Prove Domain of Identity: (-1, 1], C is Any Real Number

[NATURE.M]

Homework Statement

Prove that there is a constant C such that
arctan\sqrt{\frac{1-x}{1+x}} = C - \frac{1}{2}arcsinx for all x in a certain domain. What is the largest domain on which this identity is true? What is the value of the constant C?

The Attempt at a Solution

Now I know how to prove the initial statement (showing the derivatives are equal which implies they differ by only a constant), but I wanted to verify the largest domain and the value of C.

For the largest domain on which this identity is true I obtained (-1, 1] (since arcsinx is defined on
[-1, 1], and since -1 is not allowed).

And I believe C can be any real number.

So I'd just like to verify whether or not my interpretation is correct?

 

[scurty]

As far as I can tell, your domain is correct.

Am I right in interpreting that you are saying this equation is true for any value of C? If so, that's not true.

To solve for C, pick any value for x in the domain, plug it into the equation, and solve for C.

 

[NATURE.M]

As far as I can tell, your domain is correct.

Am I right in interpreting that you are saying this equation is true for any value of C? If so, that's not true.

To solve for C, pick any value for x in the domain, plug it into the equation, and solve for C.

Yeah your right. I made a mistake there. And thanks scurty.

 

[scurty]

Yeah your right. I made a mistake there. And thanks scurty.

Edit: Nevermind.

To make this post worthwhile, I suggest x values of 0 or 1 to solve for C.

 

[NATURE.M]

Edit: Nevermind.

To make this post worthwhile, I suggest x values of 0 or 1 to solve for C.

Yeah I used x=0 and obtained C=\pi/4.

 

[vanhees71]

It's all correct, but C is uniquely defined by the equation. Just choose a value for x, for which you know the values on both sides of the equation!