11. amplitude, phase shift and period of trig formulas-x

Amplitude, Period and Phase Shift

In this section, we construct graphs of simple waves
by stretching, compressing and shifting vertically and
horizontally the graphs of sine and cosine.

We introduce the following terms for graphs of waves.

Amplitude
The amplitude is the distance
from the waistline of a wave to
the top or bottom of the wave.

Amplitude
Amplitude
Amplitude
Waistline

Amplitude
The period is the length
the wave takes to
complete one cycle of
undulation.
Amplitude
Amplitude
Period
Period
Period Waistline

Amplitude (Vertical Extension and Compression)

Given y = sin(x), and P = (x, sin(x)) a generic point
on the graph as shown,
P = (x, sin(x))
y= sin(x)
sin(x)
x

on the graph as shown, the graph of y = 3sin(x) at x
would yield the point (x, 3sin(x))
by tripling the height of the point P.
P = (x, sin(x))
(x, 3sin(x))
y= sin(x)
sin(x)
x

P = (x, sin(x))
Hence plotting y = 3sin(x) means
to elongate the entire graph
of y = sin(x) by 3 fold.
(x, 3sin(x))
y= sin(x)
y = 3sin(x))
sin(x)
x

P = (x, sin(x))
of y = sin(x) by 3 fold while
the x–intercepts or (x, 0)’s
remain fixed because 3(0) = 0.
(x, 3sin(x))
y= sin(x)
y = 3sin(x))
sin(x)
x

P = (x, sin(x))
of y = sin(x) by 3 fold while
the x–intercepts or (x, 0)’s
remain fixed because 3(0) = 0.
Likewise setting y = (1/3)sin(x) would
compress vertically the entire graph
to a third of its original height while
the x–intercepts or (x, 0)’s remain fixed.
(x, 3sin(x))
y= sin(x)
y= sin(x)/3
y = 3sin(x))
sin(x)
x

Amplitude
Assuming A > 0, the graph of y = A sin(x)
is the vertical-stretch/compression of y = sin(x)
having amplitude A.

Amplitude
having amplitude A.
x
y= sin(x)
y = 3sin(x))
y= sin(x)/3

Amplitude
having amplitude A.
If A < 0, the graph of y = A sin(x)
is the vertical stretch/compression
of the reflection of sin(x) which is
y = –sin(x), having amplitude –A.
(x, – sin(x))
(x, –3sin(x))
x
y= –sin(x)
y = –3sin(x))
y= –sin(x)/3
y= sin(x)
y = 3sin(x))
y= sin(x)/3

Amplitude
having amplitude A.
(x, – sin(x))
(x, –3sin(x))
x
y= –sin(x)
y = –3sin(x))
y= –sin(x)/3
In summary, the graph of
y = A sin(x)
has amplitude lAl.
y= sin(x)
y = 3sin(x))
y= sin(x)/3

Amplitude
having amplitude A.
(x, – sin(x))
(x, –3sin(x))
x
y= –sin(x)
y = –3sin(x))
y= –sin(x)/3
In summary, the graph of
y = A sin(x)
has amplitude lAl.
Horizontal stretch/compression is
the same as adjusting the period.
y= sin(x)
y = 3sin(x))
y= sin(x)/3

Period

Period
y = sin(x) has period 2π
because it completes one
cycle of y values as x
varies from 0 to 2π
y
y = sin(x)
p = 2π
0 2π

Period
y
y = sin(x)
p = 2π
0 2π
To plot the y = sin(2x)
at a given x,
x

Period
y
y = sin(x)
p = 2π
0 2π
at a given x, we go to 2x,
look up its sine–value
sin(2x)
x 2x
sin(2x)

Period
y
y = sin(x)
p = 2π
0 2π
sin(2x) and use that to plot
(x, sin(2x)).
x 2x
sin(2x)

