Is there any numerical-accurracy difference on calculating sin(pi/2-A) and cos(A) in Matlab?

Question

I am reading a matlab function for calculating great circle distance written by my senior collegue. The distance between two points on the earth surface should be calculated using this formula:

d = r * arccos[(sin(lat1) * sin(lat2)) + cos(lat1) * cos(lat2) * cos(long2 – long1)]

However, the script has the code like this:

dist = (acos(cos(pi/180*(90-lat2)).*cos(pi/180*(90-lat1))+sin(pi/180*(90-lat2)).*sin(pi/180*(90-lat1)).*cos(pi/180*(diff_long)))) .* r_local;
(-180 < long1,long2 <= 180, -90 < lat1,lat2 <= 90)

Why are sin(pi/2-A) and cos(pi/2-A) used to replace cos(A) and sin(A)？ Doesn't it introduced more error source by using the constant pi? Since lat1, lat2 might be very close to zero in my work, is this a trick on the numerical accuracy of MATLAB's sin() and cos() function?

Look forward to answers that explain how trigonometric functions in MATLAB work and analyze the error of these functions when the argument is close or equal to 0 and pi/2.

Answer 1

If the purpose is to increase accuracy, this seems a very poor idea. When the angle is small, 90-A spoils any accuracy. That even makes tiny angles vanish (90-ε=90).

On the opposite, the sine of tiny angles is very close to the angle itself (radians) and for this reason quite accurately computed, while the cosine is virtually 1 or 1-A²/2. For top accuracy on tiny angles, you may resort to the versine, using versin(A):= 1-cos(A) = 2 sin²(A/2) and rework the equations in terms of 1-versin(A) instead of cos(A).

If the angle is close to 90°, accuracy is lost anyway, 90°-A will not restore it.