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

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The discussion centers on proving that cos(cos-1(x)) lies within the range of [-1, 1]. Participants clarify that arccos(x) is the inverse of cosine, not its derivative, which is -sin(x). The relationship cos(arccos(x)) = x holds true only when x is within the valid domain of arccos, specifically [-1, 1]. Understanding the graphical representation of the cosine function helps clarify the range and domain issues. Ultimately, the conclusion is that the output of cos(cos-1(x)) must indeed fall within the interval [-1, 1].
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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.
 
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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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