How do you write an equation of #y=cosx# with pi/2 units down? Trigonometry Graphing Trigonometric Functions Translating Sine and Cosine Functions 1 Answer Douglas K. Mar 13, 2017 When given an equation, #y=(A)cos(Bx+C)+D#, #D# is the vertical shift. The equation you seek is #y=cos(x)-pi/2# Answer link Related questions How do you graph sine and cosine functions when it is translated? How do you graph #y=sin ( x -frac{\pi}{2} )#? How do you draw a sketch of #y = 1 + cos (x - pi)# How do you shift and graph #y=-3+sinx#? How do you graph #y=3sin(1/3x+ pi/2)-2#? How do you graph #1/2sin(x-pi)#? How do you graph #-sinx+2#? How do you graph #y=3sin(1/2)x#? How do you graph #y=-2cos((pix)/3)#? How do you graph #y = (1/2)sin(x - pi)#? See all questions in Translating Sine and Cosine Functions Impact of this question 1561 views around the world You can reuse this answer Creative Commons License