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I can kind of understand the main direction (slope) of $y$ over the different $x$ intervals, but I can't figure out why the values of $y$ take on the shape of straight lines and not curves looking more like those of sin, cos...

enter image description here

EDIT: I understand that the derivative of Arccos(Sin(x)) gives 1 or -1 depending on the x interval, but it doesn't give me intuition into why that's the case.

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    $\begingroup$ think of $\sin(x)$ as $\cos(\frac{\pi}{2}-x)$ $\endgroup$ – costrom Nov 23 '15 at 18:37
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    $\begingroup$ Note that $\cos (2\pi + x), \cos(2\pi - x) = \cos x$. The normalization of $\cos^{-1}$ chosen by whatever program you're using makes the resulting function only piecewise linear. $\endgroup$ – anomaly Nov 23 '15 at 18:45
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    $\begingroup$ @costrom, ah yes...got it, thanks. $\endgroup$ – jeremy radcliff Nov 23 '15 at 18:57
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Well, to think about the equation $$y=\arccos(\sin(x))$$ let's first look at an easier one, obtained by taking the cosine of both sides (noting that $\cos(\arccos(x))=x$ - that is $\arccos$ is a right-inverse of $\cos$): $$\cos(y)=\sin(x).$$ If we plot this, we get the following graph:

$\hskip1.5in$enter image description here

which is just a series of diagonal lines going in two directions. This mystery unfolds when we rewrite $$\cos(y)=\cos\left(x+\frac{\pi}2\right)$$ and note that, from symmetries and periodicity, we have that $\cos(u)=\cos(v)$ whenever $u=2\pi k \pm v$ for some integer $k$. This gives us that there is a line $y=x+\frac{\pi}2$ included as well as the line $y=x-\frac{\pi}2$ and any shift of these horizontally (or vertically) by $2\pi$, giving rise to the lattice pattern.

However, the definition of $\arccos$ that you are using seems to mandate that its output falls in the interval $[0,\pi]$, hence $y$ must be in there. If we highlight the suitable range of $y$ in gray and trace the lines it covers in red, we recover the pattern you observed:

$\hskip1.5in$enter image description here

which makes it clear that the alternation is merely a small part of a larger pattern, than can't be fully represented due to the fact that $\arccos$ is limited in its range.

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  • $\begingroup$ Thank you, this is very helpful, it clarified a lot of things for me. $\endgroup$ – jeremy radcliff Nov 24 '15 at 18:05
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The inverse of the cosine is defined on $[-1, 1]$ and maps to $[0, \pi]$.
The sine function on $[-\pi/2, \pi/2]$ maps to $[-1,1]$. $$ \arccos(\sin(x)) = \arccos(\cos(\pi/2 - x)) = -x + \pi/2 $$ The sine funcion on $[\pi/2, \pi/2 + \pi]$ is $$ \sin(x) = -\sin(x-\pi) = \sin(\pi - x) $$ and maps to $[-1,1]$. We get $$ \arccos(\sin(x)) = \arccos(\cos(\pi/2 - (\pi -x)) = x - \pi/2 $$ In both cases we can add integer multiples of $2\pi$ to the argument to the cosine function. This gives

$$ \arccos(\sin(x)) = \begin{cases} -x + \pi/2 + 2\pi k & \text{for } x \in [-\pi+2\pi k, \pi/2 + 2\pi k] \\ x - \pi/2 + 2\pi k & \text{for } x \in [\pi/2+2\pi k, \pi/2 + \pi + 2\pi k] \\ \end{cases} $$ where $k \in \mathbb{Z}$.

arccos(sin(x))

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Note that $$cos(x-\pi)=sin(x)$$ $$x-\pi=arccos(sin(x))$$ $$and$$ $$cos(x)=cos(x\pm 2n\pi), n=0,1,2,3,...$$

You get solutions between $\pi$ and $0$ because of the $arccos$. This explains why your solution is linear with a slope of one.

ALSO: $$cos(x)=cos(-x)$$

This means that your solution is positive and negative. Combining all of this results in $$arccos(sin(x))=\pm(x-\pi\pm2n\pi)$$ $$=x\pm(2n-1)\pi,-x\pm(2n-1)\pi$$

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