# Proving inverse functions identity

• Sep 13th 2012, 07:35 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, 08:11 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 $\theta=\arccos(x)$, so $\cos(\theta)=\frac{x}{1}$.

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

What don't you follow?
• Sep 13th 2012, 08:18 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, 08:21 AM
MaxJasper
Re: Proving inverse functions identity
Revised(Crying)
• Sep 13th 2012, 08:28 AM
Deveno
Re: Proving inverse functions identity
(deleted due to previous post deletion)
• Sep 13th 2012, 08:34 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, 08:48 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, 11:00 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, 12:06 PM
MaxJasper
Re: Proving inverse functions identity
If you are dealing with real numbers only then :

$1-x^2\geq 0$

Hence we investigate:

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

http://mathhelpforum.com/attachment....1&d=1347575251
• Sep 13th 2012, 02:19 PM
emakarov
Re: Proving inverse functions identity
Let $\varphi(x)$ denote $\arccos x$. We have the following facts.

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

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

Therefore,

$\arctan\frac{\sqrt{1-x^2}}{x}=\arctan(\tan\varphi(x))=\varphi(x)-\pi=\arccos x-\pi$ by (6). For $0\le x\le 1$ we simialrly have $0\le\varphi(x)\le\pi/2$ and $\arctan(\tan\varphi(x))=\varphi(x)$, so $\arctan\frac{\sqrt{1-x^2}}{x}=\arccos x$.
• Sep 14th 2012, 12:40 PM
Danielc
Re: Proving inverse functions identity
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 26, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://mathhelpforum.com/trigonometry/203388-proving-inverse-functions-identity-print.html", "fetch_time": 1519617796000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2018-09/segments/1518891817999.51/warc/CC-MAIN-20180226025358-20180226045358-00564.warc.gz", "warc_record_offset": 222046957, "warc_record_length": 4702, "token_count": 1506, "char_count": 4499, "score": 3.859375, "int_score": 4, "crawl": "CC-MAIN-2018-09", "snapshot_type": "longest", "language": "en", "language_score": 0.9326489567756653, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00039-of-00064.parquet"}