Restriction of domains and inverse trigonometric functions

Question

Because I had to restrict the sine domain to $ \left[ -\frac{\pi }{2},\frac{\pi }{2}\right] $ is it correct to write, let say $$\arcsin \left( \sin \left( \frac{3\pi }{2}\right) \right) =-\frac{\pi }{2}\ ?$$

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EDIT

For example, if you change the name from "sin" to "Sin" when you restrict the domain, and then define "arcSin" or something like that, and use this new defined functions in the formula, it is clear that the argument of the "Sine" function is out of the domain. But here, the "sin" function is the original or the one I made with a restricted domain?

Maybe I had to write $\arcsin (\sin (7\pi/2))$, for example, in my question. What I am trying to argue is that with the sine domain restriction, maybe is no longer correct to write, or to justified, the input of an angle out of the new domain of the new sine function.

Answer 1

By definition, $\arcsin x = \theta$ if $\theta$ is the unique angle in the interval $[-\pi/2, \pi/2]$ such that $\sin\theta = x$. Since $$\arcsin\left[\sin\left(\frac{3\pi}{2}\right)\right] = \arcsin(-1)$$ and $-\pi/2$ is the unique angle $\theta$ in the interval $[-\pi/2, \pi/2]$ such that $\sin\theta = -1$, your answer is correct.

Note that $\arcsin(\sin\theta) = \theta$ if and only if $\theta \in [-\pi/2, \pi/2]$ since the function $g: [-1, 1] \to [-\pi/2, \pi/2]$ defined by $g(x) = \arcsin x$ is the inverse of the function $f: [-\pi/2, \pi/2] \to [-1, 1]$ defined by $f(x) = \sin x$.

If $\theta \notin [-\pi/2, \pi/2]$, $\arcsin(\sin\theta) \neq \theta$ since $\theta \notin [-\pi/2, \pi/2]$. We can still evaluate $\arcsin(\sin\theta)$ in this case since the function $h: \mathbb{R} \to [-1, 1]$ defined by $h(x) = \sin x$ is defined for every real number $x$.

Answer 2

Because I had to restrict the sine domain to $ \left[ -\frac{\pi }{2},\frac{\pi }{2}\right] $ is it correct to write, let say $$\arcsin \left( \sin \left( \frac{3\pi }{2}\right) \right) =-\frac{\pi }{2}\ ?$$

1. Reading through your comments, what you're trying to ask is this:

   • "Since arcsin has principal range $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right],$ then this means that in the given context $\sin$ accepts only $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2}\right],$ so how is $$\arcsin \left( \sin \left( \frac{3\pi }{2}\right) \right)$$ even a valid expression? Is it really correct to say that it equals $-\dfrac\pi2$ ?"

   The answer is No, Yes, Yes: the composed function $$\arcsin\left(\sin(x)\right)$$ has the same domain as $\sin$, which is $\mathbb R,$ and has principal range $\left[ -\frac{\pi }{2},\frac{\pi }{2}\right].$

   To be clear: whatever restriction to do to $\arcsin$ affects only $\arcsin$ itself, and has absolutely no impact on other trigonometric functions in the same context, or their domains or ranges. What is important to check is that the range of $\sin$ is a subset of the domain of $\arcsin;$ this is indeed satisfied.

2. \begin{align}\\ \forall x{\in}\left[-\frac\pi2,\frac\pi2\right]\;\arcsin\left(\sin(x)\right) &= x\\ \arcsin\left(\sin(x)\right) &\not\equiv x\\ \sin\left(\arcsin(x)\right)&\equiv x.\end{align}

Answer 3

Let us regard the sine as a function $$\sin : \mathbb R \to [-1,1] .$$ It maps each interval $I_k = [k\pi - \frac{\pi}{2},k\pi + \frac{\pi}{2}] $, $k \in \mathbb{Z}$, bijectively onto $[-1,1]$. Therefore each of the functions $\sin_k = \sin \mid_{I_k} : I_k \to [-1,1]$ has an inverse $\arcsin_k : [-1,1] \to I_k$.

The functions $\arcsin_k$ are the branches of the arcsine and for $k=0$ we obtain the principal value which is simply denoted by

$$\arcsin : [-1,1] \to [- \frac{\pi}{2}, \frac{\pi}{2}] .$$

However, we may also regard the functions $\arcsin_k$ as real-valued functions $$\arcsin_k : [-1,1] \to \mathbb R$$ which allows to consider the compositions $$\sin \circ \arcsin_k : [-1,1] \to [-1,1] ,\tag{1}$$ $$\arcsin_k \circ \sin : \mathbb R \to \mathbb R . \tag{2}$$

For each $k$ the function $\sin \circ \arcsin_k$ is the identity in $[-1,1]$, that is $$\sin(\arcsin_k(y)) = y \text{ for all } y \in [-1,1] .$$ This means that each branch of the arcsine is a right inverse for the sine.

In contrast no function $\arcsin_k \circ \sin$ can be the identity on $\mathbb R$ because the image is $I_k \subsetneqq \mathbb R$; we have $$\arcsin_k(\sin(x)) = x \text{ if and only if } x \in I_k .$$

In particular, for the principal branch $\arcsin$ the expression $$\arcsin(\sin(x)) \in [- \frac{\pi}{2}, \frac{\pi}{2}] $$ is defined for all $x \in \mathbb R$, but $$\arcsin(\sin(x)) = x \text{ if and only if } x \in [- \frac{\pi}{2}, \frac{\pi}{2}] .$$

In fact we get $$\arcsin(\sin(\frac{3\pi}{2})) = - \frac{\pi}{2} .$$