I learnt the maclaurin series of sin x and then i realised it cant work for both degrees and radians. Of course it should work in radians but i cant seem to understand why. In the proof of the serie, which is x - x^3/3! + x^5/5! - x^7/7! + ....(cba with latex) , you only require sin 0 = 0 and cos 0 = 1. this is true for both degree and radian so that cant be the reason why degree dont work. now i understand that you need sinx/x -> 1 as x -> 0, but this seem to be true for degree as well, because as an angle x in degree get smaller, so does sin x, so they both -> 0. so sinx/x -> 1 for degree too (surely?)

Then i found a proof that prove sinx/x = 1 without referring to degree or radian again. link

Direct Proof from Definition of Sine

This proof works directly from the definition of the sine function:

sinx =∑ n=0∞(−1)nx2n+1(2n+1)!By the definition of the sine function =(−1)0 x2⋅0+1(2⋅0+1)!+∑n=1∞(−1)nx2n+1(2n+1)! =x+∑n=1∞(−1)nx2n+1(2n+1)!

now this confuse me even more cos now we have a proof of sinx/x without referring to deg or rad, which suggest that both will work.

limx→0sinxx =lim x→0x+∑∞n=1(−1)nx2n+1(2n+1)!x =lim x→0xx+limx→0∑∞n=1(−1)nx2n+1(2n+1)!x =1+lim x→0∑∞n=1(−1)nx2n(2n)!1by Power Series Differentiable on Interval of Convergence and L'Hôpital's Rule =1+lim x→0∑n=1∞(−1)nx2n(2n)! =1+∑ n=1∞(−1)n02n(2n)!by Polynomial is Continuous =1

i know deg cant work for the power serie, cos then you could divide the 360deg angle into any number of parts and call that a measurement and it would work. but why radian? its just dividing the 360 angle into 2pi parts

sorry if anything is unclear. please ask. this is probably a noob question with a obvious answer but its bugging me for ages. i just cant find the answer or understand it