10
$\begingroup$

I recently determined that for all integers $a$ and $b$ such that $a\neq b$ and $b\neq 0$,

$ \arctan\left(\frac{a}{b}\right) + \frac{\pi}{4} = \arctan\left(\frac{b+a}{b-a}\right) $

This implies that 45 degrees away from any angle with a rational value for tangent lies another angle with a rational value for tangent. The tangent values are related.

If anyone can let me know if this has been done/shown/proven before, please let me know. Thanks!

  • 0
    +1 for research. Praise, not shame, for the (re)discovery.2016-06-05

3 Answers 3

12

As written, the formula is not true: the values of $\arctan(x)$ are always between $-\frac{\pi}2$ and $\frac{\pi}{2}$. Pick a rational number $\frac{a}{b}$ with $\frac{\pi}{4}\lt \frac{a}{b}\lt \frac{\pi}{2}$. For example, $a=11$, $b=10$. Then the left hand side, $\arctan\left(\frac{11}{10}\right)+\frac{\pi}{4}\approx 1.6184$ whereas the right hand side is negative: $\arctan\left(\frac{11+10}{10-11}\right) = \arctan(-21) \approx -1.5232.$

I think that what you mean is that if $\alpha$ is an angle such that $\tan(\alpha)$ is rational, different from $1$, $\tan(\alpha)=\frac{a}{b}\neq 1,\qquad a,b\text{ integers},$ then $\tan\left(\alpha+\frac{\pi}{4}\right) = \frac{b+a}{b-a}.$

Certainly, well done if you discovered it by yourself! However, it is not new. In fact, the result is true even if $a$ and $b$ are not integers; all you need is for $a$ to be different from $b$, that is, for $\alpha\neq\frac{\pi}{4}$.

There are well-known formulas that express the sine, cosine, and tangent of a sum of angles in terms of the sines, cosines, and tangents of the summands:

$\begin{align*} \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\\ \cos(\alpha+\beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\\ \tan(\alpha+\beta) &= \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}. \end{align*}$ Taking $\beta=\frac{\pi}{4}$, since $\tan(\frac{\pi}4) = 1$, we get $\tan\left(\alpha+\frac{\pi}{4}\right) = \frac{\frac{a}{b}+1}{1-\frac{a}{b}} = \frac{\quad\frac{a+b}{b}\quad}{\frac{b-a}{b}} = \frac{a+b}{b-a},$ giving your formula.

  • 0
    @Brian: Good; just double-checking. You'd be surprised how many students today, even in "advanced classes" like calculus, cannot tell the difference between the actual value and$a$decimal approximation given by a calculator.2011-12-14
1

If you differentiate the function $f(t)=\arctan t - \arctan\frac{1 + t}{1 - t},$ you get zero, so the function is constant in each of the two intervals $(-\infty,1)$ and $(1,+\infty)$ on which it is defined.

  • Its value at zero is $\pi/2$, so that $f(t)=-\pi/4$ for all $t<1$, so $ \arctan t + \frac\pi4 = \arctan\frac{1 + t}{1 - t},\qquad\forall t<1.$

  • On the other hand, one easily shows that $\lim_{t\to+\infty}f(t)=\frac{3\pi}{4}$, so $ \arctan t - \frac{3\pi}4 = \arctan\frac{1 + t}{1 - t},\qquad\forall t>1.$

If $t=a/b$ is a rational number smaller that $1$, then the first point is your identity. If it larger than $1$, we see that you have to change things a bit.

0

Quoting from Wikipedia's list of trigonometric identities:

BEGIN QUOTE

$ f(x) = \frac{(\cos\alpha)x - \sin\alpha}{(\sin\alpha)x + \cos\alpha}, $

[$\ldots\ldots$ some material omitted here $\ldots\ldots$]

If $x$ is the slope of a line, then $f(x)$ is the slope of its rotation through an angle of $-\alpha$.

END QUOTE

Dividing the numerator and denominator by $\tan\alpha$ may give the same result as is posted here.