Period
y
y = sin(x)
p = 2π
0 2π
(x, sin(2x)).
x 2x
sin(2x)
u

Period
y
y = sin(x)
p = 2π
0 2π
(x, sin(2x)).
x 2x
sin(2x)
2uu

Period
y
y = sin(x)
p = 2π
0 2π
(x, sin(2x)).
x 2x
sin(2x)
2uu
sin(2u)

Period
y
y = sin(x)
p = 2π
0 2π
(x, sin(2x)).
As x varies from 0 to π so that 2x varies from 0 to 2π,
the total effect is to compress horizontally the graph of
y = sin(x) to half of its 2π period.
x 2x
sin(2x)
2uu
sin(2u)

Period
y
y = sin(x)
p = 2π
0 2π
(x, sin(2x)).
y = sin(x) to half of its 2π period.
y
y = sin(2x)0 π
x
x
2x
sin(2x)
sin(2x) u
2u
sin(2u)
u
sin(2u)

Period
y
y = sin(x)
p = 2π
0 2π
(x, sin(2x)).
y = sin(x) to half of its 2π period. Hence the graph for
y = sin(2x) have a new period of π.
y
y = sin(2x)0
p = π
π
x
x
2x
sin(2x)
sin(2x) u
2u
sin(2u)
u
sin(2u)

Period
In general, with B > 0,
by setting Bx = 2π,
we’ve x = (1/B)2π so
y = sin(Bx) has period
p = (1/B)2π or x = 2π/B.
y
y = sin(x)
p = 2π
0 2π
y
y = sin(2x)
0 π
x
x
2x
sin(2x)
sin(2x)
2uu
sin(2u)

Period
p = (1/B)2π or x = 2π/B.
y
y = sin(x)
p = 2π
0 2π
y
y = sin(2x)
0
p = 2π/2 = π
π
x
x
2x
sin(2x)
sin(2x)
2uu
sin(2u)

Period
p = (1/B)2π or x = 2π/B.
y
y = sin(x)
p = 2π
0 2π
y
y = sin(2x)
0 π
x
x
2x
sin(2x)
sin(2x)
2uu
sin(2u)
(1/B) is the horizontal
stretch/compression factor.
p = 2π/2 = π

Period
p = (1/B)2π or x = 2π/B.
y
y = sin(x)
p = 2π
0 2π
y
y = sin(2x)
0 π
x
x
2x
sin(2x)
sin(2x)
2uu
sin(2u)
If B > 1, then (1/B)2π < 2π,
so y = sin(Bx) has a period shorter then 2π
and its graph is a compression the graph of y = sin(x).
p = 2π/2 = π

Period
p = (1/B)2π or x = 2π/B.
y
y = sin(x)
p = 2π
0 2π
y
y = sin(2x)
0 π
x
x
2x
sin(2x)
sin(2x)
2uu
sin(2u)
If B > 1, then (1/B)2π < 2π,
so y = sin(Bx) has a period shorter then 2π
and its graph is a compression the graph of y = sin(x).
If B < 1 then (1/B)2π > 2π,
so y = sin(Bx) has a period more than 2π and its graph
is a stretch or elongation of the graph of y = sin(x).
p = 2π/2 = π

Given the point (x, y), (x, y + 1)
is 1 units above it so the point
(x, sin(x) + 1) is 1 unit up from
(x, sin(x)).

y= sin(x)
(x, sin(x)).
(x, sin(x)+1)
x(x, sin(x))

y= sin(x)
(x, sin(x)). Hence the graph of
y = sin(x) + 1 is 1 unit up
from the graph of y = sin(x).
(x, sin(x)+1)
y= sin(x) + 1
x(x, sin(x))
2

y= sin(x)
(x, sin(x)+1)
y= sin(x) + 1
x(x, sin(x))
Similarly, the point (x, y – 1)
is 1 units below (x, y)
2

y= sin(x)
(x, sin(x)+1)
(x, sin(x)–1)
y= sin(x) + 1
x(x, sin(x))
is 1 units below (x, y) so (x, sin(x) – 1) is 1 unit down
from (x, sin(x)).
2

