# Proof for the formula of sum of arcsine functions $\arcsin x + \arcsin y$

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

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.

Using this, $\displaystyle-\frac\pi2\leq \arcsin z\le\frac\pi2$ 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 $x=-X,y=-Y$

• 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 '14 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 '14 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 '14 at 7:57
• @labbhattacharjee: Wouldn't there be $\le$ sign in your first relation? – user142971 Mar 3 '16 at 8:24
• @user36790, The equality has been catered in the second – lab bhattacharjee Mar 3 '16 at 8:49

Hope the graph explains the x,y domain ( |x|< 1, |y|< 1 ) and range.

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.

• 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 '14 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 '14 at 17:57

Let $$a=\sin^{-1}x,$$ $$b=\sin^{-1}y\implies\sin a=x,$$ $$\sin b=y$$ and $$a,b\in[-\pi/2,\pi/2]\implies a+b\in[-\pi,\pi]$$ $$\sin(a+b)=\sin a\cos b+\cos a\sin a=x\sqrt{1-y^2}+y\sqrt{1-x^2}\\=\sin\bigg[\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)\bigg]\\ \implies a+b=\color{red}{\sin^{-1}x+\sin^{-1}y=n\pi+(-1)^n\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}$$ Case 1: $$\dfrac{-\pi}{2}<\sin^{-1}x+\sin^{-1}y<\dfrac{\pi}{2}$$ $$\color{darkblue}{\sin^{-1}x+\sin^{-1}y=\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}$$ $$\cos(a+b)>0\implies \cos a.\cos b-\sin a.\sin b>0\implies \cos a.\cos b>\sin a.\sin b\\ \sqrt{1-x^2}.\sqrt{1-y^2}>xy\implies 1-x^2-y^2+x^2y^2>x^2y^2\implies x^2+y^2<1$$ Case 2 & 3: $$\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y<\pi$$ and $$-\pi<\sin^{-1}x+\sin^{-1}y<\dfrac{-\pi}{2}$$ $$\cos(a+b)<0\implies x^2+y^2>1$$

Case 2-: $$\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y<\pi$$ $$\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)=\pi-(\sin^{-1}x+\sin^{-1}y)\\ \implies \color{darkblue}{\sin^{-1}x+\sin^{-1}y=\pi-\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}$$ $$\dfrac{\pi}{2}<\sin^{-1}x+\sin^{-1}y<\pi \;\&\;\dfrac{-\pi}{2}\leq\sin^{-1}x,\;\sin^{-1}y\leq\dfrac{\pi}{2}\\ \implies 0<\sin^{-1}x,\sin^{-1}y\leq\dfrac{\pi}{2}\implies 0< x,y\leq1$$ Case 3-: $$-\pi<\sin^{-1}x+\sin^{-1}y<\dfrac{-\pi}{2}$$

$$\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)=-\pi-(\sin^{-1}x+\sin^{-1}y)\\ \implies \color{darkblue}{\sin^{-1}x+\sin^{-1}y=-\pi-\sin^{-1}\Big(x\sqrt{1-y^2}+y\sqrt{1-x^2}\Big)}$$ $$-\pi<\sin^{-1}x+\sin^{-1}y<\dfrac{-\pi}{2} \;\&\;\dfrac{-\pi}{2}\leq\sin^{-1}x,\;\sin^{-1}y\leq\dfrac{\pi}{2}\\ \implies \dfrac{\pi}{2}\leq\sin^{-1}x,\sin^{-1}y\leq 0\implies -1\leq x,y\leq 0$$