4
$\begingroup$

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

$\endgroup$
3
+50
$\begingroup$

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 $${\rm area}(D)=2+{\pi\over2}\ .$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.