Differential of $\arccos(\sin x)$

Question

By the chain rule, $$\frac{d}{dx}\arccos(\sin x) = \cos x \;\cdot\; \frac{-1}{\sqrt{1-(\sin x)^2}} = \frac{-\cos x}{\sqrt{\cos^2x}} = -1$$

Without assuming the differential of $\arccos(x)$, one can also let $y = \arccos(\sin x)$ and then differentiate $\cos y = \sin x$ implicitly to eventually reach the same result.

And yet, $$ \int -1 \;dx = -x + c $$

Does this mean that $\arccos(\sin x)$ can be expressed in the form $-x + c$? If so, what is the value of $c$ and how can this be proven? If not, where is the flaw in my logic above?

Answer 1

Actually, $$\frac{d}{dx}\left[\arccos (\sin(x))\right]=\frac{-\cos(x)}{|\cos(x)|}=\begin{cases}-1\qquad\text{if $x\in\left(-\frac{\pi}{2}\pm 2k\pi,\frac{\pi}{2}\pm 2k\pi\right)$}\\1\qquad~~~\text{if $x\in\left(\frac{\pi}{2}\pm 2k\pi,\frac{3\pi}{2}\pm 2k\pi\right)$}\end{cases}$$ and from this we can show that $\arccos (\sin(x))$ is not a linear function, but rather a piecewise "zig-zag" function (I recommend graphing the function).

Answer 2

Hint: Draw a right triangle with angle $x$, and see that $\arccos(\sin x)=\frac\pi2-x$ (for $0\leq x\leq\frac\pi2$).

Answer 3

We have

$$\sqrt{\cos^2 x}=\vert\cos x\vert=\left\{\begin{array}\\\cos x \;\text{if}\; x\in[-\frac\pi2+2k\pi,\frac\pi2+2k\pi]\\-\cos x \;\text{otherwise} \end{array}\right.$$ And by periodicity it suffices to find the expression of $\arccos(\sin x)$ on $[0,2\pi]$.

Answer 4

$\sin x = \cos (\frac \pi2 - x)\\ \arccos (\sin x) = \arccos (\cos (\frac \pi2 - x)) = \frac {\pi}{2} - x$

Or at least it does for the correct intervals of $x.$

Since $\sin x$ is perodic, and $\arccos x$ is a 1-1 function

$\arccos (\sin x)$ is a triangle wave.