What is the restriction for the sine function?

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Homework Help Overview

The discussion revolves around the restrictions of the sine function, particularly in relation to its inverse, the arcsine function. Participants explore the implications of defining inverse functions within specific domains.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of inverse functions and the necessity of restricting the sine function to ensure it is one-to-one. There are varying opinions on the appropriate interval for this restriction, with some suggesting [0, π] and others proposing [-π/2, π/2].

Discussion Status

The conversation is active, with participants presenting different viewpoints on the correct interval for restricting the sine function. There is no explicit consensus, but the discussion is exploring the nuances of function definitions and their inverses.

Contextual Notes

Participants are navigating the definitions and properties of trigonometric functions and their inverses, highlighting potential misunderstandings regarding the intervals used for these functions.

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can anybody pls give me an example of this f(f^-1) = x??
thanx...
 
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It's the definition of the inverse of a function f(x). For example, y = 2x and y^-1 = 1/2 x
 
how about if i want to write it using sin x??
 
Although I would say f(f-1(x))= x. You're missing the "x" on the left side!

"arcsine" is defined as the inverse of sine (that's why your calculator has them paired). sin(arcsin(x))= x.

There are "technical" problems. Since sin(x) is not "one-to-one" ([itex]sin(\pi)= 0= sin(0)[/itex]) there can't be a true inverse (a function can't return both 0 and [itex]\pi[/itex] for x= 0). What is normally done is restrict the sine function to x=0 to [itex]\pi[/itex] (which is really a different function than sine defined for all x) so that arcsin returns the "principal value"- the value between 0 and [itex]\pi[/itex].
 
HallsofIvy said:
What is normally done is restrict the sine function to x=0 to [itex]\pi[/itex] (which is really a different function than sine defined for all x) so that arcsin returns the "principal value"- the value between 0 and [itex]\pi[/itex].
We don't really restrict sin function to x = 0 to [itex]\pi[/itex].
We, however, restrict sin function to [tex]x = -\frac{\pi}{2}[/tex] to [tex]x = \frac{\pi}{2}[/tex]. :)
 
VietDao29 said:
We don't really restrict sin function to x = 0 to [itex]\pi[/itex].
We, however, restrict sin function to [tex]x = -\frac{\pi}{2}[/tex] to [tex]x = \frac{\pi}{2}[/tex]. :)
Oops! It's cosine that is restricted to "between 0 and [itex]\pi[/itex]!
 

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