y= sin(x)
(x, sin(x)+1)
(x, sin(x)–1)
y= sin(x) – 1
y= sin(x) + 1
x(x, sin(x))
from (x, sin(x)). Correspondingly, the graph of
y = sin(x) – 1 is the graph of y = sin(x) 1 unit lowered.
2
–2

y= sin(x)
Vertical Translations (In reference to graphs)
(x, sin(x)+1)
(x, sin(x)–1)
y= sin(x) – 1
y= sin(x) + 1
x(x, sin(x))
y = sin(x) + C is the vertical translation of y = sin(x).
If C > 0, y = sin(x) + C is C unit up from y = sin(x).
2
–2

y= sin(x)
Vertical Translations (In reference to graphs)
(x, sin(x)+1)
(x, sin(x)–1)
y= sin(x) – 1
y= sin(x) + 1
x(x, sin(x))
y = sin(x) + C is the vertical translation of y = sin(x).
If C > 0, y = sin(x) + C is C unit up from y = sin(x).
If C < 0, y = sin(x) + C is C unit down from y = sin(x).
2
–2

Because the graph of y = sin(x) + C is the
vertical translation of y = sin(x) by C unit,
whose waistline y = 0 is moved to y = C
which is the waistline of y = sin(x) + C.
y
x
Move by C
waistline
y = 0

y
x
Move by C
waistline
y = 0
waistline
y = C

Amplitude lAl
waistline
y = C
In particular since y = Asin(x) + C has amplitude lAl,
y = Asin(x) + C
y
x
y = Asin(x)
Move by C
waistline
y = 0

Amplitude lAl
waistline
y = C
In particular since y = Asin(x) + C has amplitude lAl,
so y ranges from y = C – lAl to y = C + lAl.
y = C+lAl
y = C–lAl
y = Asin(x) + C
y
x
y = Asin(x)
Move by C
waistline
y = 0

Example B.
Sketch graph of y = –3sin(x/2) + 5.
List the period and the amplitude. Draw the waistline
and the range of y.

The amplitude is 3 and the period is [1/(1/2)] 2π = 4π.
Example B.
and the range of y.

The waistline is y = 5, hence y ranges from 2 to 8.
Example B.
and the range of y.

The waistline is y = 5, hence y ranges from 2 to 8. To
sketch two periods of its graph, we start by drawing
two periods of the negative sine–wave
then label the above information as shown.
Example B.
and the range of y.

y
Example B.
and the range of y.

y = –3sin(x/2) + 5
y
y = 5
Example B.
and the range of y.

y = –3sin(x/2) + 5
y
y = 5 Amplitude = 3
Example B.
and the range of y.

2π 4π 6π 8π
y = –3sin(x/2) + 5
y
y = 5 Amplitude = 3
Example B.
and the range of y.

Example B.
and the range of y.
y = 5
y = 8
y = 2
2π 4π 6π 8π
y = –3sin(x/2) + 5
y
Amplitude = 3

The graphs of y = sin(x + C) are the horizontal shifts
of y = sin(x).

of y = sin(x). To plot y = sin(x + 1) at a generic x,
xx

we go to (x + 1),
x x+1x

we go to (x + 1), find its sine-value sin(x +1),
and bring it back to x to plot (x, sin(x + 1)).
x x+1
sin(x+1)
y= sin(x)
x

x x+1
sin(x+1)
(x, sin(x+1))
y= sin(x)
x

x x+1
sin(x+1)
(x, sin(x+1))
y= sin(x)
x
u

x x+1
sin(x+1)
(x, sin(x+1))
y= sin(x)
x
u u+1

x x+1
sin(x+1)
(x, sin(x+1))
y= sin(x)
x
u u+1
v

x x+1
sin(x+1)
(x, sin(x+1))
y= sin(x)
x
u u+1
v v+1

x x+1
sin(x+1)
v v+1
(x, sin(x+1))
y= sin(x)y= sin(x + 1)u u+1
(u, sin(u+1))
x

x x+1
sin(x+1)
v v+1
(x, sin(x+1))
So the graph of y = sin(x + 1) is the horizontal left
shift by 1 of y = sin(x).
(u, sin(u+1))
x

