Why is pi = 16 arctan(1/5) - 4 arctan( 1/139)

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The equation \(\pi = 16\arctan\left(\frac{1}{5}\right) - 4\arctan\left(\frac{1}{239}\right)\) is derived from Machin's formula, which utilizes the series expansion of the arctangent function. The series expansion for \(\tan^{-1}(x)\) is \(\frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} - \frac{1}{5x^5} + \ldots\). This relationship is exact and can be verified through manipulation of the series. The discussion emphasizes the importance of understanding series expansions in trigonometric identities.

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joex444
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Not actually a HW question, but, is there a quick explanation that could show why:

\pi = 16\arctan(\frac{1}{5}) - 4\arctan(\frac{1}{239})

I read that somewhere, and it turned out to be true, so I was just wondering how that came about...
 
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We have the series expansion:
Tan^{-1}(x) = \frac {\pi} 2 - \frac 1 x + \frac 1 {3x^3} - \frac 1 {5x^5} ...
By working with this you should be able to verify your claim. It is not clear to me whether your relationship is exact or just good to more decimal places then are being displayed.
 
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http://mcraefamily.com/MathHelp/GeometryTrigEquiv.htm (see Machin's formula)
 
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