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.