How to go about proving that $\cos^2 x$ is everywhere differentiable? My first line of reasoning was to try directly evaluating $$\lim\limits_{h \to 0}\frac{\cos^2 (x+h) - \cos^2 (x)}{h}$$ and showing such a limit existed for any x, but when $\cos^2(x)$ evaluates to zero (e.g. when $x = \frac{\pi}{2}$), then directly evaluating the limit yields the indeterminate form $\frac{0}{0}$. The same thing happens even after transforming $\cos^2 x$ to $\frac{1}{2} \cos (2x)$, so I don't think this is the correct strategy. I considered using L'Hopitals rule, but don't think that is the route that the question intended, as it is suggested to use the identity $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$ along with standard trigonometric identities. I am not sure where to begin. What would be a step in the right direction?
 A: Use the trigonometric identity
$$
\cos A - \cos B = -2 \sin\frac{A+B}2\sin\frac{A-B}2,
$$
either directly or for the $\cos 2x$ form.
A: $$\begin{align}L &= \lim\limits_{h \to 0}\frac{\cos^2 (x+h) - \cos^2 (x)}{h} \\ &= \lim\limits_{h \to 0}\frac{(\cos(x)\cos(h)-\sin(x)\sin(h))^2 - \cos^2 (x)}{h}\\ &= \lim\limits_{h \to 0}\frac{\cos^2(x)\cos^2(h) + \sin^2(x)\sin^2(h) - 2\cos(x)\cos(h)\sin(x)\sin(h) - \cos^2 (x)}{h} \\ &= \lim\limits_{h \to 0} \dfrac{\cos^2(x)(\cos^2(h)-1)}{h} + \lim\limits_{h \to 0} \sin^2(x)\sin(h)\dfrac{\sin(h)}{h}- \lim\limits_{h \to 0} \dfrac{\sin(2x)\sin(2h)}{2h} \\ &= \cos^2(x)\lim\limits_{h \to 0} \dfrac{\cos^2(h)-1}{h} + \sin^2(x)\lim\limits_{h \to 0}\sin(h)- \sin(2x)\lim\limits_{h \to 0} \dfrac{\sin(2h)}{2h}\\ &= \cos^2(x)\lim\limits_{h \to 0} \dfrac{(\cos(h)-1)}{h}\lim\limits_{h \to 0}(\cos(h)+1) - \sin(2x)\\ &= 2\cos^2(x)\lim\limits_{h \to 0} \dfrac{(\cos(h)-1)}{h} - \sin(2x) \\ &= -\sin(2x)\end{align}$$
A: HINT:
$$\cos^2B-\cos^2A=1-\sin^2B-(1-\sin^2A)$$
Using  Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $, this becomes  $$\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$$
So, $$\cos^2(x+h)-\cos^2x=\sin(2x+h)\sin(-h)=-\sin(2x+h)\sin(h)$$
