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For the addition of inverse trigonometric functions, $\sin^{-1} x+\sin^{-1} y = \sin^{-1} ({x\sqrt{1-y^2} +y\sqrt{1-x^2}})$

But for $\sin^{-1} ({x\sqrt{1-y^2} +y\sqrt{1-x^2}}) = \left\{\begin{matrix} \sin^{-1} x +\sin^{-1} y, \text{ if } -1\leq x,y<1 \text{ and } x^2+y^2 \leq 1 \text{ or, if } xy<0 \text{ and } x^2+y^2 > 1 \\ \pi -(\sin^{-1} x +\sin^{-1} y), \text{ if } 0<x,y\leq 1 \text{ and } x^2 +y^2 >1 \\ -\pi-(\sin^{-1} x + \sin^{-1} y), \text{ if } -1\leq x,y<0 \text{ and } x^2+y^2 > 1 \end{matrix}\right.$

Now what I don't understand is that, " Where did the conditions come from?". Is there proof or anything for the conditions? Please explain.

rash
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  • https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y – lab bhattacharjee Jun 04 '19 at 06:40
  • @labbhattacharjee yeah well this is the inverse of that post. Still might help if I get answers. – rash Jun 04 '19 at 06:41
  • The point is that $-\pi/2\le \sin^{-1} t\le\pi/2$, but $\sin^{-1}x+\sin^{-1}y$ may lie outside that interval. – Angina Seng Jun 04 '19 at 06:42
  • @LordSharktheUnknown I know you have to $\pi -$ cuz it may lie out of the interval. But how would know like if $xy \leq 1$ you have to do as follows? How do you get that specific condition? – rash Jun 04 '19 at 06:46

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