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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 ?

Vishu
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1 Answers1

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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 $b\in [0,1]$.

Mark Viola
  • 179,405
  • We don't study all this is there an other method ? –  May 08 '20 at 09:52
  • Do you understand that the arcsine is an increasing function (I used calculus, but you can show this another way)? If you know that $a\sqrt{1-b^2}+b\sqrt{1-a^2}$ is restricted to have a maximum value of $1$ (Why?), can you see that the maximum occurs when $a^2+b^2=1$ (Just set $a\sqrt{1-b^2}+b\sqrt{1-a^2}$ to $1$ and turn the crank)? So when $a^2+b^2=1$, $a\sqrt{1-b^2}+b\sqrt{1-a^2}=1$. This means that $arcsin(x)$ must be $\pi/2$. And since $\arcsin(a)+\arcsin(b)$ increase in both $a$ and $b$, at any values of $a$ and $b$ outside the circle $a^2+b^2=1$ are not permitted. That's all there is. – Mark Viola May 09 '20 at 22:28
  • Hi Linda. I've edited to provide a way forward without calculus. I've left sections ("ASIDES") that are for those familiar with calculus, but you can gloss over those. Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola May 12 '20 at 16:09
  • i do an other method I found $ab\leq0$ or $a^2+b^2\leq 1$ –  May 12 '20 at 16:52
  • Hi Linda. What was your method? Was this answer useful? – Mark Viola May 13 '20 at 16:38
  • https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y/674096?noredirect=1#comment7540541_674096 –  May 14 '20 at 21:38