[VA] SRC#004- Fun with Sexagesimal Trigs

[Albert Chan]

B = 1/(s45s46) + 1/(s47s18) + ... 1/(s89s90) + 1/(s91s92) + ... + 1/(s133s134)
   = 1/(c44c45) + 1/(c42c43) + ... 1/(c0c1) + 1/(c1c2) + ... + 1/(c43c44)
   = 1/(c0c1) + 1/(c1c2) + 1/(c2c3) + ... + 1/(c44c45)

Proof below statement by induction. If success, it proved B = tan(45°) / sin(1°) = 1/sin(1°)

sum(1/(cos(k°)*cos((k+1)°)), k=0 to n-1) = tan(n°) / sin(1°)

For n=1, 1/(c0c1) = 1/c1 = tan(1°) / sin(1°)

Assume this work for n=k, add 1 more term, we got:

tan(k°) / sin(1°) + 1/(ck c(k+1)) = (sk c(k+1) + s1) / (s1 ck c(k+1))

Since s1 = s((k+1) - k) = s(k+1) ck - c(k+1) sk, we got:

Sum of k+1 terms = (s(k+1) ck) / (s1 ck c(k+1)) = tan((k+1)°) / sin(1°)

QED

Update: above general formula work in any angle unit, not just degrees.