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The problem

The task is to find the intervals for which the function $f \left( x \right) =\arcsin \left( x \right) -\arcsin \left( \sqrt{1-x^{2}} \right)$ is constant and for which it takes the form of:

$f \left( x \right) =a \cdot \arcsin \left( x \right) +b, \\a,b=const, a \neq 0$

Constants a,b are also to be calculated.

What I have done so far

I've defined the functions domain, calculated its derivative and the derivatives domain. $D_{f} x \in \left\langle -1,1\right\rangle\\ f' \left( x \right) =\frac{\sqrt{x^{2}}+x}{\sqrt{x^{2}-x^{4}}}\\ D_{f'} x \in \left(-1,0\right) \cup \left(0,1\right)\\ f' \left( x \right) =0 \Leftrightarrow \sqrt{x^{2}}+x=0\\ \left|x\right|=-x \Leftrightarrow x \le 0$

Which indicates that our function is constant for x'es from the interval (-1,0). That's where I currently am. I don't know how to go about the second part of the task, that is finding the intervals for which the function behaves as: $f \left( x \right) =a \cdot \arcsin \left( x \right) +b$ and calculating a,b.

Any help, greatly appreciated.

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As $f$ is defined on $[-1,\,1]$, you may, for any $x\in[-1,\,1]$ take a unique $\alpha\in[-\pi/2,\,\pi/2]$ such that $\sin\alpha=x$. This number is by definition called $\alpha=\arcsin x$. Now you know that $\sin^2\alpha+\cos^2\alpha=1$ and $\cos\alpha\geq0$. This gives $\cos^2\alpha=1-\sin^2\alpha=1-x^2$ and therefore $\cos\alpha=\sqrt{1-x^2}$. Since $\sin(\pi/2-\alpha)=\cos\alpha$, you can conclude that $$\arcsin(\cos\alpha)=\pi/2-\alpha \quad\text{if}\quad\pi/2-\alpha\in[-\pi/2,\,\pi/2],$$ but if $\pi/2-\alpha\notin[-\pi/2,\,\pi/2]$ this means that $\alpha<0$, and thus that $\alpha\in[-\pi/2,\,0)$ you should then use $\sin(\pi/2+\alpha)=\cos\alpha$ and get $$\arcsin(\cos\alpha)=\pi/2+\alpha \quad\text{if}\quad\alpha\in[-\pi/2,\,0].$$ Altogether, you have, for $x\geq0$, $f(x)=\alpha-\pi/2+\alpha=2\alpha-\pi/2$ and for $x\leq0$ $f(x)=\alpha-\pi/2-\alpha=-\pi/2$.

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I really like your approach. One question though, why do we "know" that $\cos\alpha\ \geq0$? – Tom Feb 27 '14 at 14:20
We have $\cos\alpha\geq0$ because $\alpha\in[-\pi/2,\,\pi/2]$. – Tom-Tom Feb 27 '14 at 14:32

I do not know what you are allowed to use. From Abramowitz/Stegun 4.3.45 and 4.4.2 (or and we have $$\arcsin(\sqrt{1-x^2}) = \arccos x, \quad 0 \le x \le 1$$ $$\arccos x = \frac{\pi}{2} - \arcsin x$$ Combining these results you get for $0 \le x \le 1$ $$\arcsin x-\arcsin(\sqrt{1-x^2}) = \arcsin x -\left(\frac{\pi}{2} - \arcsin x\right) = 2\arcsin x -\frac{\pi}{2}$$ and therefore $a=2$ and $b=-\frac{\pi}{2}.$

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the second equality found in Abramowitz is valid only for positive $x$. For negative $x$, it is $\arccos x=\pi/2+\arcsin x$. – Tom-Tom Feb 27 '14 at 12:30
@V. Rossetto: No, the formula I gave is correct. Look for example at $x=-\frac{1}{2}:$ $$\arccos(-\frac{1}{2}) = \frac{2\pi}{3},\quad \arcsin(-\frac{1}{2}) = -\frac{\pi}{6}.$$ Your formula gives the wrong result $$\frac{\pi}{2}+\arcsin(-\frac{1}{2})=\frac{\pi}{2}-\frac{\pi}{6} = \frac{\pi}{3},$$ while the other formula is correct $$\frac{\pi}{2}-\arcsin(-\frac{1}{2})= \frac{\pi}{2}+\frac{\pi}{6}= \frac{2\pi}{3}$$ – gammatester Feb 27 '14 at 12:48
You are right, I mixed up the propagation of the minus sign. The correct derivation for $x<0$ is $\arcsin\sqrt{1-x^2}=\arcsin\sqrt{1-(-x)^2}=\arccos(-x)=\pi-\arccos x=\pi/2+\arcsin x$. – Tom-Tom Feb 27 '14 at 14:45


Let us consider $$f \left( x \right) =\arcsin \left( x \right) -\arcsin \left( \sqrt{1-x^{2}} \right)$$ and $$g \left( x \right) =a \cdot \arcsin \left( x \right) +b$$ In order they be identical (and this can only happen in the range [$0<x<1$]), a minimum requirement is that there will be two points where their values should be equal. Let us take the two limiting points $x=0$ and $x=1$. So, $f(0)=-\frac{\pi }{2}$ and $g(0)=b$, $f(1)=\frac{\pi }{2}$ and $g(1)=a \frac{\pi }{2}+b$; you then have two very simple linear equations to solve for $a$ and $b$.

I am sure that you can take from here.

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In your answer you implicitely use that one of the intervals is $[0,\,1]$. – Tom-Tom Feb 27 '14 at 11:09
@V.Rossetto. Yes, indeed. I do not see how $f$ and $g$ could be identical in [$-1,0$] since $f$ is constant over that range and that the OP precised that $a \neq 0$. Please tell me if I am wrong. Thanks. – Claude Leibovici Feb 27 '14 at 11:14
In his question, the OP asked for a constant solution or a solution of the form you use. – Tom-Tom Feb 27 '14 at 12:28

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