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Using multiple triangles to find the values of some arctan angles in a trigonometry question so that I could use an exact expression I wondered how could I prove that the following relation applies to all possible rational numbers without using geometry to prove individual case by case:

$\frac{\pi}{4}-\arctan \frac{m}{n}=\arctan \frac{1}{m+n}$

Where the original geometrical thought was

$\frac{\pi}{4}-\arctan \frac{1}{2}=\arctan \frac{1}{3}$

$\frac{\pi}{4}-\arctan \frac{2}{3}=\arctan \frac{1}{5}$

And lastly, could I use this more widely with complex numbers and if so with what limitations?

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  • $\begingroup$ As $\arctan\frac12>\arctan\frac13$, you first equation is clearly false. Come to think of it, so is your second. $\endgroup$
    – Angina Seng
    Dec 25, 2017 at 11:02
  • $\begingroup$ Oh yes, I just noticed I typed + instead of a - sign. Thanks for noticing! So it should be pi per 4 minus the arctan $\endgroup$
    – Fr. Eon
    Dec 25, 2017 at 11:06

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I'm not sure exactly what you are asking here; your equations are wrong. But if $0<m<n$ then $$\frac\pi4-\arctan\frac mn=\arctan\frac{n-m}{m+n}.$$ In particular, $$\frac\pi4-\arctan\frac m{m+1}=\arctan\frac{1}{2m+1}$$ so that $$\frac\pi4-\arctan\frac12=\arctan\frac{1}{3},$$ $$\frac\pi4-\arctan\frac23=\arctan\frac{1}{5},$$ $$\frac\pi4-\arctan\frac34=\arctan\frac{1}{7}$$ etc. Is this what you were aiming at?

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  • $\begingroup$ Yes.So in proving it analytically is it all about the relation between m and n? $\endgroup$
    – Fr. Eon
    Dec 25, 2017 at 11:09
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Notice that $$\tan \left(\frac{\pi}{4} - \arctan(\frac{m}{n}) \right) = \frac{n-m}{n+m} $$

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