Now I got a result , is it correct ?

set \(\displaystyle t_1 t_2 =n \) and \(\displaystyle A_n = \dfrac{ \sin(n) }{n} \; , \: 2\leq n=t_1 t_2 \leq 24\).

\(\displaystyle \dfrac{A_{n+1}}{A_n} -1= \dfrac{n \sin(n+1) }{(1+n)\sin(n) }-1=\dfrac{n \sin(n+1)-\sin(n)-n\sin(n) }{(1+n)\sin(n) }<0\; , n\leq 24\).

since \(\displaystyle A_n \) is decreasing in the given domain then \(\displaystyle \sup A_n = A_2 = \dfrac{\sin 2}{2}\).