Transcript Graphs of Sine and Cosine

Graphs of Sine and Cosine
Section 4.5
Sine Curve
1

2
π
3
2
2π
-1

2
3
2
Key Points:0 π
2π
Value: 0 1 0 -1 0
Cosine Curve
1

2
π
3
2
2π
-1

2
3
2
Key Points:0 π
2π
Value: 1 0 -1 0 1
Equations
 For the rest of this section, we will be graphing:
y = a Sin (bx – c) + d
y = a Cos (bx – c) + d
y = Sin x
a=1 c=0
b=1 d=0
Graph the equation y = 2 Sin x
2
1
-1

2
π
3
2
2π
-2

2
3
2
Key Points:0 π
2π
Value: 0 2 0 -2 0
Amplitude (a)
 Half the distance between the maximum and minimum
values of the function
 Given by the value of │a │
 Graph the functions:
y = 4 Sin x
y = ½ Cos x
y = -2 Sin x
y = 4 Sin x
y = ½Cos x
y = -2Sin x
4
3
2
1
-1
-2
-3
-4

2
π
3
2
2π
y = a Sin (bx – c) + d
b gives us the period of the curve
Period =
2
b
y = 4 Sin 2x
Amplitude = 4
2
Period = 2 = π
Key Points
Would having a period of π change the
key points of the curve?
1
-1

2
π
3
2
2π
Finding Key Points
In General
1)
Find the period of the
curve
2)
Divide the period into 4
equal parts
3)
From your starting
point, add this distance 4
times for each period
For Y = 4Sin 2x
1)
2)
3)
Period = π

Distance =
4
  3
0,
,
,
, 
4 2 4
y = 4Sin 2x
4
1
-1
-4

4

2
3
4
π
Graph the following curves
y = 4 Cos 8x
y = ½ Cos 2πx
y = -2 Sin 6x
y = 4Cos 8x

b = 8 → Period =
4
Amplitude = 4

→ Distance =
16
4

16
-4

8
3
16

4
y = ½Cos 2πx
b = 2π → Period = 1
Amplitude = ½
1
→ Distance =
4
½
-½
1
4
1
2
3
4
1
y = -2Sin 6x
Amplitude = 2

b = 6 → Period =
3

→ Distance =
12
2

12
-2

6

4

3
y = a Sin (bx – c) + d
 a = amplitude
 b = Find the period →
2
b
 c = Find the “phase shift”
→ horizontal shift
→
c
b
y = ½ Sin (x -

3
)
a = ½
 b = 1 → Period =
c =

3
→ P. S. =
c
b
2
1

 2

3


3
1
y = -3 Cos (2πx + 4π)
a = 3
 b = 2π → Period =
 c =  4 → P. S. =
c
b
2
2
1
 4
 2

2
y = a Sin (bx – c) + d
 a = amplitude
 b = Find the period →
2
b
 c = Find the “phase shift” →
 d = Vertical Shift
c
b
y=
 
 x
2 Cos  
3
4 
 2
a = 2
b =
c =
1
2
→ Period =

4
c
b
d = 3
→ P. S. =
2
1
2
 4

2

4



1
4
2
2
y
2 
x

2
=4 Sin 
3 
 3
a = 4
b =

3
→ Period =
2
c = 
3
 d = -2
→ P. S. =
2

c

b
6
3
 2

3
3
 2
y
2 
x

2
= 4 Sin 
3 
 3
2
-2
1

2
-2
-6
1
5
2
4