x x+1
sin(x+1)
v v+1
(x, sin(x+1))
shift by 1 of y = sin(x). In general, the graph of
(u, sin(u+1))
y = sin(x + C) is the horizontal translation of y = sin(x).
x

x x+1
sin(x+1)
v v+1
(x, sin(x+1))
(u, sin(u+1))
If C > 0, it is the C units left shift of y = sin(x).
x

x x+1
sin(x+1)
v v+1
(x, sin(x+1))
(u, sin(u+1))
If C > 0, it is the C units left shift of y = sin(x).
If C < 0, it is the C units right shift of y = sin(x).
x

Here is the summary of all the observations above.
The graph of the formula y = Asin(B(x + C)) + D
is the transformation of the graph of y = sin(x) with
A = its amplitude
(1/B)2π = 2π/B = its period
C = the amount of left/right shift from sin(x)
D = vertical shift placing the waistline at y = D.

A = its amplitude
In physic, the left/right shift is called the phase–shift.

Example C. a. List the amplitude, period, waistline, and
the phase shift of the graph of y = 3sin(2x – π) – 4.
A = its amplitude

A = its amplitude
a. First, arrange the formula in the correct format,
y = 3sin(2x – π) – 4 = 3sin(2(x – π/2)) – 4.

A = its amplitude
a. First, arrange the formula in the correct format,
y = 3sin(2x – π) – 4 = 3sin(2(x – π/2)) – 4.
So the amplitude = 3, the period is 2π/2 = π, the
waistline is y = –4 and the phase shift is π/2 to the right.

b. Sketch two periods of the graph of
y = 3sin(2x – π) – 4 = 3sin(2(x – π/2)) – 4

y = 3sin(2x – π) – 4 = 3sin(2(x – π/2)) – 4
Again, we draw with two periods of the sine wave
then label all the information from part a.
amplitude = 3,
the period is 2π/2 = π,
the waistline is y = –4
phase shift = π/2 to the right.

y = 3sin(2x – π) – 4 = 3sin(2(x – π/2)) – 4
amplitude = 3,
y = 3sin(2x – π) – 4

y = 3sin(2x – π) – 4 = 3sin(2(x – π/2)) – 4
amplitude = 3,
y = –4
y = 3sin(2x – π) – 4
x

y = 3sin(2x – π) – 4 = 3sin(2(x – π/2)) – 4
amplitude = 3,
y = –4
y = –1
y = –7
y = 3sin(2x – π) – 4
x

y = 3sin(2x – π) – 4 = 3sin(2(x – π/2)) – 4
amplitude = 3,
y = –4
y = –1
y = –7
π 3π/2 2π 5π/2
y = 3sin(2x – π) – 4
π/2
y
x

c. Label the points on the waistline, the high points,
and the low points on the graph.
y = –4
y = –1
y = –7
π 3π/2 2π 5π/2
y = 3sin(2x – π) – 4
π/2
y
x
(π/2, –4) (π,–4) (3π/2,–4) (2π,–4)
(3π/4,–1) (7π/4,–1)
Amplitude = 3,
Period is 2π/2 = π,
Waistline is y = –4
Phase Shift = π/2 to the right.
(5π/4,–7) (9π/4,–7)

