1. y = | a | • f (x) by horizontal and vertical translations
GRAPHING SINE AND COSINE FUNCTIONS
In previous chapters you learned that the graph of y = a • f (x – h) + k is related to the graph of
y = | a | • f (x) by horizontal and vertical translations
and by a reflection when a is negative.
This also applies to sine, cosine, and tangent functions.

2. GRAPHING SINE AND COSINE FUNCTIONS
TRANSFORMATIONS OF SINE AND COSINE GRAPHS
To obtain the graph of
y = a sin b (x – h) + k or y = a cos b (x – h) + k
Transform the graph of y = | a | sin bx or
y = | a | cos bx as follows.

3. GRAPHING SINE AND COSINE FUNCTIONS
TRANSFORMATIONS OF SINE AND COSINE GRAPHS
VERTICAL SHIFT
Shift the graph k units
vertically. k
y = a • sin bx + k
y = a • sin bx

4. GRAPHING SINE AND COSINE FUNCTIONS
TRANSFORMATIONS OF SINE AND COSINE GRAPHS
HORIZONTAL SHIFT
Shift the graph h units
Vertically. h
y = a • sin b(x – h)
y = a • sin bx

5. GRAPHING SINE AND COSINE FUNCTIONS
TRANSFORMATIONS OF SINE AND COSINE GRAPHS
y = a • sin bx + k
REFLECTION
If a < 0, reflect the
graph in the line
y = k after any
vertical and horizontal
shifts have been
performed.
y = – a • sin bx + k

6.  Graph y = – 2 + 3 sin 4 x. SOLUTION 2 4 2
Graphing a Vertical Translation
Graph y = – sin 4 x.
Because the graph is a transformation of the graph of y = 3 sin 4x, the amplitude is 3 and the period is = 2 4  2
SOLUTION
By comparing the given equation to the general equation
y = a sin b(x – h) + k, you can see that h = 0, k = – 2, and a > 0. 3 8  4 2
Therefore translate the graph of
y = 3 sin 4x down two units.

7. Graph y = – 2 + 3 sin 4 x. SOLUTION
Graphing a Vertical Translation
Graph y = – sin 4 x.
The graph oscillates 3 units up and down from its center
line y = – 2.
SOLUTION
Therefore, the maximum value of the function is – = 1 and the minimum value of the function is – 2 – 3 = –5 3 8  4 2 3 8  4 2
y = – 2

8.   Graph y = – 2 + 3 sin 4 x. SOLUTION 4 2 8 3 8
Graphing a Vertical Translation
Graph y = – sin 4 x.
SOLUTION
The five key points are:
On y = k : (0, 2); , – 2 ; , – 2  4 2 3 8  4 2
Maximum: , 1  8
Minimum: , – 5 3 8

9. Graphing a Vertical Translation
Graph y = – sin 4 x.
CHECK 3 8  4 2
You can check your graph with a graphing calculator. Use the Maximum, Minimum and Intersect features to check the key points.

10. Graphing a Vertical Translation
Graph y = 2 cos x –  4 2 3 3
Because the graph is a transformation of the graph of y = 2 cos x, the amplitude is 2 and the period
is = 3 . 2 2
SOLUTION

11. Graphing a Vertical Translation
Graph y = 2 cos x – π 4 2 3
By comparing the given equation to the general equation
y = a cos b (x – h) + k, you can see that h = ,
k = 0, and a > 0.  4
SOLUTION
Therefore, translate the
graph of y = 2 cos x
right unit. 2 3  4
Notice that the maximum
occurs unit to the right of the y-axis.  4

12.     Graph y = 2 cos x – . 4 2 3 SOLUTION 4 1 • 3 + , 0 = , 0 5 2
Graphing a Horizontal Translation
Graph y = 2 cos x –  4 2 3
SOLUTION
The five key points are:  4 1
On y = k : • 3 , 0 = (, 0);
• 3 , 0 = , 0 5 2 3
Maximum: , 2 = , 2 ;
13 4
3 , 2 = , 2 ; 
Minimum: • 3 , – 2 = , – 2 7 4  1 2

13. When you plot the five points on the graph, note that the
Graphing a Reflection
Graph y = – 3 sin x.
Because the graph is a reflection of the graph of y = 3 sin x, the amplitude is 3 and the period is 2.
SOLUTION
When you plot the five points on the graph, note that the
intercepts are the same as they are for the graph of
y = 3 sin x.

14. However, when the graph is reflected in the x-axis, the
Graphing a Reflection
Graph y = – 3 sin x.
SOLUTION
However, when the graph is reflected in the x-axis, the
maximum becomes a minimum and the minimum
becomes a maximum.

15.  Graph y = – 3 sin x. SOLUTION 1 • 2, 0 = (, 0) 2 1 2 4 3 3 2
Graphing a Reflection
Graph y = – 3 sin x.
SOLUTION
The five key points are:
On y = k : (0, 0); (2, 0); 1 2
• 2, 0 = (, 0)
Minimum: • 2, – 3 = , – 3 1 4  2
Maximum: • 2, 3 = , 3 3 4 3 2

16. Modeling Circular Motion
FERRIS WHEEL You are riding a Ferris wheel. Your height h
(in feet) above the ground at any time t (in seconds) can be
modeled by the following equation:
h = 25 sin t –  15
The Ferris wheel turns for 135 seconds before it stops
to let the first passengers off.
Graph your height above the ground as a
function of time.
What are your minimum and maximum heights above the ground?

17. Modeling Circular Motion
h = 25 sin t –  15
SOLUTION
The amplitude is 25 and the period is = 30. 2  15
The wheel turns = 4.5 times in 135 seconds, so the graph shows 4.5 cycles.
130 30

18. Modeling Circular Motion
h = 25 sin t –  15
SOLUTION
The key five points are (7.5, 30), (15, 55), (22.5, 30),
(30, 5) and (37.5, 30).

19. Modeling Circular Motion
h = 25 sin t –  15
SOLUTION
Since the amplitude is 25 and the graph is shifted up 30 units, the maximum height is = 55 feet.
The minimum height is 30 – 25 = 5 feet.

20. • Shift the graph k units vertically and h units horizontally.
GRAPHING TANGENT FUNCTIONS
TRANSFORMATIONS OF TANGENT GRAPHS
To obtain the graph of y = a tan b (x – h) + k transform
the graph of y = a tan bx as follows. |
• Shift the graph k units vertically and h units horizontally.
• Then, if a < 0, reflect the graph in the line y = k.

21.  Graph y = – 2 tan x + . 4 SOLUTION
Combining a Translation and a Reflection
Graph y = – 2 tan x  4
SOLUTION
The graph is a transformation of the graph of y = 2 tan x, so the period is .

22.    Graph y = – 2 tan x + . 4 SOLUTION
Combining a Translation and a Reflection
Graph y = – 2 tan x  4
SOLUTION
By comparing the given equation to
y = a tan b (x – h) + k, you can see
that h = – , k = 0, and a < 0.  4
Therefore translate the graph of
y = 2 tan x left unit and then reflect it in the x-axis.  4

23.     Graph y = – 2 tan x + . 4 2 •1 4 3 4 4 •1 4 2
Combining a Translation and a Reflection
Graph y = – 2 tan x  4
Asymptotes:
x = – – = – ; x = – = 
2 •1 4 3
On y = k:
(h, k) = – , 0  4
Halfway points:
– – , 2 = – , 2 ; – , – 2 = (0, – 2) 
4 •1 4 2