Spivak, Ch. 15, Trigonometric Functions", Proof of $\arcsin{\alpha}+\arcsin{\beta}=\arcsin{(\alpha \sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$. The following problem is from Chapter 15 "Trigonometric Functions" from Spivak's Calculus



*Prove that

$$\arcsin{\alpha}+\arcsin{\beta}=\arcsin{(\alpha
 \sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$$
indicating any restrictions on $\alpha$ and $\beta$.

My question is about the restrictions on $\alpha$ and $\beta$. I will show the solution from the solution manual first, and then specify my question.
Here is the solution manual solution

From the addition formula for $\sin$ we obtain, for $|\alpha|\leq 1$
and $|\beta|\leq 1$,
$$\sin{(\arcsin{\alpha}+\arcsin{\beta})}=\sin{(\arcsin{\alpha})\cos{(\arcsin{\beta})}}+\cos{(\arcsin{\alpha})\sin{(\arcsin{\beta})}}$$
$$=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}$$

Note that though it is not mentioned a significant step is taken in showing that
$$\cos{(\arcsin{x})}=\sqrt{1-x^2}$$
This is done by computing the derivative of $\sin(x)$ as the reciprocal of the derivative of $\arcsin$ at $\sin{x}$.

Consequently
$$\arcsin{\alpha}+\arcsin{\beta}=\arcsin{(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}\tag{1}$$
provided that $-\pi/2\leq\arcsin{\alpha}+\arcsin{\beta}\leq\pi/2$. [If
$\pi/2<\arcsin{\alpha}+\arcsin{\beta}\leq \pi$, the right side must be
replaced with
$\pi-\arcsin{(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$, and if
$-\pi\leq \arcsin{\alpha}+\arcsin{\beta}\leq -\pi/2$, replaced with
$-\pi-\arcsin{(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$.]

My question is about the last paragraph.
Let's start at the point where we have
$$\sin{(\arcsin{\alpha}+\arcsin{\beta})}=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}$$
and we want to take the $\arcsin$ of each side to obtain $(1)$.
Now, $\arcsin{\alpha}$ and $\arcsin{\beta}$ are each in $(-\pi/2, \pi/2)$. Their sum is in $(-\pi, \pi)$, and the whole left expression is thus in $[-1,1]$.
$\pm 1$ occur when $\arcsin{\alpha}+\arcsin{\beta}=\pm \frac{\pi}{2}$.
If one of these two cases occurs, then $\arcsin$ isn't defined for $\sin{(\arcsin{\alpha}+\arcsin{\beta})}=\pm 1$.
EDIT: the sentence above is is incorrect. I confused $\arcsin$ with its derivative $\arcsin'$.
So my first question is about the following snippet of the solution manual solution

provided that $-\pi/2\leq\arcsin{\alpha}+\arcsin{\beta}\leq\pi/2$

where the values $\pi/2$ and $-\pi/2$ are included in the possibilities. Are these values really allowed?
EDIT: given the first edit above, yes, these values are allowed.
My second question is about the discussion when $\pi/2<\arcsin{\alpha}+\arcsin{\beta}\leq \pi$ or $-\pi\leq \arcsin{\alpha}+\arcsin{\beta}\leq -\pi/2$. What is happening in these cases, I don't know what Spivak's solution manual did there.
 A: I'd start from the conditions first.
When it exists, the right-hand side is an angle in the interval $[-\pi/2,\pi/2]$, so the left-hand side must as well.
The existence condition is (with $a=\alpha,b=\beta$ for ease of typing)
$$
\textstyle \lvert\, a\sqrt{1-b^2}+b\sqrt{1-a^2}\,\rvert\le 1\tag{*}
$$
In particular, $|a|\le1$ and $|b|\le1$, but this was already known from examining the left-hand side.
Condition (*) is satisfied for $|a|\le 1$ and $|b|\le 1$ by using Cauchy-Schwarz with the vectors
$$
\textstyle v=(a,\sqrt{1-a^2}\,)\qquad w=(\sqrt{1-b^2},b)
$$
and the standard CS inequality $|v\cdot w|\le\|v\|\,\|w\|$ becomes
$$
\textstyle \lvert\, a\sqrt{1-b^2}+b\sqrt{1-a^2}\,\rvert\le\sqrt{a^2+1-a^2}\sqrt{1-b^2+b^2}=1
$$
OK, this leaves only the condition that
$$
-\frac{\pi}{2}\le \arcsin a+\arcsin b\le\frac{\pi}{2} \tag{**}
$$
Under this condition, we can take the sine of both sides. From the left-hand side we get
$$
\textstyle
\sin(\arcsin a)\cos(\arcsin b)+\cos(\arcsin a)\sin(\arcsin b)
=a\sqrt{1-b^2}+b\sqrt{1-a^2}
$$
because, for $|x|\le 1$, we have
$$
\textstyle\cos(\arcsin x)=\sqrt{1-\sin^2(\arcsin x)}=\sqrt{1-x^2}
$$
because $-\pi/2\le\arcsin x\le\pi/2$ and so $\cos\arcsin x\ge0$.
Thus we conclude that, for $|a|\le1$ and $|b|\le 1$,

