It is known that the following holds good: $$ \sin^{-1} x + \sin^{-1}y \\ \begin{align} &=\sin^{-1}( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 \le 1 \;\text{ or }\; x^2+y^2 > 1, xy< 0\\ &=\pi - \sin^{-1}( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 > 1, 0< x,y \le 1\\ &=-\pi - \sin^{-1}( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 > 1, -1< x,y \le 0\\ \end{align} $$ Ok I know how to prove the basic formulas by taking different ranges for x.But how to proove that these conditions : $-1< x,y \le 0\\$ and $-1< x,y \le 0\\$ must hold in the second and third cases ?
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what is $n, in \sin^{-1} x + \sin^{-1} ny ?$ – abel Apr 14 '15 at 21:40
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Typo...corrected :-) – Apr 14 '15 at 21:42
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I can't understand why should x and y be restricted by such conditions ? – Apr 14 '15 at 21:52
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@SanchayanDutta, Related : http://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y – lab bhattacharjee Apr 15 '15 at 03:32
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you need $-1 \le x \le x$ and $-1 \le y \le 1$ for $\sin^{-1} x$ and $\sin^{-1} y$ to have real values. the range of $\sin^{-1}$ is $[-\pi/2, \pi/2]$ so that $ -\pi \le \sin^{-1} x + \sin^{-1} y \le \pi.$
i will deal with the restricted case $0 \le x \le 1, 0 \le y \le 1.$ fix an $x.$ now break the rectangle $[0,x] \times [0,1]$ into two rectangles $[0,x] \times [0,\sqrt{1-x^2}] \cup[0,x]\times [\sqrt{1-x^2},1] = A\cup B$
for points $(x,y) \in A, 0 \le \sin^{-1} x + \sin^{-1}y \le \pi/2.$ therefore $$\sin^{-1} x + \sin^{-1} y = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right).$$
for points $(x,y) \in B, \sin^{-1} x + \sin^{-1}y = \pi- \left((\pi/2 - \sin^{-1}y) + (\pi/2 - \sin^{-1}x)\right) > \pi/2.$ therefore $$\sin^{-1} x + \sin^{-1} y = \pi - \sin^{-1}\left(y\sqrt{1-x^2} + x\sqrt{1-y^2}\right).$$
abel
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Not clear as to what you mean by rectangle.Please explain a bit more clearly if possible. – Apr 15 '15 at 04:59