y_(1) = 4 sin (omega t + kx), y_(2) =-4 cos(omega t + kx), the phase difference is

y_(1) = 4 sin (omega t + kx), y_(2) =-4 cos(omega t + kx), the phase d

1.	y_(1) = 4 sin (omega t + kx), y_(2) =-4 cos(omega t + kx), the phase difference is
Answer» `PI/2` `(3pi)/2` `pi` ZERO Answer :B