So I have this identity: arctan(1/2) + arctan(1/3) = arctan(1)
I want to prove the identity using linear algebra: First, I will draw a line with slope 1/2. Then, I will change to the coordinate system represented by that line, and draw a line from that coordinate system with a slope 1/3. When I go back to my original coordinate system, I should find a line with slope 1.
I wanted to verify the result first using the formula $B\vec{x} = \vec{r}$, where B is the new coordinate system with basis [2, 1] and [-1, 2], (representing the coordinate system from the line with slope 1/2), x is the vector in the original coordinate system ([1, 1]), and r is the vector relative to the B coordinate system.
I calculated, and got this:
$$ \begin{bmatrix} 2 & -1 \\\\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 \\\\ 1 \end{bmatrix} =\begin{bmatrix} 2-1 \\\\ 1+2 \end{bmatrix} =\begin{bmatrix} 1 \\\\ 3 \end{bmatrix} $$
The result [1, 3] seems like it represents the line 1/3, but that is not correct: it represents the line with slope 3 instead. It somehow got flipped upside down, and I have no idea why. It's been a while since I have done linear algebra, and I fear that I had made the most unintelligent mistake known to man, but I cannot spot why I am receiving this wrong answer.
Follow up: I solved arctan(1/3) + arctan(1/4) + arctan(1/5) + arctan(1/n) = arctan(1) the same way, this time chaining together the B matrixes. I got [1, 47], which again, was almost correct but instead it was flipped and is supposed to be [47, 1].
Thanks @TomKern
My formula $B\vec{x}=\vec{r}$ was incorrect. It is actually suppose to be $B\vec{r}=\vec{x}$, where B is the new coordinate system, r the vector relative to the new coordinate system, and x is the resulting vector in the original coordinate system.
I should've multiplied B by [3, 1] instead of [1, 1], which gives me [5, 5], corresponding to a line with slope 1.