# Math Help - Prove this identity

1. ## Prove this identity

Prove this identity:
arcsin(x-1/x+1) = 2arctan(sqrt(x)) - pi/2

First of all to prove this do i take the derivative of one side?
My book does a horrible job at explaining any of this and then they throw problems like this in there.

I tried taking the derivative of the right side and got

arcsin(x-1/x+1) = 1/[sqrt(x) + (x)sqrt(x)]

but im not sure what to do from here or even if i am doing it correctly.

any help would be great.

thank you

2. Originally Posted by mybrohshi5
Prove this identity:
arcsin(x-1/x+1) = 2arctan(sqrt(x)) - pi/2

First of all to prove this do i take the derivative of one side?
My book does a horrible job at explaining any of this and then they throw problems like this in there.

I tried taking the derivative of the right side and got

arcsin(x-1/x+1) = 1/[sqrt(x) + (x)sqrt(x)]

but im not sure what to do from here or even if i am doing it correctly.

any help would be great.

thank you
Is the identity true - put $x = 1$!

3. Originally Posted by Danny
Is the identity true - put $x = 1$!
Perhaps he meant to write:

$\arcsin\left(\frac{x-1}{x+1}\right)=2\arctan(\sqrt{x})-\frac{\pi}{2}$

because that does hold when $x=1$ (and everywhere in fact, because it's true!).

We are going to show that $\frac{x-1}{x+1}=\sin\left(2\arctan(\sqrt{x})-\frac{\pi}{2}\right)$

Using the sum identity, the right side becomes:

$\sin\left(2\arctan(\sqrt{x})\right)\cos(\pi/2)-\cos\left(2\arctan(\sqrt{x})\right)\sin(\pi/2)=-\cos\left(2\arctan(\sqrt{x})\right)$

Using the double angle formula ( $\cos2\theta=\cos^2\theta-\sin^2\theta$, with $\theta=\arctan(\sqrt{x})$ in this case), we can draw out the triangle and find:

$-\cos\left(2\arctan(\sqrt{x})\right)=-\left(\left(\frac{1}{\sqrt{x+1}}\right)^2-\left(\frac{\sqrt{x}}{\sqrt{x+1}}\right)^2\right)= \frac{x}{x+1}-\frac{1}{x+1}=\frac{x-1}{x+1}$

as desired.

4. Originally Posted by mybrohshi5
Prove this identity:
arcsin(x-1/x+1) = 2arctan(sqrt(x)) - pi/2
let $\theta = \arctan(\sqrt{x})$

take the sine of the right side ...

$\sin \left(2\theta - \frac{\pi}{2}\right) =$

$\sin(2\theta)\cos\left(\frac{\pi}{2}\right) - \cos(2\theta)\sin\left(\frac{\pi}{2}\right) =$

$-\cos(2\theta) = 2\sin^2{\theta} - 1$

if $\theta = \arctan(\sqrt{x})$ , $\sin{\theta} = \frac{\sqrt{x}}{\sqrt{1+x}}$

$2\sin^2{\theta} - 1 = \frac{2x}{1+x} - \frac{1+x}{1+x} = \frac{x-1}{x+1}$

so ...

$\sin \left(2\theta - \frac{\pi}{2}\right) = \frac{x-1}{x+1}$

and ...

$\arcsin\left(\frac{x-1}{x+1}\right) = 2\theta - \frac{\pi}{2} = 2\arctan(\sqrt{x}) - \frac{\pi}{2}
$