[VA] SRC #009 - Pi Day 2021 Special

[robve]

(03-17-2021 02:32 AM)Albert Chan Wrote:  

(03-16-2021 09:09 PM)robve Wrote:  $$ \arctan u - \arctan v = \arctan\frac{u-v}{1+uv} $$

This is not quite right, LHS has possible range of ±pi, RHS is limited to ±pi/2

Right. I did not include the two necessary conditions since these hold, note the (mod π):

$$ \arctan u - \arctan v = \arctan\frac{u-v}{1+uv}\quad \pmod \pi,\quad uv\neq 1 $$

EDIT: the arctan identity comes from

$$ \tan \alpha \pm \tan \beta = \frac{\tan \alpha \pm \tan \beta}{1\mp \tan \alpha\, \tan \beta} $$

Since 0<=e<π and 0<=(e-1)/(e+1)<π we have (e.g. verify numerically)

$$\tan\arctan\mathrm{e}=\mathrm{e}; \quad \tan\arctan\frac{\mathrm{e}-1}{\mathrm{e}+1} = \frac{\mathrm{e}-1}{\mathrm{e}+1}$$

Then

$$ \alpha=\arctan\mathrm{e};\quad \beta=\arctan \frac{\mathrm{e}-1}{\mathrm{e}+1};\quad \frac{\tan \alpha - \tan \beta}{1+ \tan \alpha\, \tan \beta} = \frac{\mathrm{e} - \frac{\mathrm{e}-1}{\mathrm{e}+1}}{1+\mathrm{e}\frac{\mathrm{e}-1}{\mathrm{e}+1}} = 1 $$

Generally, tan has period π

$$ \tan(k\pi+\theta) = \theta $$

for any integer k.

- Rob