Both #y = A sint + k # and #y = A cos t + k# have the same midline as #y = 0#. The period is #2pi#, for both. Amplitude is #abs A# , for both.
Changing #y# to #y- 2#, the midline becomes #y - 2 = 0#. So, #k = 2#, for both.
The minimum amplitude #= 0#, in the midline #y = 2#. At #t = 0#, #y = A sin t + 2 = 2#. OK.
It is not so for .#y = A cos t + 2 = 3#, with amplitude, the maximum #A#. So change #t# to #(t +- pi/2)#. Then
# y = A cos(t +- pi/2) + 2 = 2#, at #t = 0 rArr# the amplitude is #0#.
See graphs, with #A = 1#.
Graph of #y = sin t + 2#:
graph{(y-sin x -2)(y-2) = 0}
Graph of #y = cos ( t - pi/2 ) + 2#:
graph{(y - cos ( x - pi/2 )-2)(y-2) = 0}
Graph of #y = cos ( t + pi/2 ) + 2#:
graph{(y - cos ( x + pi/2 )-2)(y-2) = 0}
Observe that the first two are the same and the third graph is different.
