Connect the curve 5

$acosax=0$

at x=5pi/2

$cos(a\frac{5\pi}{2})$

a=1/5 for minimum as cospi/2 is zero

The curve of $sin(ax)$ is tangent to the curve of $sin(x)$ at $x=\frac{5\pi}{2}$ . This means that the derivative of the curves are equal at $x=\frac{5\pi}{2}$ . This means that $acos(ax)=cos(x)$ and since $cos(\frac{5\pi}{2})=0$ then $acos(\frac{5\pi}{2}a)=0$

So $a=0$ or $cos(\frac{5\pi}{2}a)=0$

a cannot be zero to be written in form of co-prime A and B. so $cos(\frac{5\pi}{2}a)=0$

and since $cos(\frac{\pi}{2})=0.$ Then $\frac{5\pi}{2}a=\frac{\pi}{2}$

$a=\frac{1}{5}$

$1+5=6$

Let $f(x):=\sin(x),\quad g(x):=\sin(ax)$ . The demand " $f$ is tangent to $g$ at $x=\frac{5\pi}{2}$ " yields two equations:

$\begin{aligned} f(x)&\overset{!}{=}g(x)&\Rightarrow&&1&=\sin\left(\frac{5\pi}{2}\right)\overset{!}{=}\sin(ax)=\sin\left(\frac{5a\pi}{2}\right)&\Rightarrow &&\frac{5a\pi}{2}&=\frac{\pi}{2}+2\pi n,\quad \cos(ax)=0\\\\ f'(x)&\overset{!}{=}g'(x)&\Rightarrow&&0&=\cos\left(\frac{5\pi}{2}\right)\overset{!}{=}a\cos(ax)=a\cdot0=0&\text{true} \end{aligned}$

The second expression true for any $a$ . Solving the first expression we have $a=\frac{1}{5}+\frac{4n}{5}$ , the minimal positive value being $\boxed{a=\frac{1}{5}}$