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Where does an absolute function should appear and why?
Having the following equation:

$$ \cos (\arcsin(\cos(\theta))) = \lvert \sin (\theta) \rvert $$ With $ -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$.

I draw a triangle and easily found the following:

  1. Defining: $\cos(\theta) \equiv \frac{a}{\sqrt{a^2 + 1}} \\$
  2. then $\arcsin(\cos(\theta)) = \frac{\pi}{2} - \theta$
  3. thus $\cos (\arcsin(\cos(\theta))) = \sin(\theta)$

But why should the absolute value appear and on which phase?
How does the absolute value ultimately surrounds the entire expression?

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If we ignored domain and range of functions, we'd end up something like this:

$$\begin{split} \cos(\theta) &= \sin( \frac{\pi}2 + \theta) \\ \arcsin(\cos(\theta)) &= \frac{\pi}2 + \theta (*)\\ \cos(\arcsin(\cos(\theta))) &= \cos(\frac{\pi}2 + \theta) \\ &= -\sin \theta \end{split}$$

However the $(*)$ line isn't quite right. The range of $\arcsin$ is $[-\frac{\pi}2,\frac{\pi}2]$, and the range of the right hand side is $[0,\pi]$. So for positive $\theta$, it's incorrect, it should really be:

$$ \begin{split} \arcsin(\cos(\theta)) &= \left\{ \begin{array}{ll} \frac{\pi}2 + \theta & \mbox{if } \theta < 0 \\ \frac{\pi}2 - \theta & \mbox{if } \theta \geq 0 \end{array} \right. \\ &= \frac{\pi}2 - |\theta| \\ \cos(\arcsin(\cos(\theta))) &= \cos(\frac{\pi}2 - |\theta|) \\ &= \left\{ \begin{array}{ll} -\sin(\theta) & \mbox{if } \theta < 0 \\ \sin(\theta) & \mbox{if } \theta \geq 0 \end{array} \right. \\ \end{split} $$

Since $\sin \theta < 0$ for $-\frac{\pi}2 \leq \theta \leq 0$, that simply reduces to $\cos(\arcsin(\cos(\theta))) = |\sin(\theta)|$.

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  • $\begingroup$ Thank you, this is very clear to me now. $\endgroup$ – Dor Jun 4 '15 at 16:15
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the left hand side of your equation is an even function of $\theta,$ therefore the right hand side must also be an even function of $\theta.$ that is the reason you see the absolute value sign on the right hand side.


p.s. if you take $0 < t < \pi/2,$ then $\sin^{-1}(\cos t) = \pi/2 - t$ and $\cos(\sin^{-1}(\cos t) ) = \cos(\pi/2 -t) = \sin t = |\sin t|.$

if you take $-\pi/2< t < 0,$ then $\sin^{-1}(\cos t) = \pi/2 + t$ and $\cos(\sin^{-1}(\cos t) ) = \cos(\pi/2 +t) = -\sin t = |\sin t|.$

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  • $\begingroup$ Does the absolute function appear as a result of $\arcsin(cos(\theta))$? Whatever happen on those phases - is obscure.. $\endgroup$ – Dor Jun 4 '15 at 15:57
  • $\begingroup$ @Dor, i have edited my post. see if that helps. $\endgroup$ – abel Jun 4 '15 at 16:07
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Because $\cos$ is positive between $-\frac{\pi}{2}\le \theta \le \frac{\pi}{2}$. Note further that sine is negative between $-\frac{\pi}{2}\le \theta \lt 0$.

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  • $\begingroup$ Could you please elaborate? $\endgroup$ – Dor Jun 4 '15 at 15:45
  • $\begingroup$ @Dor, on left side we have even function, and odd is on left $\endgroup$ – Michael Galuza Jun 4 '15 at 15:47
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$1$. Defining: $\cos(\theta)\equiv\dfrac{a}{\sqrt{a^2+1}}$

This is your first mistake, since we can also have $\cos(\theta)\equiv-~\dfrac{a}{\sqrt{a^2+1}}$

$2$. Then $\arcsin(\cos(\theta))=\dfrac\pi2-\theta$

This is your second mistake, since we can also have $\arcsin(\cos(\theta))=\dfrac\pi2+\theta$

$3$. Thus $\cos(\arcsin(\cos(\theta)))=\sin(\theta)$

This is your third mistake, which follows from the previous two.

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