The graph of sin inverse (sin x) after the domain of (- pi/2, pi/2)

[esha]

the graph of sin inverse (sin x) between the domain of ( -pi/2,pi/2) is y = x. but after it crosses that domain of course the expression won't be the same anymore because sin inverse has its principle value as ( - pi/2, pi/2) due to sin x many to one natured function. now the way these expressions change is what doesn't seem intuitive to me. can anybody please tell me how to derive those expressions logically?

 

[Mark44]

the graph of sin inverse (sin x) between the domain of ( -pi/2,pi/2) is y = x.

No, it isn't. The graphs of the two functions are close together when x is in the interval [-.5, .5], but they aren't identical.

but after it crosses that domain of course the expression won't be the same anymore because sin inverse has its principle value as ( - pi/2, pi/2) due to sin x many to one natured function. now the way these expressions change is what doesn't seem intuitive to me. can anybody please tell me how to derive those expressions logically?

What expression? Are you trying to understand why the graph of $y = \sin^{-1}(x)$ looks the way it looks? Are you asking how the values on this graph are calculated? If you have studied calculus, one of the topics presented later is infinite series. One such series is the expansion for the arcsine function. See https://math.stackexchange.com/questions/197874/maclaurin-expansion-of-arcsin-x.

 

[esha]

its not sin inverse graph..
that's pretty straight forward..
. m talking bout sin inverse ( sin x) graph

 

[esha]

they are different...

 

[Mark44]

its not sin inverse graph..
that's pretty straight forward..
. m talking bout sin inverse ( sin x) graph

It wasn't clear to me what you were asking about, which is the graph of $y = \sin^{-1}(\sin(x))$. On the interval $[-\pi/2, \pi/2]$, the graph of this function is the same as that of y = x, which is what you said.

On the interval $[\pi/2, 3\pi/2]$, the graph of $y = \sin^{-1}(\sin(x))$ goes from $(\pi/2, \pi/2)$ down to (0, 0), and then to $(3\pi/2, -\pi/2)$, so this line segment has a slope of -1 with an equation of $y = -(x - \pi)$ (in other words, the same as the graph of y = -x, but shifted to the right by $\pi$ units.

You have to go through this kind of analysis on each of the intervals $[-\pi/2 + 2n\pi, \pi/2 + 2n\pi]$. Doing this you get a sawtooth graph like this one:http://www.wolframalpha.com/input/?i=plot+y+=+arcsin(sin(x))

 

[esha]

thanks... for the explanation