If $\arcsin x+\arcsin y=\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})$

Then the area represented by the locus of point $(x,y)$

if it is given that $|x|,|y|\leq 1$

My Try: Put $x=\sin \alpha$ and $y=\sin \beta $

and $\alpha,\beta \in [-90^\circ,90^\circ]$ and $\alpha+\beta \in [-180^\circ,180^\circ]$

$\alpha+\beta = \arcsin(\sin (\alpha+\beta))$ which is possible

when $\alpha+\beta\in [-90^\circ,90^\circ]$

Could some help me what is the area enclosed by its locus.Thanks


We are given the square $Q:=[-1,1]^2$ in the $(x,y)$-plane and are told to determine the domain $D$ consisting of all points $(x,y)\in Q$ satisfying the equation $$\arcsin x+\arcsin y=\arcsin\bigl(x\sqrt{1-y^2}+y\sqrt{1-x^2}\bigr)\ .\tag{1}$$ To this end we draw in a second figure the square $\hat Q:=\bigl[-{\pi\over2},{\pi\over2}\bigr]^2$ in the $(\alpha,\beta)$-plane and consider the map $$\psi:\quad \hat Q\to Q,\qquad(\alpha,\beta)\mapsto\left\{\eqalign{x&=\sin\alpha \cr y&=\sin\beta\cr}\right.$$ which maps the square $\hat Q$ bijectively onto $Q$. We have $$\psi^{-1}:\quad Q\to\hat Q,\qquad (x,y)\mapsto\left\{\eqalign{\alpha&=\arcsin x \cr \beta&=\arcsin y\cr}\right.\quad.$$ Furthermore one has $$\sqrt{1-x^2}=\cos\alpha,\quad \sqrt{1-y^2}=\cos\beta\ .$$ The equation $(1)$ reads in the variables $(\alpha,\beta)\in\hat Q$ as follows: $$\alpha+\beta=\arcsin\bigl(\sin\alpha\cos\beta+\sin\beta\cos\alpha\bigr)=\arcsin\bigl(\sin(\alpha+\beta)\bigr)\ ,$$ and this can be rewritten as $$\alpha+\beta=\left\{\eqalign{\alpha+\beta\qquad&\bigl(|\alpha+\beta|\leq{\pi\over2}\bigr)\cr \pi-(\alpha+\beta)\quad&\bigl(\alpha+\beta\geq{\pi\over2})\cr -\pi-(\alpha+\beta)\quad&\bigl(\alpha+\beta\leq-{\pi\over2}\bigr)\ .\cr}\right.$$ When $|\alpha+\beta|\leq{\pi\over2}$ this requires nothing. If $\alpha+\beta\geq{\pi\over2}$ this says that $\alpha+\beta={\pi\over2}$, and if $\alpha+\beta\leq-{\pi\over2}$ this says that $\alpha+\beta=-{\pi\over2}$.

We therefore obtain the desired domain $\hat D\subset\hat Q$ by cutting off the two triangles from $\hat Q$ on which $|\alpha+\beta|>{\pi\over2}$. In the original $(x,y)$-figure we obtain the desired domain $D\subset Q$ by cutting off the $\psi$-images of these triangles, which are the points in the first and third quadrants outside the circle $x^2+y^2=1$. Therefore one has $${\rm area}(D)=2+{\pi\over2}\ .$$


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