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

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We don't know what level he/she is at and what kind of help is most appropriate.In summary, the conversation discusses the concept of cos(cos^-1(x)) and its range of values. The attempt at a solution involves using the derivative of cos(x) as arccos(x), but this is incorrect as the derivative of cos(x) is -sin(x). The correct approach is to understand the inverse relationship between cos(x) and arccos(x) and consider the possible inputs for x in order to determine the range of cos(cos^-1(x)).
  • #1
SteliosVas
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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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  • #2
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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  • #3
SteliosVas said:
I know the derivative of cos(x) is arcos(x).
You might want to check again.
 
  • #4
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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  • #5
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
 
  • #6
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.
 
  • #7
Hint:arccos (x)=y means cos(y) = x.
 
  • #8
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.
 
  • #9
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.
 

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

What is the definition of cos(cos-1(x))?

The expression cos(cos-1(x)) refers to the composition of two trigonometric functions, cosine and inverse cosine. In other words, it is the cosine of the angle whose cosine is x. This can also be written as cos-1(x), arccos(x), or cos-1(x).

Why is proving that cos(cos-1(x)) lies in the interval [-1,1] important?

It is important because this interval represents the range of possible values for the cosine function. If we can prove that cos(cos-1(x)) lies within this interval, we are essentially verifying that the composition of these two functions is valid and follows the rules of trigonometry.

How can we prove that cos(cos-1(x)) lies in the interval [-1,1]?

To prove that cos(cos-1(x)) lies in the interval [-1,1], we can use the fundamental trigonometric identity cos2(x) + sin2(x) = 1. By substituting cos-1(x) for x, we get cos2(cos-1(x)) + sin2(cos-1(x)) = 1. Since the range of sin and cos is [-1,1], this identity shows that the value of cos(cos-1(x)) must also lie within that interval.

Can we use a graph to visualize the proof of cos(cos-1(x)) lying in the interval [-1,1]?

Yes, we can use a graph of the cosine function to visualize this proof. The graph of cos(cos-1(x)) is essentially the same as the graph of cos(x), but with the x-axis representing the cosine of the angle instead of the angle itself. This graph shows that the values of cos(cos-1(x)) are bounded by the interval [-1,1] and cannot exceed it.

Are there any other methods for proving that cos(cos-1(x)) lies in the interval [-1,1]?

Yes, there are other methods for proving this. One approach is to use the unit circle and the definition of cosine and inverse cosine to show that the range of cos(cos-1(x)) is [-1,1]. Another method is to use calculus and the derivative of the inverse cosine function to prove that cos(cos-1(x)) has a maximum and minimum value of -1 and 1 respectively, and therefore must lie in the interval [-1,1].

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