# Proving inverse functions identity

• Sep 13th 2012, 06:35 AM
Danielc
Proving inverse functions identity
Hello everybody and thanks for reading.
I have been trying to solve the following problem without success, so I am looking for some advices..

I have the following equation

arccos(x) = arctan( root(1 - x^2) / x)

So I have to prove if that 'identity' is valid and say for what values of x is valid.

Even when I am able to construct the triangle from the data, I don't know how to find the valid values for x.

So far I have constructed a triangle with opposite root(1-x^2), adjacent x and hypotenuse 1.

I also restricted the domain so 1-x^2 >= 0, x != 0 and -1 <= x <= 1 (for the root, denominator, and arccos respectively).

I also tried to prove the identity using the Pythagorean theorem from the data of the arctan ( opposite and adjacent), and I get that the hypotenuse is 1, which is the same the arccos says.

But I have seen the plot of both functions and I realized they are only equal for 0 < x <= 1 and I don't know how I could get there. Intersecting the restrictions I said I only get [-1,1] - {0}.

I have even thought I should intersect the ranges of those functions, but I don't think that is valid.

Thanks in advance and sorry for the long post
• Sep 13th 2012, 07:11 AM
Plato
Re: Proving inverse functions identity
Quote:

Originally Posted by Danielc
I have the following equation
arccos(x) = arctan( root(1 - x^2) / x)
So I have to prove if that 'identity' is valid and say for what values of x is valid.
Even when I am able to construct the triangle from the data, I don't know how to find the valid values for x.
So far I have constructed a triangle with opposite root(1-x^2), adjacent x and hypotenuse 1.

Let $\displaystyle \theta=\arccos(x)$, so $\displaystyle \cos(\theta)=\frac{x}{1}$.

Thus $\displaystyle \tan(\theta)=\frac{\sqrt{1-x^2}}{x}$.

What don't you follow?
• Sep 13th 2012, 07:18 AM
Danielc
Re: Proving inverse functions identity
Yes, that is what I did.. But if you try the identity for, say, -1, it is not true... So what I need to do is to find (hopefully algebraically) the values that make that identity true.

I constructed the triangle doing the same as you did, but I don't know how should i proceed.

Thank you very much
• Sep 13th 2012, 07:21 AM
MaxJasper
Re: Proving inverse functions identity
Revised(Crying)
• Sep 13th 2012, 07:28 AM
Deveno
Re: Proving inverse functions identity
(deleted due to previous post deletion)
• Sep 13th 2012, 07:34 AM
MaxJasper
Re: Proving inverse functions identity
How can posts be deleted? There is no delete button at the top of my edit window(Talking)
• Sep 13th 2012, 07:48 AM
Plato
Re: Proving inverse functions identity
Quote:

Originally Posted by MaxJasper
How can posts be deleted?

At the very top of the edit box there is a delete option.
• Sep 13th 2012, 10:00 AM
Danielc
Re: Proving inverse functions identity

Yes, I can see where it is valid graphically, but I would like to know how to get to that conclusion algebraically

Thank you very much
• Sep 13th 2012, 11:06 AM
MaxJasper
Re: Proving inverse functions identity
If you are dealing with real numbers only then :

$\displaystyle 1-x^2\geq 0$

Hence we investigate:

$\displaystyle \cos ^{-1}(-1\leq x\leq 1)=?=\tan ^{-1}\left(\frac{\sqrt{1-(-1\leq x\leq 1)^2}}{-1\leq x\leq 1}\right)$

http://mathhelpforum.com/attachment....1&d=1347575251
• Sep 13th 2012, 01:19 PM
emakarov
Re: Proving inverse functions identity
Let $\displaystyle \varphi(x)$ denote $\displaystyle \arccos x$. We have the following facts.

If $\displaystyle -1\le x\le 1$, then $\displaystyle \cos(\arccos x)=\cos\varphi(x)=x$ (1)
If $\displaystyle -1\le x\le 0$, then $\displaystyle \pi/2\le \varphi(x)\le\pi$ (2)
From (2), if $\displaystyle -1\le x\le0$, then $\displaystyle \sin\varphi(x)\ge0$ (3)
The function arctan is the inverse of tan on $\displaystyle [-\pi/2,\pi/2]$ (4)
$\displaystyle \tan(\alpha)=\tan(\alpha-\pi)$ (5)
From (3) and (4), if $\displaystyle \pi/2\le\alpha\le\pi$, then $\displaystyle -\pi/2\le\alpha-\pi\le0<\pi/2$, so $\displaystyle \arctan(\tan\alpha)=\arctan(\tan(\alpha-\pi))=\alpha-\pi$ (6)

So, if $\displaystyle -1\le x\le 0$, then
$\displaystyle \frac{\sqrt{1-x^2}}{x} = \frac{\sqrt{1-\cos^2\varphi(x)}}{\cos\varphi(x)}$ by (1)
$\displaystyle = \frac{\sqrt{\sin^2\varphi(x)}}{\cos\varphi(x)} = \frac{|\sin\varphi(x)|}{\cos\varphi(x)} = \frac{\sin\varphi(x)}{\cos\varphi(x)}=\tan\varphi( x)$ by (3).

Therefore,

$\displaystyle \arctan\frac{\sqrt{1-x^2}}{x}=\arctan(\tan\varphi(x))=\varphi(x)-\pi=\arccos x-\pi$ by (6). For $\displaystyle 0\le x\le 1$ we simialrly have $\displaystyle 0\le\varphi(x)\le\pi/2$ and $\displaystyle \arctan(\tan\varphi(x))=\varphi(x)$, so $\displaystyle \arctan\frac{\sqrt{1-x^2}}{x}=\arccos x$.
• Sep 14th 2012, 11:40 AM
Danielc
Re: Proving inverse functions identity