Question

in the geometric proof for derivative of sine, the prof says that it is for all values of theta, and in the remarks he says that it is for when theta = 0, how is it that the easier proof is only for when theta =0

Answer 1

I believe for the easier proff you mean differential of Sin x when X=0, he gave that as an example because it is easier to see that it is equals to 1 but that is only for x=0, as he is only evaluating the answer for that and not the specific formula.

Answer 2

sinx =0 when x= n*[\pi\] ; where n is an integer.
at x=0 , sinx =0 not 1.