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I have a question.

Is $\arcsin(\sin (x))$ or $\sin(\arcsin(x))$ always equal to $x$?

And also for all other trigonometric ratios?

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4 Answers

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Only on $\Big[-\frac\pi2,\frac\pi2\Big]$. Since sine is a periodic function, with $\sin x=\sin\big(\pi-x\big)=\sin\big(2\pi+x\big)$, $\arcsin t$ can be any number of the form $x,(2k+1)\pi-x,2k\pi+x$, with $k\in\mathbb Z$ and $t=\sin x$.

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There's a little caveat though, while $\sin\circ\arcsin$ is equal to $\mathrm{id}_{[-1,+1]}$, the function $\arcsin\circ\sin$ is defined on all of $\Bbb R$ but takes its values in $[-\frac\pi2,+\frac\pi2]$. It is actually a piecewise linear function, linear on every interval $I_k=[k\pi-\frac\pi2,k\pi+\frac\pi2]$ that takes on the value $(-1)^k\frac\pi2$ at $k\pi+\frac\pi2$ for every $k\in\Bbb Z$. It looks like a triangular wave function.

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For a function $f \,:\, A \to B$ to have a left-inverse, i.e. a $g \,:\, B \to A$ such that $g(f(x)) = x$ for all $x \in A$, it is necessary and sufficient for $f$ to be injective, i.e. it must not map two points in $A$ to the same point in $B$.

For a function $f \,:\, A \to B$ to have a right-inverse, i.e. a $g \,:\, B \to A$ such that $f(g(x)) = x$ for all $x \in B$, it is necessary and sufficient for $f$ to be surjective, i.e. it must reach every point in $B$.

$\sin \,:\, \mathbb{R} \to \mathbb{R}$ is neither injective (since $\sin(x + 2\pi n) = \sin x$) nor surjective (since $\sin x \in [-1,1]$), so $\sin$ neither has a left nor a right inverse.

$\sin$ can, however, be made surjective by interpreting it as a function $\mathbb{R} \to [-1,1]$ instead. This $\sin$ function then has an right-inverse, but only on $[-1,1]$, and the right-inverse is $\arcsin \,:\, \mathbb[-1,1] \to \mathbb{R}$.

Similarly, $\sin$ can be made injective by interpreting it as a function $[-\frac{\pi}{2},\frac{\pi}{2}] \to \mathbb{R}$. This restriction then has a left-inverse, which is $\overline{\arcsin} \,:\, \mathbb{R} \to [-\frac{\pi}{2},\frac{\pi}{2}]$, which is the usual $\arcsin$ function on $[-1,1]$, and just maps the rest of $\mathbb{R}$ to some arbitrary value.

By combining the two restrictions, $\sin$ can even be made bijective, as a function $\sin \,:\, [-\frac{\pi}{2},\frac{\pi}{2}]  \to [-1,1]$, and it's (left- and right-) inverse is $\arcsin \,:\, \mathbb[-1,1] \to [-\frac{\pi}{2},\frac{\pi}{2}]$.

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This is true FOR EVERY x that is in between -1 and 1. If you have sin(arcsine(x)), the x is the ratio that you get for the angle arcsine(x), so it's given right there. arcsine(sin(x)) is where you are limited to angles from -pi/1 to pi/2, because arcsine only spits out angles on that interval, so if you put in, say, (2pi)/3, which is in Q2, you can't say that the answer is (2pi)/3 because arcsine doesn't output that angle, you would have to find the angle in Q4 that has the same negative ratio that (2pi)/3 has.

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