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.

  • $\begingroup$ 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 $\cos(\sin^{-1}(a))=\pm\sqrt{1-a^2}$. $\endgroup$
    – Nicolas
    Dec 16, 2020 at 12:33
  • $\begingroup$ 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. $\endgroup$ Dec 16, 2020 at 12:55
  • $\begingroup$ $$\left|\sin^{-1}(x) + \sin^{-1}(y)\right|\le\frac{\pi}{2} , \{|x|,|y|\le1\}.$$ is not true. $\endgroup$
    – Cesareo
    Dec 16, 2020 at 13:48
  • $\begingroup$ It is not a statement. It is a condition which is satisfied by some values of x and y less than 1. $\endgroup$ Dec 16, 2020 at 13:52

1 Answer 1


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

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

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