opl. vergelijk

sin(A)=sin(B) dan A = B + k*2pi of A = pi - B + k*2pi
cos(A)=cos(B) dan A = B + k*2pi of A = -B + k*2pi
sin(A)=sin(B) da
n A = B + k*2pi
of A = pi - B +
k*2pi
cos(A)=cos(B) da
n A = B + k*2pi
of A = -B + k*2p
i