Problem of the Week #51 - March 18th, 2013

[Chris L T521]

Thanks to those who participated in last week's POTW!! Here's this week's problem!

-----

Problem: Use calculus to prove the identity $\arcsin\left(\dfrac{x-1}{x+1}\right)=2\arctan(\sqrt{x})-\dfrac{\pi}{2}$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Chris L T521]

This week's problem was correctly answered by MarkFL, Siron, and Sudharaka. They each approached the problem differently, so I will post all of their solutions.

Here's MarkFL's solution:

Spoiler

Let:

$\displaystyle f(x)=\sin^{-1}\left(\frac{x-1}{x+1} \right)$

$\displaystyle g(x)=2\tan^{-1}\left(\sqrt{x} \right)-\frac{\pi}{2}$

We may observe that:

i) $\displaystyle f(0)=-\frac{\pi}{2}=g(0)$

ii) $\displaystyle f'(x)=\frac{1}{\sqrt{1-\left(\frac{x-1}{x+1} \right)^2}}\cdot\frac{2}{(x+1)^2}=\frac{1}{\sqrt{x}(x+1)}$

$\displaystyle g'(x)=\frac{2}{x+1}\cdot\frac{1}{2\sqrt{x}}=\frac{1}{\sqrt{x}(x+1)}$

By the existence and uniqueness theorem, we may then conclude that:

$f(x)=g(x)$.

Here's Siron's solution:

Spoiler

To prove:$$\arcsin\left(\frac{x-1}{x+1}\right) = 2\arctan(\sqrt{x}) -\frac{\pi}{2}$$
Proof:
Taking the sinus of both sides gives us
1. LHS
$$\sin\left[\arcsin\left(\frac{x-1}{x+1}\right)\right] = \frac{x-1}{x+1}$$

2. RHS
$$\sin\left[ 2\arctan(\sqrt{x}) -\frac{\pi}{2}\right] = \sin[2\arctan(\sqrt{x})]\cos\left(-\frac{\pi}{2}\right)-\sin\left(-\frac{\pi}{2}\right)\cos\left[2\arctan(\sqrt{x})\right] = \cos[2\arctan(\sqrt{x})] = 2\cos^2[\arctan(\sqrt{x})]-1$$

Since
$$\cos^2[\arctan(x)] = \frac{1}{1+\tan^2[\arctan(x)]} = \frac{1}{1+x^2}$$

we obtain
$$RHS = \frac{2}{1+x}-1 = \frac{2-(1+x)}{1+x} = \frac{1-x}{1+x}$$

Here's Sudharaka's solution:

Spoiler

Let, \(\displaystyle f(x)=2\, tan^{-1}\left( \sqrt{x}\right) -sin^{-1}\left( \frac{x-1}{x+1}\right) \). Differentiate this with respect to \(x\) and we get,

\[f'(x)=0\]

Therefore \(f\) is a constant function. That is,

\[2\, tan^{-1}\left( \sqrt{x}\right) -sin^{-1}\left( \frac{x-1}{x+1}\right)=C\]

where \(C\) is a constant. To find \(C\) substitute \(x=1\). Then we get, \(\displaystyle c=\frac{\pi}{2}\).

\[\therefore 2\, tan^{-1}\left( \sqrt{x}\right) -sin^{-1}\left( \frac{x-1}{x+1}\right)=\frac{\pi}{2}\]

\[\Rightarrow \arcsin\left(\dfrac{x-1}{x+1}\right)=2\arctan(\sqrt{x})-\dfrac{\pi}{2}\]