Translating Sine and Cosine Functions
Add yours
Key Questions

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(x4)# , we can think of it as graphing#y=f(x4)# .
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(x4) [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(xh)# 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]} 
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 unitsThe graph
#y = cos(theta)  1# is a graph of#cos# shifted down the yaxis by 1 unitA 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 rightFor 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!

Start with the equation
#f(x)=asin(b(xh))+k# . The h is the horizontal translation and the k is the vertical translation.Notice that the the sign in front of the h is negative. That means that you do the inverse of h. The sign in front of the k is positive, meaning you do exactly what it says.
Let me give you an example:
#f(x)=2sin(3(xpi/3))5# . You will translate the function#pi/3# to the right (the inverse of#pi/3# ) and then you will translate the function down 5 units.If the b is not already factored out, you must do that before you translate horizontally. For example, if the above function had been written as:
#f(x)=2sin(3xpi)5# , you would have had to factor out the 3 before you translated.Cosine is done exactly the same.

If you are identifying the maximum, center (i.e., midline), and minimum from an equation , you start with the vertical shift, which gives you your midline. From the midline you will add and subtract the amplitude to get the maximum (midline + amplitude) and the minimum (midline  amplitude).
The vertical shift is the number (i.e., constant) that is added to or subtracted from the term with the trig function.
The amplitude, also referred to as the wave height, is the number (i.e., constant) that is the coefficient of the term with the trig function. Wave heights are distances, so amplitude is ALWAYS positive. If there is a negative sign, this reflects the graph over the midline, but it is not part of the amplitude.
Remember that we are dealing with VERTICAL maximum, VERTICAL center, and VERTICAL minimum. Hence, you will find answers about max, min, and midline (i.e., center) from looking at the VERTICAL axis (i.e., yaxis).
If you are identifying the maximum, center (i.e., midline), and minimum from a graph , you see how high the graph goes (max) and how low it goes (min). You can then determine the midline, which is exactly in the middle between the maximum and the minimum. One way to find the midline (i.e., center) is by averaging the maximum and minimum (i.e., (max + min)/2).