# Equivalent inequality of $\left|\sin^{-1}(x) + \sin^{-1}(y)\right|\le\frac{\pi}{2}$ only in terms of x and y?

I was working with the inequality, $$\left|\sin^{-1}(x) + \sin^{-1}(y)\right|\le\frac{\pi}{2} , \{|x|,|y|\le1\}.$$

I was trying to find an inequality only in terms of x and y without containing any trigonometric functions. Then after some hit and trial I found following beautiful inequality, $$\biggl|x|x| + y|y| \biggr|\le 1 , \{|x|,|y|\le1\}.$$ But the problem is I couldn't derive this inequality and couldn't show their equivalence. I know they are equivalent by plotting them in Desmos(Function Plotting Program). Can you derive the proof showing the equivalence of these two inequalities ?

Note : The above inequality is the condition for the following relation to satisy, $$\sin^{-1}(x) + \sin^{-1}(y) = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right)$$ I had thought this formula works for every value of $$|x|,|y|\le1$$ but turns out it doesn't. It only works when x and y satisfy above inequality.

• You shall consider to take the sine of $-\frac{\pi}{2}\leq\sin^{-1}(x)+\sin^{-1}(y)\leq\frac{\pi}{2}$ -- take care of the fact that sine can be non-increasing sometimes. You will have to use $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$ as well as $\cos(\sin^{-1}(a))=\pm\sqrt{1-a^2}$. Dec 16, 2020 at 12:33
• I still can't do anything. Can you please explain with steps ? Actually the first equality was derived from $\sin(a+b) = \sin(a)\cos(b) +sin(b)\cos(a)$ and I don't think proceeding in reverse direction will yield anything. Dec 16, 2020 at 12:55
• $$\left|\sin^{-1}(x) + \sin^{-1}(y)\right|\le\frac{\pi}{2} , \{|x|,|y|\le1\}.$$ is not true. Dec 16, 2020 at 13:48
• It is not a statement. It is a condition which is satisfied by some values of x and y less than 1. Dec 16, 2020 at 13:52

We need $$\cos(\sin^{-1}x+\sin^{-1}y)\ge0$$

$$\implies\sqrt{(1-x^2)(1-y^2)}\ge xy$$

which is obvious $$xy<0$$

What if $$xy=0?$$

If $$xy>0,$$ $$(1-x^2)(1-y^2)\ge x^2y^2\iff 1\ge x^2+y^2$$

• It worked for xy > 0 but I couldn't understand how does it work for xy < 0. Dec 16, 2020 at 13:31
• @RohitJoshi, As for real $x,y$ with $-1\le x^2,y^2\le1$ $$\sqrt{(1-x^2)(1-y^2)}\ge0$$ Dec 16, 2020 at 13:44
• How does this leads to my 2nd inequality ? I want to prove my that 1st inequality implies the 2nd one. Dec 16, 2020 at 14:17
• @RohitJoshi, If $xy<0,$ either $y<0,x>0$ then $$x|x|+y|y|=x^2-y^2\le1$$ as $$x^2\le1,y^2\ge0$$ Similarly for $x<0,y>0$ Dec 16, 2020 at 15:13