The problem is, as stated:
$$\lim_{x\rightarrow 0}\frac{\cos(ax) - \cos(bx)}{\sin^2(x)}$$ Of course, a and b are real numbers.
I tried implementing the trig identity:
$$\cos(ax)-\cos(bx) = -2\sin\frac{(ax-bx)}{2}\sin\frac{(ax+bx)}{2}$$
But that didn't really take me anywhere. In my book, similar problems were often solved this way but here that doesn't seem to work.
One obviously has to try and use the known limit to reduce this to a more easy limit:
$$\lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1$$
Any help would be much appreciated.