How do you graph sine and cosine functions when it is translated?

2 Answers
Mar 22, 2015

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]}

Mar 22, 2015

Horizontal Translation

One way to think about horizontal translations of a function is to think about the value of #x# that will cause us to find #f(0)#.

We know the graph of #y=f(x)=sin(x)#.

To graph #y=sin(x-4)#, we can think of it as graphing #y=f(x-4)#.
Now, what value of #x# will make me find #f(0)=sin(0)#? Clearly, it is #x=4#.
So "#4# is the new #0#". Everything moves #4# to the right.

graph{y=sin(x-4) [-0.498, 7.295, -2.302, 1.596]}

To graph #y=sin(x+ pi/3)#, we ask ourselves, "What value of #x# will cause us to find #sin(0)#?
That will be #x=- pi/3#
So, "#- pi/3# is the new #0#". Everything moves #- pi/3# to the right. Wait a minute, surly it's more clear to say: Everything moves #pi/3# to the left.

(For the graph below, remember that #pi/3# is a little more than #1#)

graph{y=sin(x+pi/3) [-3.02, 1.845, -1.192, 1.241]}

To start the graph of #y=sin(bx- c)# , we'll need to change the period and also translate.

Cosine

The reasoning for graphing #y=cos(x-h)# is the same. The difference is that
#sin(0)=0#, so the point corresponding to the 'new #0#' goes on the #x#-axis
#cos0=1#, so the point corresponding to the 'new #0#' goes at #y=1#

#y = cos (x+ pi/3)#
graph{y=cos(x+pi/3) [-3.02, 1.845, -1.192, 1.241]}