List the amplitude, the period and draw two periods of the
following formulas. Label the high points and the low points
on the graph.
1. y = 3sin(2A)
2. y = 2sin(3A) 3. y = –3sin(2A) + 1
4. y = –2sin(3A) – 2 5. y = 5sin(A/2) – 3
6. y = –½ sin(2A) – 3/2 7. y = (–1/3)sin(2A) – 2/3
8. y = –2sin(πA) – 2 9. y = 5sin(2πA) – 3
10. y = –½ sin(2πA) – 3/2 11. y = (–1/3)sin(πA/2) – 2/3
Write the following formulas in the form of
y = Asin(B(x + C)) + D. List the amplitude, the period and draw
two periods of the following formulas. Label the points on the
waistline, the high points, and the low points on the graph.
12. y = 3sin(2A + π) – 1 13. y = 2sin(2A – π) + 3
14. y = –2sin(A/2 – π/2) – 2 15. y = 5sin(Aπ/2 + π/2) – 4
16. y = –½ sin(2πA – π) – 3/2 17. y = (–1/3)sin(Aπ/2 – π) – 2/3

Use the fact that cos(–A) = cos(A), sin(–A) = –sin(A),
tan(–A) = –tan(A) in terms of sin(A), cos(A) and tan(A).
22. sin(π – A) =
23. cos(π – A) =
26. tan(π – A) =
24. cos(π/2 – A) =
25. sin(π/2 – A) =
27. cot(π/2 – A) =
Interpret the following identities from section 9 in terms
of shifts of their graphs.
19. sin(A ± π) = –sin(A), cos(A ± π) = –cos(A)
20. sin(A + π/2) = cos(A), sin(A – π/2) = –cos(A)
21. cos(A – π/2) = sin(A), cos(A + π/2) = –sin(A)

y = 3
π
y
1. y = 3sin(2A)
π/2
3π/2 2π
y = –3
(3π/4,–3)
(π/4,3) ( 5π/4,3)
(7π/4,–3)
y = 4y
3. y = –3sin(2A) + 1
y = –2
(π/4,–2)
( 0,1)
(3π/4,4)
(5π/4,–2)
(7π/4,4)
amp = 3 p = π amp = 3 p = π
(2π,1)
5. y = 5sin(A/2) – 3
y = 2y
y = –8
(3π,–8)
(π,2) ( 5π,2)
(7π,–8)
amp = 5 p = 4π
( 0,–3)
(8π,–3)
7. y = (–1/3)sin(2A) – 2/3
y = –1/3y
y = –1
(π/4,–1)
( 0,–2/3)
(3π/4,–1/3)
(5π/4,–1)
(7π/4,–1/3)
amp = 1/3
p = π
(2π,–2/3)
9. y = 5sin(2πA) – 3
y = 2y
y = –8
(3/4,–8)
(1/4,2) ( 5/4,2)
(7/4,–8)
( 0,–3)
(2,–3)
amp = 5 p = 1
y = –1/3y
y = –1
(1,–1)
( 0,–2/3)
(3,–1/3)
(5,–1)
(7,–1/3)
amp = 1/3
p = 4
(8,–2/3)
11. y = (–1/3)sin(πA/2) – 2/3
x
x
x
x
x
x

13. y = 2sin(2(A – π/2)) + 3
15. y = 5sin(π/2(A+1)) – 4
17. y = (–1/3)sin(π/2(A–2)) – 2/3
y = 5
y = 1
(5π/4,1)
(3π/4,5) (7π/4,3)
(9π/4,1)
amp = 2
p = π
R–shift=π/2
(π/2,3)
(5π/2,3)
y = 1
y = –9
(2,–9)
(0,1) (4,1)
(6,–9)
amp = 5
p = 4
L–shift=1
(–1,–4) (7,–4)
amp = 1/3
p = 4
R–shift= 2
y = –1/3y
y = –1
(3,–1)
(2 ,–2/3)
(5,–1/3)
(7,–1)
(9,–1/3)
(10,–2/3)
x
x
x
19. If y = sin(x) is shifted
right or left by π, we get
y = –sin(x). The same
applies to cosine curve.
21. If y = cos(x) is shifted
right by π/2, we get
y = sin(x).
If y = cos(x) is shifted left
by π/2, we get y = –sin(x).
23. –cos(A)
25. cos(A)
27. tan(A)

11. amplitude, phase shift and period of trig formulas-x

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