Thread: 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

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 , so .

Thus .

What don't you follow?

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

Revised

(deleted due to previous post deletion)

How can posts be deleted? There is no delete button at the top of my edit window

Originally Posted by MaxJasper

How can posts be deleted?

EDIT your post.
At the very top of the edit box there is a delete option.

Hi and thanks for your answer.

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

If you are dealing with real numbers only then :



Hence we investigate:

Let denote . We have the following facts.

If , then (1)
If , then (2)
From (2), if , then (3)
The function arctan is the inverse of tan on (4)
(5)
From (3) and (4), if , then , so (6)

So, if , then
by (1)
by (3).

Therefore,

by (6). For we simialrly have and , so .

Thank you very much for your answers.

I wanted something like what emakarov posted, so it helped a lot. I tried to put it in simple terms so concluded:

- tan (theta) = opposite / adjacent, but as we are talking about lengths, negative values for any of both have no sense. In this case, x was in the denominator, si if it is negative it has no sense.
- So tan(theta) = sin(theta) / cos(theta) . As the numerator is always non-negative, theta must be an angle between [0, PI] so its sin is positive. Then, for theta to be positive, x >= 0, but as it can't be 0, it must be between (0, 1], considering the restrictions on the domain.

Thank you very much