# problem with some inverse trigo function graphing - need some help! [duplicate]

Possible Duplicate:
Why aren't the graphs of $\sin(\arcsin x)$ and $\arcsin(\sin x)$ the same?

I faced this equation while dealing with some inverse trigo functions.

$\arcsin(\sin(x)) + \arcsin(\sin(y)) = x + y$

$x,y \in [-\pi;\pi]$

I have been trying to simplify/graph this equation for a while but cannot figure it out ! I'm not sure which methods I should use to solve this problem, so could somebody show me a way and give explanation? Thank you in advance!

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## marked as duplicate by Cameron Buie, Ross Millikan, William, tomasz, J. M.Sep 25 '12 at 14:46

This question was marked as an exact duplicate of an existing question.

Can you graph $\arcsin(\sin x)$? – Thomas Andrews Sep 5 '12 at 13:28
yup i guess this is it math.stackexchange.com/questions/148679/… – Keon LEE Sep 5 '12 at 13:42

$\sin^{-1}(\sin x)$ is a multi-valued function reduces to $n\pi+(-1)^nx$ where $n$ is any integer.

So, $n\pi+(-1)^nx+m\pi+(-1)^mx=x+y$

(1)If $m,n$ both are even, $(m+n)\pi=0\implies m+n=0$

We can not find any relation between $x,y$ here.

(2)If $m$ is odd$=2M+1$(say), and $n$ is even$=2N$, $2N\pi+x+(2M+1)\pi-y=x+y\implies 2y=(2N+2M+1)\pi\implies y=\frac{(2N+2M+1)\pi}{2}=(N+M)\pi+\frac{\pi}{2}\implies y=-3\frac{\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$

(3)Similarly if $m$ is even, $n$ is odd, $x=-3\frac{\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$

(4) If both are odd, let $m=2M+1,n=2N+1, (2N+1)\pi-x+(2M+1)\pi-y=x+y$ $\implies x+y=(M+N+1)\pi$ Now, $-2\pi ≤x+y ≤2\pi$

So, $x+y=±2\pi, ±\pi, 0$.

Alternatively, the principal value of $sin^{-1}(\sin x)$ must lie $\in [-\frac{\pi}{2};\frac{\pi}{2}]$

Let $sin^{-1}(\sin x)=z\implies \sin z=\sin x$

$z=x$ if $-\frac{\pi}{2}≤x≤\frac{\pi}{2}$(as Thomas Andrews has observed),

$z=\pi-x$ if $\frac{\pi}{2}<x≤\pi$

$z=-x-\pi$ if $-\pi≤x<-\frac{\pi}{2}$

So, there are three regions for $x$.

So, there are $3\cdot 3=9$ regions for $x,y$.

For example,

If $-\frac{\pi}{2}≤x≤\frac{\pi}{2}$ and $-\frac{\pi}{2}≤y≤\frac{\pi}{2}$, any values of $x,y$ in the above ranges will satisfy the given equation.

If $-\frac{\pi}{2}≤x≤\frac{\pi}{2}$ and $\frac{\pi}{2}<y≤\pi$ $x+\pi-y=x+y\implies y=\frac{\pi}{2}$ which is impossible as $\frac{\pi}{2}<y≤\pi$

If $-\frac{\pi}{2}≤x≤\frac{\pi}{2}$ and $-\pi≤y<-\frac{\pi}{2},$ $x-y-\pi=x+y\implies y=-\frac{\pi}{2}$ which is again impossible

and so on.

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Most people take $\arcsin x$ as a single-valued function, not a multi-valued function. – Thomas Andrews Sep 5 '12 at 13:50
If that single-value is taken i.e., $arcsin(\sin x)=x$ then, the left hand side becomes x+y, so resulting in an identity, right? – lab bhattacharjee Sep 5 '12 at 13:54
No, there is still a value for $\arcsin(\sin x)$ for all $x\in[-\pi,\pi]$, and it is only the identity on $[-\frac{\pi}2,\frac{\pi}2]$ – Thomas Andrews Sep 5 '12 at 15:18
Agreed & modified the answer – lab bhattacharjee Sep 6 '12 at 13:03