# Proof of Arccos

1. Oct 16, 2014

### SteliosVas

1. The problem statement, all variables and given/known data

Show cos(cos-1(x) lies in range [-1,1]
2. Relevant equations

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

3. 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]..

2. Oct 16, 2014

### mamort

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. Oct 16, 2014

### rtsswmdktbmhw

You might want to check again.

4. Oct 16, 2014

### mamort

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: Oct 16, 2014
5. Oct 16, 2014

### SteamKing

Staff Emeritus
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. Oct 16, 2014

### RUber

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. Oct 16, 2014

### RUber

Hint:arccos (x)=y means cos(y) = x.

8. Oct 16, 2014

### vela

Staff Emeritus
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. Oct 17, 2014

### Ray Vickson

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. Oct 19, 2014

### GFauxPas

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. Oct 20, 2014

### RUber

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. Oct 20, 2014

### Staff: Mentor

Let's hold off on any more replies until the OP comes back.