Show that arcsin is the inverse of sin

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    Inverse Sin
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Homework Help Overview

The discussion revolves around the properties of the arcsine function and its relationship with the sine function, specifically addressing the equation arcsin(sin(x)) = x. Participants are exploring the conditions under which this equation holds true and the implications of the inverse function composition.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the conditions for the equality arcsin(sin(x)) = x and questioning the range of values for which this holds. There is also discussion about the graphical representation of the function and its behavior across different intervals.

Discussion Status

The discussion is ongoing, with participants providing insights into the definitions and properties of inverse functions. Some have pointed out the specific intervals where the equations are valid, while others are seeking further clarification on the graphical aspects and the implications of using complex arcsin.

Contextual Notes

There is mention of the need to consider specific intervals, such as -1 ≤ x ≤ 1 for arcsin(sin(x)), and the potential inclusion of complex numbers in the discussion, which adds complexity to the understanding of the functions involved.

gymko
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Homework Statement



I have a problem, I don't know to substantiate, why arcsin(sin(x)) = sin(arcsin(x)) = x ?
Thank you very much for each advice.



Homework Equations


The Attempt at a Solution

 
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i didnt exactly get your question. Are you trying to find the range of values of x for which the above eq is valid?

what are u asking? can u be clearer.
 


My question is: why arcsin(sin(x)) is possible to regulate for x. Why arcsin(sin(x)) = x?

Why graph for y = arcsin(sin(x)) is y = x?

Thank you.
 


Do you know about the composition of inverse functions?
Also, arcsin(sin(x)) = sin(arcsin(x)) = x only for -1≤x≤1 for all three.
 


[itex]\sin(\arcsin(x)) = x[/itex] for all real [itex]x[/itex] but [itex]\arcsin(\sin(x))[/itex] only for [itex]-\pi/2 \le x \le \pi/2[/itex] . That's with one standard way of defining [itex]\arcsin[/itex]
 


well look at it this way :

sin( asin(x ) )

let x = 1;

the asin(1) = P;
then sin(P) = 1;
 


g_edgar said:
[itex]\sin(\arcsin(x)) = x[/itex] for all real [itex]x[/itex]
Did you mean all x in [-1,1]? (Or are you making an assertion about the complex Arcsin function?)
 


Thank you very much for all. :)
 


y=arcsin(sin(x))

sin(y) = sin(x)

y = x
 
  • #10


Gregg said:
y=arcsin(sin(x))

sin(y) = sin(x)

y = x
:confused:
 
  • #11


Hurkyl said:
Did you mean all x in [-1,1]? (Or are you making an assertion about the complex Arcsin function?)

You are right. For my equation you have to use complex arcsin. Wherever arcsin is defined, we have sin(arcsin(x)) = x , that is what we mean by arcsin.
 
  • #12


Hurkyl said:
:confused:

gymko said:
My question is: why arcsin(sin(x)) is possible to regulate for x. Why arcsin(sin(x)) = x?

Why graph for y = arcsin(sin(x)) is y = x?

Thank you.

y=arcsin[sin[x]] implies sin[y] = sin[x] and y=x. Forgetting the intervals, I don't the question had much to do with the interval which this valid for. It was just how does the graph look like y=x.
 

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