Exercise arcsin I have this exercise:
Consider the equation $\arcsin(x)=\arcsin(a)+\arcsin(b)$, $a,b\in\mathbb{R}$
1)Study according to the values ​​of a and b the existence of solution of the equation
I say, for $x,a,b\in [-1,1]$ I have:
$x=\sin(\arcsin(a)+\arcsin(b))=\sin(\arcsin(a))\cos(\arcsin(b))+\cos(\arcsin(a))\sin(\arcsin(b))$
then
$x=a\sqrt{1-b^2}+b\sqrt{1-a^2}$
the other question 
draw the family of the point $M(a,b)$ for which the equation has a solution 
How to do this ?
 A: Let $f(a,b)=\arcsin(a)+\arcsin(b)$.  We assume in our analysis that $0\le a\le 1$ and $0\le b\le 1$.  Analysis for $a$ and $b$ in the other quadrants are left as exercises.

We wish to restrict $f(a,b)$ such that $0\le f(a,b)\le \pi/2$.  That is, $f(a,b)=\arcsin(x)$ for some $x\in [0,1]$.

First note that the arcsine function increases monotonically.  Hence, $f(a,b)=\arcsin(a)+\arcsin(b)$ increases in each of its arguments $a$ and $b$.


ASIDE:  Use of Calculus to Show that $\arcsin(x)$ Increases Monotonically
One way to see that $f(a,b)$ increases monotonically in $a$ and  $b$ to note  its first partial derivatives given by 
$$\frac{\partial f(a,b)}{\partial a}=\frac1{\sqrt{1-a^2}}\\\\
\frac{\partial f(a,b)}{\partial b}=\frac1{\sqrt{1-b^2}}$$
are strictly positive.    


Next, note that $\sin(f(a,b))=a\sqrt{1-b^2}+b\sqrt{1-a^2}$.  Inasmuch as the value of the sine function cannot exceed $1$, the maximum value of $a\sqrt{1-b^2}+b\sqrt{1-a^2}$ is $1$.  
It is straightforward to show that the locus of points for which $a\sqrt{1-b^2}+b\sqrt{1-a^2}=1$ is given by the set $\{(a,b)|a^2+b^2=1\}$.


ASIDE:  Use of Calculus to Find Where $\sin(f(a,b))$ Is a Maximum Value
As easy way to show that $\sin(f(a,b))=1$ when $a^2+b^2=1$ is to set its first partial derivatives, given by
$$\frac{\partial \sin(f(a,b))}{\partial a}=\sqrt{1-b^2}-\frac{ab}{\sqrt{1-a^2}}\\\\
\frac{\partial \sin(f(a,b))}{\partial b}=\sqrt{1-a^2}-\frac{ab}{\sqrt{1-b^2}}$$
  to $0$ and solve the resulting equations.


So, we now have found that $\sin(f(a,b))$ has its maximum value on the circle $a^2+b^2=1$ for which $\sin(f(a,b))=1$.  Therefore, $f(a,b)=\pi/2$ on the circle $a^2+b^2=1$.

Inasmuch as $f(a,b)$ is increasing in both $a$ and $b$, we find that $f(a,b)>\pi/2$ when $a^2+b^2>1$.  Hence, the solution space is given by 
$$a^2+b^2\le 1$$
for $a\in[0,1]$ and $b\in [0,1]$.

