Proving John Machin's Formula for Arctan Difference

[DivGradCurl]

I'm a bit puzzled by this problem, which involves the John Machin's formula. Here it goes:

Show that for xy \neq -1,

\arctan x - \arctan y = \arctan \frac{x-y}{1+xy}

ïf the left side lies between -\pi /2 and \pi /2.

By the way, John Machin's formula is:

4 \arctan \frac{1}{5} - \arctan \frac{1}{239} = \arctan \frac{\pi}{4}

He used this to find \pi correct to 100 decimal places.

Anyhow, here is what I've done so far:

If

\arctan x = \sum _{n=0} ^{\infty} \left( -1 \right) ^n \frac{x^{2n+1}}{2n+1}

\arctan y = \sum _{n=0} ^{\infty} \left( -1 \right) ^n \frac{y^{2n+1}}{2n+1}

\arctan \frac{x-y}{1+xy} = \sum _{n=0} ^{\infty} \left( -1 \right) ^n \frac{\left( \frac{x-y}{1+xy} \right)^{2n+1}}{2n+1}

Then

\sum _{n=0} ^{\infty} \left( -1 \right) ^n \frac{\left( x^{2n+1} - y^{2n+1}\right)}{2n+1} = \sum _{n=0} ^{\infty} \left( -1 \right) ^n \frac{\left( \frac{x-y}{1+xy} \right)^{2n+1}}{2n+1}

Therefore

x^{2n+1} - y^{2n+1} = \left( \frac{x-y}{1+xy} \right)^{2n+1}

I'm not sure I am on the right path. Maybe, I simply need to pick a couple of values to demonstrate that the given relationship is true, but it feels like it might not be enough. In other words, "show" up there looks a bit ambiguous. Any help is highly appreciated.

Thanks

 

[dextercioby]

My friend,why use calculus to get nowhere,when u can use (circular) trigonometry to prove in an elegant way?
Take "tan"out both sides of your identity and use what i'll write below:
\tan(a-b)=:\frac{\sin(a-b)}{\cos(a-b)}=\frac{\sin a\cos b-\sin b\cos a}{\cos a\cos b+\sin a\sin b}

Simplify the last fraction through the product of 'cosines' to find the celebrated formula
\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}
Substitute in the last identity:
a\rightarrow \arctan x;b\rightarrow \arctan y
,make use of the fact that "tan",on the interval give in the problem is a uniform function and get exactly the formula u would get if taking "tan" from the identity u need to prove.

Daniel.

PS.Trigonometry is beauty...

 

[DivGradCurl]

Thanks for your input, Daniel.

Let me see if I understand it...

\arctan x - \arctan y = \arctan \frac{x-y}{1+xy} \qquad (1)​

x - y = \frac{x-y}{1+xy} \qquad (2)​

\tan \left( x - y \right) = \tan \left( \frac{x-y}{1+xy} \right) \qquad (3)​

\tan \left( x - y \right) = \frac{\sin \left( x - y \right)}{\cos \left( x - y \right)} \qquad (4)​

\tan \left( x - y \right) = \frac{\sin x \cos y - \cos x \sin y}{\cos x \cos y + \sin x \sin y} \qquad (5)​

\tan \left( x - y \right) = \left( \frac{\frac{1}{\cos x \cos y}}{\frac{1}{\cos x \cos y}} \right) \frac{\sin x \cos y - \cos x \sin y}{\cos x \cos y + \sin x \sin y} \qquad (6)​

\tan \left( x - y \right) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \qquad (7)​

\arctan x - \arctan y = \arctan \frac{x-y}{1+xy} \qquad (8)​

 

[dextercioby]

U didn't. I can't explain in another way the bunch of c*** at numbers (2) and (3)...

Formulas (4) pp.(7) give a proof for the trigonometrical identity that can be used to prove your identity.I stated that proof just to let u know i didn't invent it nor rediscovered it hundreds of years later.

I said to apply tangent on both sides of your identity:
\tan(\arctan x-\arctan y) =\tan[\arctan(\frac{x-y}{1+xy})]

`Work the left hand side using the formula I've given proof:
\tan(\arctan x-\arctan y) =\frac{\tan(\arctan x)-\tan(\arctan y)}{1+\tan(\arctan x)\tan(\arctan y)}=\frac{x-y}{1+xy} (1)

Work the right hand side:
\tan[\arctan(\frac{x-y}{1+xy})] =\frac{x-y}{1+xy} (2)

Interpretation of the relations (1) and (2) is that you have shown that:
\tan A=C (1') \tan B=C (2')(3)
From (3) it follows immediately
\tan A=\tan B (4)
From (4) and from tha fact that on the interval (-\frac{\pi}{2},\frac{\pi}{2}) the function "tan" is a uniform/surjective function,u get that
A=B (5)
,which is nothing but your identity in symbolic form.

Daniel.

 

[DivGradCurl]

Oh... I now see what you mean. Thank you.