Can you get an infinite number of derivatives from $\frac{d}{dx}\sin(x)$? I know that the first derivative of sine of x is cosine of x, but I'm really facing a problem trying to understand the other derivatives of the sine function
The following 2 statements are both true:
$$\frac{d}{dx}\sin(x) = \cos(x)$$
$$\frac{d}{dx}\cos(x) = -\sin(x)$$
And then, according to the constant multiple rule, we can see that the following is also true:
$\frac{d}{dx}-\sin(x) = -1 * \frac{d}{dx}\sin(x) = -1 * \cos(x) = -\cos(x)$
Hence:
$\frac{d}{dx}-\cos(x) = -1*\frac{d}{dx}\cos(x) = -1*-\sin(x) = \sin(x) ....$
can we keep repeating this process to infinity? and what does it really mean that the fourth derivative of sin(x) is sin(x), I've seen a visualisation that explains why the derivative of sin(x) is cos(x), so I know why the first derivative of sin(x) is cos(x), at least intuitively, but I can't really get the idea of the rest of the derivatives of sin(x), or why they are repeating, and can we really keep repeating this with no problems?

Yes, you get the derivative as $$\dfrac{d^n \sin(x)}{dx^n} = \sin \left(\frac{\pi  n}{2}+x\right)$$ for $n \ge 0$ and $n$ integer.
This just produces a repeating pattern of $$\sin (x),\cos (x),-\sin (x),-\cos (x) \ldots$$
