It is known that $$\sin^{−1}x+\sin^{−1}y = \sin^{-1}\left[x\sqrt{1 – y^2} + y\sqrt{1 – x^2}\right] $$ if $x, y ≥ 0$ and $x^2+y^2 ≤ 1.$
I know that the given condition makes sure that $\sin^{−1}x+\sin^{−1}y$ lies in the range $[-\pi/2,\pi/2].$
But I can't figure out how the condition is derived?
It is known that $$\sin^{−1}x+\sin^{−1}y = \sin^{-1}\left[x\sqrt{1 – y^2} + y\sqrt{1 – x^2}\right] \tag1$$ if $x, y ≥ 0 \quad \text{and}\quad x^2+y^2 ≤ 1.\tag2$
But I can't figure out how the condition is derived?
Condition $(2)$ cannot be derived from formula $(1).$ In other words, $$(1)\kern.6em\not\kern-.6em\implies(2),$$ and it is false that the given formula is applicable only under the given condition.
For example, $(x,y)=(1,-1)$ satisfies $(1)$ but not $(2).$
The answers to this post, and the fact that the three cases are disjoint, shows that $$(1)\iff \Big(x^2+y^2 \le 1 \quad\text{or}\quad(x^2+y^2 > 1 \quad\text{and}\quad xy< 0)\Big).$$
Actually the range is not restricted, but can be unbounded depending on integer value chosen for $k$:
$$\sin^{−1}x+\sin^{−1}y = \sin^{-1}\left(x\sqrt{1 – y^2} + y\sqrt{1 – x^2}\right)+ (-1)^k k \pi $$
By the given conditions you can say 0 ≤ x,y ≤ 1 Therefore 0 ≤ arcsinx , arcsiny ≤ π/2 and this implies that 0 ≤ arcsinx + arcsiny ≤ π . But right side is also arcsin of positive value and that implies it should be between 0 and π//2 .Therefore here you have a contradiction and to avoid it you need to consider the formula Sinθ = Sin ( π - θ) when you derive the main result. Therefore the correct formula should be, arcsinx + arcsiny = the given result or π-the given result.
One counter example is $x=\frac{\sqrt 3}{2}$ and $y=\frac{-\sqrt 3}{2}$.
– Li Kwok Keung Jul 09 '22 at 01:05