I've got this: $$\sin{(A*B)}\approx \frac{Si(B^2)-Si(A^2)}{2(\ln{B}-\ln{A})}$$, whenever the RHS is defined and B is close to A ( I don't know how close). Here $Si(x)$ is the integral of $\frac{\sin{x}}{x}$ But, to check it, I need to evaluate the $Si(x)$ function. I'm new with Taylor series, so, how am I supposed to do it? The only confusion is that should I work with the Taylor series of $Si(x)$ around $x=A^2$ or $x=B^2$ to check it? How do I evaluate $Si(B^2)-Si(A^2)$ approximately? And, have I found something new because it's an expression for $sin(A*B)$ and that it relates $sinx$ to $Si(x)$ without integrals?
EDIT: I think this result is true for any function $f(x)$. To get the approximate expression of $f(A*B)$, just replace $Si(x)$ by the integral of $\frac{f(x)}{x}$. I think it can be converted into a (maybe) useful result as it would mean: $$\int_a^b\frac{f(x)}{x}dx\approx f(\sqrt{ab})*\ln{\frac{b}{a}}$$
This is too long for a comment.
I do not know how we can arrive to the result but we can show that the result is correct starting from the RHS.
Around $x=a$, Taylor series is $$\text{Si}\left(x\right)=\text{Si}(a)+\frac{(x-a) \sin (a)}{a}+O\left((x-a)^2\right)$$
Then, composing series $$\text{Si}\left(B^2\right)=\text{Si}\left(A^2\right)+\frac{2 (B-A) \sin \left(A^2\right)}{A}+O\left((B-A)^2\right)$$ $$\log(B)=\log (A)+\frac{B-A}{A}+O\left((B-A)^2\right)$$ which make
$$\frac{\text{Si}\left(B^2\right)-\text{Si}\left(A^2\right)}{2 (\log (B)-\log (A))}=\sin \left(A^2\right)+A (B-A) \cos \left(A^2\right)+O\left((B-A)^2\right)$$ while $$\sin(AB)=\sin \left(A^2\right)+A (B-A) \cos \left(A^2\right)+O\left((B-A)^2\right)$$