$\arcsin a+\arcsin b=\arcsin\bigl(a\sqrt{1-b^2}+b\sqrt{1-a^2}\,\bigr)$ if and only if $\lvert\arcsin a+\arcsin b\rvert\le \pi/2$

A: Let's start at the point where we have
$$\sin{(\arcsin{\alpha}+\arcsin{\beta})}=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}\tag{1}$$
and we want to take the $\arcsin$ of each side to obtain $(1)$.
First, let's just note that $\alpha$ and $\beta$ are each in $[-1,1]$, and thus $\arcsin{\alpha}$ and $\arcsin{\beta}$ are each in $[-\pi/2,\pi/2]$. The sum $(\arcsin{\alpha}+\arcsin{\beta})$ is thus in $[-\pi,\pi]$.
We need to consider what happens for all such possible values of $(\arcsin{\alpha}+\arcsin{\beta})$.
Case 1: $-\pi/2\leq \arcsin{\alpha}+\arcsin{\beta}\leq\pi/2 $
The equation
$$\arcsin{(\sin{(\arcsin{\alpha}+\arcsin{\beta})})}=\arcsin{\alpha}+\arcsin{\beta}\tag{2}$$
is true.
We can thus take the $\arcsin$ of both sides of $(1)$ and end up with
$$\arcsin{\alpha}+\arcsin{\beta}=\arcsin{(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$$
Case 2: $\pi/2\leq \arcsin{\alpha}+\arcsin{\beta}\leq\pi $
Equation $(2)$ is no longer true (indeed, it doesn't even make sense because of the domain of $\arcsin$).
However, it is true that $\sin{x}=\sin{\pi-x}$. Thus,
$$(\pi-\arcsin{\alpha}-\arcsin{\beta}) \in (-\pi/2,0]$$
and
$$\sin{(\pi-\arcsin{\alpha}-\arcsin{\beta})}=\sin{(\arcsin{\alpha}+\arcsin{\beta})}=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}\tag{3}$$
Now it is true that
$$\arcsin{(\sin{(\pi-\arcsin{\alpha}-\arcsin{\beta})})}=\pi-\arcsin{\alpha}-\arcsin{\beta}$$
So we can take the $\arcsin$ of the first and last expressions in $(3)$ to obtain
$$\pi-\arcsin{\alpha}-\arcsin{\beta}=\arcsin{(\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2})}$$
and the desired
$$\arcsin{\alpha}+\arcsin{\beta}=\pi-\arcsin{(\alpha\sqrt{1-\beta^2}-\beta\sqrt{1-\alpha^2})}$$
Case 3: $-\pi\leq \arcsin{\alpha}+\arcsin{\beta}<-\pi/2$
This case is similar to case 2. Since
$$\sin{(-\pi-x)}=\sin{x}$$
we have that
$$(-\pi-\arcsin{\alpha}-\arcsin{\beta}) \in (-\pi/2,0]$$
$$\sin{(-\pi-\arcsin{\alpha}-\arcsin{\beta})}=\alpha\sqrt{1-\beta^2}+\beta\sqrt{1-\alpha^2}\tag{3}$$
and taking the $\arcsin$ of each side we obtain
$$-\pi-\arcsin{\alpha}-\arcsin{\beta}=\arcsin{(\alpha\sqrt{1-\beta^2}-\beta\sqrt{1-\alpha^2})}$$
$$\arcsin{\alpha}+\arcsin{\beta}=-\pi-\arcsin{(\alpha\sqrt{1-\beta^2}-\beta\sqrt{1-\alpha^2})}$$
$\blacksquare$
Spivak's solution manual sure does skip a lotta steps.
A: HINT

*

*$\arcsin(\sin\alpha)=\alpha$ only when $\alpha\in[-\frac\pi2,\frac\pi2]$


*$\arcsin(\sin\alpha)=\pi-\alpha$ when $\alpha\in[\frac\pi2,\pi]$


*$\arcsin(\sin\alpha)=-\pi-\alpha$ when $\alpha\in[-\pi,-\frac\pi2]$
