Rightarrow \left\{ \begin{array}{l}
\sin \dfrac{\alpha }{2} = \dfrac{1}{{\sqrt 5 }}\\
\cos \dfrac{\alpha }{2} = \dfrac{2}{{\sqrt 5 }}
\end{array} \right.\\
\sin \dfrac{{5\alpha }}{2} = \sin \left( {2\alpha  + \dfrac{\alpha }{2}} \right) = \sin 2\alpha .\cos \dfrac{\alpha }{2} + \cos 2\alpha .\sin \dfrac{\alpha }{2}\\
 = \dfrac{{24}}{{25}}.\dfrac{2}{{\sqrt 5 }} + \dfrac{{ - 7}}{{25}}.\dfrac{1}{{\sqrt 5 }} = \dfrac{{41}}{{25\sqrt 5 }}
\end{array}\)

$0<\alpha<\dfrac{\pi}{2}$

$\Rightarrow \cos\alpha>0$, $\sin\dfrac{\alpha}{2}>0,\cos\dfrac{\alpha}{2}>0$

$\Rightarrow \cos\alpha=\sqrt{1-\sin^2\alpha}=\dfrac{3}{5}$

$\sin^2\dfrac{\alpha}{2}=\dfrac{1-\cos\alpha}{2}=\dfrac{1}{5}$

$\Rightarrow \sin\dfrac{\alpha}{2}=\dfrac{1}{\sqrt5}$

$\Rightarrow \cos\dfrac{\alpha}{2}=\dfrac{2}{\sqrt5}$

$\sin 2\alpha=2\sin\alpha.\cos\alpha=\dfrac{24}{25}$

$\cos 2\alpha=\cos^2\alpha-\sin^2\alpha=\dfrac{-7}{25}$

Vậy ta có:

$\sin\dfrac{5\alpha}{2}=\sin(\dfrac{\alpha}{2}+2\alpha)$

$=\sin\dfrac{\alpha}{2}.\cos2\alpha+\cos\dfrac{\alpha}{2}\sin2\alpha$

$=\dfrac{41}{125}$