My book says:

$$\arcsin x + \arcsin y = \pi- \arcsin\left\{x\sqrt{1-y^2} + y\sqrt{1-x^2}\right\}$$ when $$0\lt x,y\le 1 \,\,\textrm{and}\,\, x^2+ y^2\gt 1$$

Mr. Author's description:

Let $x= \sin A$ and $y= \sin B$ and $A,B\in \left(0,\frac{\pi}{2}\right]$

$$\sin(A+B) = x\sqrt{1-y^2} + y\sqrt{1-x^2}$$ and $$\cos(A+B)= \sqrt{1-x^2}\sqrt{1-y^2}- xy$$

$$(A+B)\in (0,\pi]$$ and

from $x^2 + y^2 \gt 1$ we get \begin{align}1-x^2 &\lt y^2\\ 1-y^2 &\lt x^2 \\ \implies \sqrt{1-x^2}\sqrt{1-y^2} &\lt xy\\ \implies \sqrt{1-x^2}\sqrt{1-y^2}- xy &\lt 0\\ \implies \cos(A+ B)\lt 0\\ \implies (A+B) \,\, \textrm{lies either in II quadrant or in III quadrant.}\\ \implies \frac{\pi}{2}\color{red}{\le} A+B\le \pi\\ \implies 0\le \pi - (A+B) \le \frac{\pi}{2}\\ \implies \sin(\pi-(A+B)) &= x\sqrt{1-y^2} + y\sqrt{1-x^2} \\ \implies \sin^{-1}x+ \sin^{-1}y &= \pi-\sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right) \end{align}

My thoughts:

Quite an easy deduction if seen superficially; but has anyone noticed the $\color{red}{\le}\;?$

How could there be any possibility of equality as reflected by $\ge\;?$ Equality of $A+B$ with $\pi/2$ would mean $\sin(A+B)= 1$ and $\cos(A+B)\color{red}{=}0$ but that is what wouldn't happen as the author himself deduced earlier that $\cos(A+B)\lt 0\;.$ So, it is wrong to include $\pi/2$ in the set that is $A+B \in \left(\frac{\pi}{2}, \pi\right]$ and not $\left[\frac{\pi}{2}, \pi\right]\,$ isn't it?

But, it should be bore in mind that the range of $\arcsin$ for $A+B$ lying in the second quadrant and hence $\pi -(A+B)$ should be $\left[\frac{\pi}{2},\pi\right]\;;$ for this, the constraint should be $x^2 + y^2 \color{red}{\ge} 1 $ and not $\gt 1\;.$ Then $\cos(A+B)\le 0$ and all the later deduction would be valid.

So, can anyone tell me whether my thinking is correct? For $\arcsin x + \arcsin y$ to be equal to $\arcsin\left\{x\sqrt{1-y^2} + y\sqrt{1-x^2}\right\}\,,$ would the constraint be $0\lt x,y \le 1\,\,\textrm{and}\,\, x^2 + y^2 \gt 1$ as stated by the book or $0\lt x,y \le 1\,\,\textrm{and}\,\, x^2 + y^2 \ge 1$ as my thinking predicts?

  • $\begingroup$ See math.stackexchange.com/questions/672575/… $\endgroup$ – lab bhattacharjee Mar 3 '16 at 7:05
  • $\begingroup$ @lab bhattacharjee: Thanks, really; but that's not what I wanted; my book already deduced that; what I wanted to is whether my line of thinking is correct or not. $\endgroup$ – user142971 Mar 3 '16 at 7:47

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