I'm sorry, if this question seems too simple..
Is it possible to reduce
sin 8x/cos 4x it like this:
4 (sin 2x)/ 4(cos x) sin 2x/cos x
Or this trig. expression cannot be simplified any further?
We have $\sin(8x)=2\sin(4x)\cos(4x)$. This is a particular case of the double-angle identity $\sin 2u=2\sin u\cos u$ (we just take $u=4x$).
So our numerator can be written as $2\sin(4x)\cos(4x)$, and our denominator is $\cos(4x)$. There is cancellation, and we obtain $2\sin(4x)$.
Remark: Your proposed simplification was not correct. It seems to assume, for example, that $\cos 4x=4\cos x$. That is not true in general. Take almost any specific value of $x$, like $x=\pi/2$ ($90$ degrees). We have $\cos 4x=1$ and $4\cos x=0$. There was a similar incorrect assertion about $\sin 8x$.