Translations of Sine and Cosine Curves
We already know how
translate a graph from our study of functions. A translation is
the form y - k = f(x - h), where it is translated k units
and h units horizontally. This fits right into our study of the
and Cosine curves.
Sine and Cosine
If the graphs of y
A Sin Bx and y = A Cos Bx are translated h units horizontally and k
vertically, then the resulting graphs have the equations:
y - k = A Sin B(x -
and y - k = A Cos B(x -h)
of Graph of y = A sin B(x - C) (Manipula Math)
1) Graph the
y - 3 = 2 sinp(x - 1)
Solution: This graph has the same
and period length without the translation. A = 2 and period
= 2p/ p= 2
Graph this one. Points on the graph are: zeros (0, 0), (1,
0), (2, 0)
maximum at (.5, 2) and minimum at (1.5, -2)
Now translate the
by moving the five points above, three units up and one unit to the
This makes the
(1, 3), (2,
(3, 3), (1.5, 5), and (2.5, 5)
Both graphs are
on the next grid. The green
one is the final graph of the function.
2) Graph the function: y + 2 = 4 cos p/4(x
Solution: This is the graph of y
= cos (p/4)x translated 2 units down
and one unit to the left. The amplitude is 4 and the period
= 8. Here is the graph of the function in
Maximum points at: (0, 4), (8, 4)
Minimum point at (4, -4)
Zeros at: (2, 0) and (6, 0)
Now translate the graph by moving each
2 units down and 1 unit left!
The points are translated to: (-1,
(7, 2), (3, -6), (1, -2) and (5, -2). These points are marked on
the green graph with black dots!
green graph is the final solution!
3) Graph y = 3 cos 4(x - p)
maximum points at: (0, 3), (p/2,
Minimum point at ( p/4,
zeros at: (p/8,
0) and (3p/8, 0)
Now shift all the points pi units to the
Maximum points at: (p,3),
minimum point at (5p/4,
zeros at: (9p/8,
0), (11p/8, 0)
This is actually the same graph as the one
we translated it. Why?
has A = 3 with period length 2p/
4 = p/ 2. It is shifted punits
to the right. Graph it first without the translation.
4) Give an equation for
This graph has many different solutions
ones for both the sine and cosine functions. In either case, the
amplitude and period length are the same. Look at the graph and
the maximum and minimum points.
They happen at 1 and -3. Remember
formula to find the amplitude of a graph? That's right, its
- minimum divided by two.
A = [1 - (-3)]/2 = 2
The period length is 6. Look at the
and count between maximum points or minimum points. Now solve for
B in the equation
Period Length = 2p/B
6 = 2p/B
6B = 2p
B = p/3
Our equation has the form y = 2 sin (p/3)x
or y = 2 cos (p/3)x before the translation.
For the sine curve, the graph starts at a
When you translate a graph, the "zero" becomes the line midway between
the max and min values. Look at the graph:
The black dot represents a "zero" of the
before the translation. Now determine where the new point
It was moved 1 unit right and 1 unit down. You now have h and
Put them in the formula and voila!!
y + 1 = 2 sin [(p/3)(x
You can do a
translation for the cosine curve, but remember, the cosine starts a
at a maximum point!!
5) Find an equation
the following graph:
Let's find a cosine function for this
A = [5-(-3)]/2= 8/2 = 4
Period length is 4pcounting
from the maximum point to the next maximum point.
4pB = 2p
B = 1/2
Before the translation, the equation is y
4 Cos [(1/2)x]
Now determine the translation.
the maximum point is the starting point for the cosine function.
Study this graph:
Look at the red
lines. The maximum point is moved punits
to the right and one unit up from the origin. The purple
graph is before the translation/
The equation becomes:
y - 1 = 4 cos
this will give you a small glimpse into graphing with translations!!