[VA] SRC#004- Fun with Sexagesimal Trigs

[Albert Chan]

(02-11-2019 09:26 PM)J-F Garnier Wrote:  the identification of the B value stopped me for a while.

Here is a good way to estimate B.

Middle term angle is around 90°, any angles bigger than that can be used to fill the "holes":

Example, 1/(sin(133°)sin(134°)) = 1/(sin(47°)sin(46°))

So, B = 1/(sin(45°)sin(46°)) + 1/(sin(46°)sin(47°)) + ... + 1/(sin(89°)sin(90°))

I remember derivative of tan(x) is sec(x)^2 = 1/cos(x)^2

So, change all the sines to cosines, angles goes from 0° to 45°
The x's are in degree, so need to scale the area, like this:

sum ~ (180/Pi) * integrate(sec(x)^2, x, 0, Pi/4) = (180/Pi) * (1 - 0) = 180/Pi

Actual sum is not exactly like this, but should be close.
My guess for true sum is 1/sin(1°), but to prove it is hard ...