Proving cos(cos-1(x)) Lies in [-1,1]

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Homework Help Overview

The discussion revolves around proving that cos(cos-1(x)) lies within the range of [-1, 1]. The subject area includes trigonometric functions and their inverses, particularly focusing on the properties of cosine and arccosine.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between cosine and its inverse, questioning how the properties of these functions lead to the conclusion about the range. Some participants attempt to clarify misconceptions about derivatives and the definitions of the functions involved.

Discussion Status

The discussion is ongoing, with participants providing hints and questioning assumptions. There is a mix of clarifications regarding the definitions of derivatives and inverse functions, and some participants suggest examining graphical representations to aid understanding.

Contextual Notes

There are mentions of potential misinformation regarding derivatives, and participants emphasize the importance of understanding the domain of arccos(x) to resolve the problem. The original poster expresses confusion about proving the range of cos(arccos(x)).

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Homework Statement



Show cos(cos-1(x) lies in range [-1,1]

Homework Equations



I know the derivative of cos(x) is arcos(x).

The Attempt at a Solution



D/dx(cos(x)) = arcos(x)

However i am stuck how to prove cos(arcos(x)) has a domain of [-1,1]..
 
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I think it is taken in context of a trig function in which cosine is described by a circle with a radius equal to one. Then since the cos and it's inverse cancel each other out leave the answer is as follows

cos(cos-1(x)) = x

This would simply mean that the line would run along the x axis, in terms of trig terminology this would be [-1,1].
 

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SteliosVas said:
I know the derivative of cos(x) is arcos(x).
You might want to check again.
 
Sorry I didn't read the rest, arcos(x) is the inverse of cos(x) not the derivative. Derivative is as follows

d/dx cos(x) = -sin(x)
d/dx sin(x) = cos(x)

This is because sin(x) and cos(x) are essentially the same function except the phase between the two is pi/2 (90 degrees or 1/4 revolution of a full circle), hence one can be derived by the other.
 
Last edited:
SteliosVas said:

Homework Equations



I know the derivative of cos(x) is arcos(x).

You should forget this bit of mathematical misinformation immediately.

The derivative of cos(x) = -sin(x)

Here is a list of other derivatives:

http://en.wikipedia.org/wiki/Derivative

You should check what you 'know' about derivatives of other functions against this list.

The 'arccos (x)' is the inverse of cosine (x), not the derivative, which has a specific meaning in mathematics.

arccos(cosine(x)) = x
 
For what values of x is ##\cos^{-1} x ## defined ( domain ).
X can only be in that interval, other than that--
## f(f^{-1}(x))=x## all the time by definition.
 
Hint:arccos (x)=y means cos(y) = x.
 
SteamKing said:
The 'arccos (x)' is the inverse of cosine (x), not the derivative, which has a specific meaning in mathematics.

arccos(cosine(x)) = x
This should really be the other way around: cos(arccos(x)) = x. It's possible for arccos(cos(x)) to differ from x, when x isn't in the range of the arccosine function.
 
SteliosVas said:

Homework Statement



Show cos(cos-1(x) lies in range [-1,1]

Homework Equations



I know the derivative of cos(x) is arcos(x).

The Attempt at a Solution



D/dx(cos(x)) = arcos(x)

However i am stuck how to prove cos(arcos(x)) has a domain of [-1,1]..

Have you looked at a graph of ##y = \cos(\theta)## over a broad range of ##\theta##? I suggest you do that first, before trying to use fancy identities or derivatives and the like. Then, just remember what ##\arccos(x)## stands for.
 
  • #10
In the problem you're analyzing the image of cos(arccos(x)).
But then you say you're trying to figure out the domain of cos(arccos(x)). Which is it?
 
  • #11
Before you can know the range (image) of a function, you must know which inputs would be valid.
For example, you could look at cos(arccos(x)) and say, hey, that's a function and its inverse, so the output is just x.
Then, you would say, hey, that doesn't make sense, since I know cosine has a range between -1 and 1...
So, you need to know the possible input values for x (domain of arccos(x)) in order to properly resolve this apparent contradiction.
 
  • #12
Let's hold off on any more replies until the OP comes back.
 

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