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It is known that the following holds good: $$ \arcsin x + \arcsin y \\ \begin{align} &=\arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 \le 1 \;\text{ or }\; x^2+y^2 > 1, xy< 0\\ &=\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 > 1, 0< x,y \le 1\\ &=-\pi - \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) \;\;;x^2+y^2 > 1, -1< x,y \le 0\\ \end{align} $$

But I couldn't find a proof for the above. I tried to prove this myself, but failed. I have no clue how to bring in $x\sqrt{1-y^2} + y\sqrt{1-y^2}$ from the conditions like $x^2 + y^2 < 1$. Please understand that I don't have any problem in getting the 'crux' part of the RHS : $ \arcsin( x\sqrt{1-y^2} + y\sqrt{1-x^2}) $. I face trouble only in checking the range of that 'crux' under the given conditions.

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2 Answers 2

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Using this, $\displaystyle-\frac\pi2\arcsin z\le\frac\pi2 $ for $-1\le z\le1$

So, $\displaystyle-\pi\le\arcsin x+\arcsin y\le\pi$

Again, $\displaystyle\arcsin x+\arcsin y= \begin{cases} \\-\pi- \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})& \mbox{if } -\pi\le\arcsin x+\arcsin y<-\frac\pi2\\ \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2}) &\mbox{if } -\frac\pi2\le\arcsin x+\arcsin y\le\frac\pi2 \\ \pi- \arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})& \mbox{if }\frac\pi2<\arcsin x+\arcsin y\le\pi \end{cases} $

and as like other trigonometric ratios are $\ge0$ for the angles in $\left[0,\frac\pi2\right]$

So, $\displaystyle\arcsin z\begin{cases}\text{lies in } \left[0,\frac\pi2\right] &\mbox{if } z\ge0 \\ \text{lies in } \left[-\frac\pi2,0\right] & \mbox{if } z<0 \end{cases} $

Case $(i):$ Observe that if $\displaystyle x\cdot y<0\ \ \ \ (1)$ i.e., $x,y$ are of opposite sign, $\displaystyle -\frac\pi2\le\arcsin x+\arcsin y\le\frac\pi2$

Case $(ii):$ If $x>0,y>0$ $\displaystyle \arcsin x+\arcsin y$ will be $\displaystyle \le\frac\pi2$ according as $\displaystyle \arcsin x\le\frac\pi2-\arcsin y$

But as $\displaystyle\arcsin y+\arccos y=\frac\pi2,$ we need $\displaystyle \arcsin x\le\arccos y$

Again as the principal value of inverse cosine ratio lies in $\in[0,\pi],$ $\displaystyle\arccos y=\arcsin(+\sqrt{1-y^2})\implies \arcsin x\le\arcsin\sqrt{1-y^2}$

Now as sine ratio is increasing in $\displaystyle \left[0,\frac\pi2\right],$ we need $\displaystyle x\le\sqrt{1-y^2}\iff x^2\le1-y^2$ as $x,y>0$

$\displaystyle\implies x^2+y^2\le1 \ \ \ \ (2)$

So, $(1),(2)$ are the required condition for $\displaystyle \arcsin x+\arcsin y\le\frac\pi2$

Case $(iii):$

Now as $\displaystyle-\frac\pi2\arcsin(-u)\le\frac\pi2 \iff -\frac\pi2\arcsin(u)\le\frac\pi2$

$\arcsin(-u)=-\arcsin u$

Use this fact to find the similar condition when $x<0,y<0$ setting $x=-X,y=-Y$

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@ParthThakkar, how about this? –  lab bhattacharjee Feb 12 at 18:27
    
@ParthThakkar, is there is any further confusion. You may remove the queries which have been removed by yourself –  lab bhattacharjee Feb 13 at 7:25
    
If there is any, I'll surely ask. I'll work it out myself first so that there is no scope of doubt. What I asked was just by going through your answer. But I think it's pretty fine now. Thanks! –  Parth Thakkar Feb 13 at 7:28
    
In the first case, by the condition $xy<0$ alone we have the proper range. That is, $\arcsin(x\sqrt{1-y^2} + y\sqrt{1-x^2}) \in [-\pi/2, \pi/2]$. Then why is that additional condition $x^2 + y^2 > 1$ required? –  Parth Thakkar Feb 13 at 7:34
    
@ParthThakkar, its not required, the condition is either $x^2+y^2\le1$ or ($xy<0$ where $x^2+y^2$ is obviously $>1$) –  lab bhattacharjee Feb 13 at 7:57

Take the sine of both sides, and use the angle addition formula, then further simplify it by using the fact that $\cos\arcsin t=\sqrt{\cos^2\arcsin t}=\sqrt{1-\sin^2\arcsin t}=\sqrt{1-t^2}$. Then apply the $\arcsin$ function to both sides, and you're done.

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As I have said in my question, I don't have any problem in getting the main part: $ \arcsin(x\sqrt{1-x^2} + y\sqrt{1-y^2}) $. I don't know how to do the range checking so that I can adjust the RHS with $\pi$. –  Parth Thakkar Feb 12 at 11:38
    
@ParthThakkar: $\sin t=\sin(\pi-t)$. (Always picture the unit circle in your mind). And since $\pi$ and $-\pi$ are the same (again, visualize), QED. –  Lucian Feb 12 at 17:57

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