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I want to show $\arcsin (\sin^2x) $ is not differentiable at $\pi/2+\pi k$. (if its true).

So far I have: $$ \frac{\arcsin (\sin^2(\pi/2+\pi k +h))-\arcsin (\sin^2(\pi/2+\pi k))}{h}=\frac{\arcsin (\cos^2h)-\ \pi/2}{h} = \frac {\arccos (\cos^2 h)}{h} $$

so it's sufficient to wish to show this limit exists or doesn't exist:

$$\lim_{h \to 0} \frac {\arccos(\cos^2h)} {h}$$

Not sure how to proceed.

EDIT: I edited the question to cover $-\pi/2 +2\pi k$ aswell. I feel the question didn't get enough attention.

I have been told by a friend it's solveable using l'hopitals rule looking at the one sided limits, however l'hopitals is 2 months ahead in my course. a little strange this problem appears now.

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  • $\begingroup$ $\lim_{h\rightarrow 0} \arccos(\cos^2 h)$ is $0$. Your logic above didn't follow. $\endgroup$ – Narut Sereewattanawoot Nov 21 '13 at 22:16
  • $\begingroup$ care to propose a way to approach your statement? $\endgroup$ – user7610 Nov 21 '13 at 22:19
  • $\begingroup$ Graphing might be looked down upon but it seems the limit doesn't exist: graph from wolfram $\endgroup$ – user7610 Nov 21 '13 at 22:32
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    $\begingroup$ Right. The limit of $\arccos(\cos^2h)/h$, not of $\arccos(\cos^2h)$. You dropped the $h$ in the denominator in your last line and that's why I said it didn't follow. $\endgroup$ – Narut Sereewattanawoot Nov 21 '13 at 22:35
  • $\begingroup$ Ah. sorry. editing now. $\endgroup$ – user7610 Nov 21 '13 at 22:36
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Since the function has period $\pi$, computing the limit at $\pi/2$ is sufficient. Now, where's the error?

The computation $$ \arcsin\left(\sin^2\left(\frac{\pi}{2}+h\right)\right)-\arcsin\sin^2\frac{\pi}{2}= \arcsin\cos^2 h-\frac{\pi}{2}=-\arccos\cos^2h $$ is correct. Now your limit is exactly computing the derivative of $f(x)=\arccos\cos^2x$ at $0$.

The function is continuous and its derivative, except possibly at zero, exists: $$ f'(x)=\frac{2\sin x\cos x}{\sqrt{1-\cos^2x}}= \frac{2\sin x\cos x}{|\sin x|} $$ and the limit of this at $0$ is different when computed from the left and from the right: $$ \lim_{x\to 0+}f'(x)=1,\quad \lim_{x\to 0-}f'(x)=-1. $$

Now, I know that this uses l'Hôpital's theorem, but this is a guide. If $f$ were differentiable at $0$, the derivative should be $0$, because the function is even and non negative. So we want to show that $$ \frac{f(x)}{x}\ge1 $$ holds for $0<x<k$ (for some $k>0$). This means $$ \arccos\cos^2x>x $$ or, since the cosine is decreasing in $[0,\pi/2]$, $$ \cos^2x<\cos x $$ which is true for $0<x<\pi/2$.

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  • $\begingroup$ you found the limit of it's alleged derivative doesn't exist, which lemma is used to infer the limit doesn't exist in the original function? $\endgroup$ – user7610 Nov 24 '13 at 11:56
  • $\begingroup$ because even the derivative of $|x|$ is not continuous at $0$ and $|x|$ is still continuous... $\endgroup$ – user7610 Nov 24 '13 at 12:04
  • $\begingroup$ @Yaniv.Fish Yes, of course. Your problem is completely equivalent to $f(x)=\arccos\cos^2x$ being differentiable at $0$. I've added a proof independent from considering the limit of the derivatives. $\endgroup$ – egreg Nov 24 '13 at 12:06

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