Translating Sine and Cosine Functions

Key Questions

  • How do you identify the vertical and horizontal translations of sine and cosine from a graph and an equation?

    For an equation:

    A vertical translation is of the form:
    #y = sin(theta) + A# where # A!=0#
    OR #y = cos(theta) + A#

    Example: #y = sin(theta) + 5# is a #sin# graph that has been shifted up by 5 units

    The graph #y = cos(theta) - 1# is a graph of #cos# shifted down the y-axis by 1 unit

    A horizontal translation is of the form:
    #y = sin(theta + A)# where #A!=0#

    Examples:
    The graph #y = sin(theta + pi/2)# is a graph of #sin# that has been shifted #pi/2# radians to the right

    For a graph:
    I'm to illustrate with an example given above:

    For compare:
    #y = cos(theta)#
    graph{cosx [-5.325, 6.675, -5.16, 4.84]}

    and

    #y = cos(theta) - 1#
    graph{cosx -1 [-5.325, 6.675, -5.16, 4.84]}
    To verify that the graph of #y = cos(theta) - 1# is a vertical translation, if you look on the graph,

    the point where #theta = 0# is no more at #y = 1# it is now at # y = 0#

    That is, the original graph of #y= costheta# has been shifted down by 1 unit.

    Another way to look at it is to see that, every point has been brought down 1 unit!

    Antoine · 1 · Apr 12 2015
  • How do you graph sine and cosine functions when it is translated?

    I think you'll find a useful answer here: http://socratic.org/trigonometry/graphing-trigonometric-functions/translating-sine-and-cosine-functions

    Vertical translation

    Graphing #y=sinx+k# Which is the same as #y=k+sinx#:

    In this case we start with a number (or angle) #x#. We find the sine of #x#, which will be a number between #-1# and #1#. The after that, we get #y# by adding #k#. (Remember that #k# could be negative.)

    This gives us a final #y# value betwee #-1+k# and #1+k#.

    This will translate the graph up if #k# is positive (#k>0#)
    or down if #k# is negative (#k<0#)

    Examples:

    #y=sinx#
    graph{y=sinx [-5.578, 5.52, -1.46, 4.09]}

    #y=sinx+2 = 2+sinx#
    graph{y=sinx+2 [-5.578, 5.52, -1.46, 4.09]}

    #y=sinx-4=-4+sinx#
    graph{y=sinx-4 [-5.58, 5.52, -5.17, 0.38]}

    The reasoning is the same for #y=cosx+k=k+cosx#, but the starting graph looks different, so the final graph is also different:

    #y=cosx#
    graph{y=cosx [-5.578, 5.52, -1.46, 4.09]}

    #y=cosx+2 = 2+cosx#
    graph{y=cosx+2 [-5.578, 5.52, -1.46, 4.09]}

    Jim H · 1 · Mar 22 2015

Questions

Graphing Trigonometric Functions