Section 1– 5 B:
!
Changing the Frequency and Period of Trigonometric Functions
Shorter Periods!
Longer Periods!
If f (x) is a trionometric function then
compared to f (x)
f (b • x) where b > 1
the graph of y = f (b • x) has a SHORTER period.
The roots are closer together and the
basic shape repeats more frequently.
If f (x) is a trionometric function then
compared to f (x)
f (b • x) where 0 < b < 1
the graph of y = f (b • x) has a LONGER period.
The roots are farther apart and the
basic shape repeats less frequently.
The period of f (b • x) is found by
setting the period of f(x) equal to b • x
and solving for x
The period of f (b • x) is found by
setting the period of f(x) equal to b • x
and solving for x
The roots of f (b • x) are found
by setting the original roots of f(x)
equal to b • x and solving for x
The roots of f (b • x) are found
by setting the original roots of f(x)
equal to b • x and solving for x
!
!
Example 1!
Example 2
The graph below in black is the graph !
of the standard 1 period of y = sin(x) !
The graph below in black is the graph
of the standard 1 period of y = sin(x)
with roots at 0,π ,2π and a period of 2π !
with roots at 0,π ,2π and a period of 2π
The graph below in red is the graph of !
The graph below in red is the graph of
⎛π⎞
1 period of y = sin
with roots at
⎝ 2⎠
1 period of y = sin(2x) with roots at!
0,
π
, π and a period of π !
2
y
1
0
π
2
π
2π
0,2π ,4π and a period of 4π
y
1
0
x
–1
π
2π
4π
x
–1
!
Set the period of y = sin(x)
equal to x / 2 and solve for x
x
= 2π → x = 4π
2
⎛ x⎞
the period for sin
is 4π
⎝ 2⎠
Set the period of y = sin(x)
equal to 2x and solve for x
2x = 2π
x= π
the period for y = sin(2x) is π !
Math 370 Section 1 – 5B !
!
Page 1 of 7!
© 2016 Eitel
How to find the period for y = sin (b • x )
1. To find the period of y = sin (b • x ) set the period of y = sin ( x ) which is 2π
equal to b• x and solve for x.
How to graph y = sin (b • x )
1. To find the roots of y = sin (b • x ) set the roots of y = sin ( x ) which are 0 , π , 2π
equal to b • x and solve for
2. Graph and label the roots for y = sin (b • x )
3. Graph the sine function using these roots.
4. The max and min are at 1 and – 1
Example 1!
Example 2
⎛ x⎞
y = sin
⎝ 3⎠
y = sin (2x ) !
The roots for 1 period of !
The roots for 1 period of
y = sin ( x ) are x = 0,π ,2π !
!
Set these Roots 0,π ,2π !
y = sin ( x ) are x = 0,π ,2π
Set these Roots 0,π ,2π ! !
x
equal to
and solve for x
3
x
= 0 , π , 2π
3
⎛ x⎞
3•
= 3• 0 , 3• π , 3• 2π
⎝ 3⎠
equal to 2x and solve for x!
2x = 0 , π , 2π
2x 0 π 2π
= ,
,
2 2 2
2
⎛ x⎞
The roots for y = sin
are
⎝ 3⎠
The roots for y = sin(2x) are !
π
x=0 ,
, π
2
the new period is π
x = 0 , 3π , 6π
the new period is 6π
π
,π !
2
Graph the sine function using these roots!
New Roots = 0 ,
New Roots = 0 , 3π , 6π
Graph the sine function using these roots
y
y
1
1
π
π
2
6π
x
3π
–1
x
–1
!
Math 370 Section 1 – 5B !
!
Page 2 of 7!
© 2016 Eitel
Example 3!
y = sin
Example 4
⎛ 2x ⎞
⎝ 3⎠!
y = sin
⎛ x⎞
⎝ 4π ⎠
The roots for 1 period of !
The roots for 1 period of
y = sin ( x ) are x = 0,π ,2π !
!
Set these Roots 0,π ,2π !
2x
equal to
and solve for x!
3
2x
= 0 , π , 2π
3
3 ⎛ 2x ⎞ 3
3
3
•
= • 0 , • π , • 2π
2 ⎝ 3⎠ 2
2
2
y = sin ( x ) are x = 0,π ,2π
Set these Roots 0,π ,2π ! !
x
equal to
and solve for x
4π
πx
= 0 , π , 2π
4
4 ⎛π x⎞ 4
4
4
•
= • 0 , • π , • 2π
π ⎝ 4 ⎠ π
π
π
⎛ 2x ⎞
The roots for y = sin
are
⎝ 3⎠
x=0 ,
⎛π x⎞
The roots for y = sin
are
⎝ 4 ⎠
3π
, 3π
2
x=0 , 4 , 8
!
!
Graph the sine function using these roots!
Graph the sine function using these roots
y
y
1
1
3π
–1
3π
2
8
x
4
x
–1
!
The period for y = sin (x) is 2π
The period for y = sin (x) is 2π
⎛ 2 x⎞
The period for y = sin
is
⎝ 3⎠
The period for y = sin
2x
= 2π
3
3 2x 3
is •
= • 2π
2 3 2
→ 1 period = 3π
πx
= 2π
4
4 πx 4
•
= • 2π
π 4 π
→ 1 period = 8
Math 370 Section 1 – 5B !
!
!
Page 3 of 7!
⎛π x⎞
is
⎝ 4 ⎠
© 2016 Eitel
Example 5!
⎛ 3x ⎞
y = cos
⎝ 2 ⎠!
Example 6
y = cos ( 4 x )
The roots for 1 period of !
−π π 3π
y = cos ( x ) are x =
, ,
!
2 2 2
The asymptotes for 1 period of
−π π 3π
y = cos ( x ) are x =
, ,
!
2 2 2
Set these Roots!
3x
equal to
and solve for x!
2
3x
−π π 3π
=
,
,
2
2
2
2
2 ⎛ 3x ⎞ 2 −π 2 π 2 3π
•
= •
, • , •
3 ⎝ 2⎠ 3 2
3 2 3 2
Set these asymptotes!
x
equal to
and solve for x
2
−π π 3π
,
,
2
2
2
1
1 −π 1 π 1 3π
• (4 x ) = •
, • , •
4
4 2
4 2 4 2
4x =
⎛ 3x ⎞
The roots for y = sin
are
⎝ 2⎠
x=
−π π
,
, π
3
3
The roots for y = cos ( 4 x ) are
−π π 3π
x=
,
,
8
8
8
!
!
Graph y = cos
1
⎛ 3x ⎞
using these roots!
⎝ 2⎠
Graph y = cos ( 4x ) using these roots
y
1
−π
8
π
−π
3
–1
x
π
3
y
π
8
3π
8
x
–1
!
The period for y = cos (x) is 2π
The period for y = cos
⎛ 3 x⎞
is
⎝ 2⎠
3x
= 2π
2
2 3x 2
•
= • 2π
3 2
3
4π
→ 1 period =
3
Math 370 Section 1 – 5B !
The period for y = cos (x) is 2π
The period for y = cos ( 4 x ) is
4 x = 2π
1
1
• 4 x = • 2π
4
4
π
→ 1 period =
2
!
!
Page 4 of 7!
© 2016 Eitel
Example 7!
Example 8
y = sec ( 3x ) !
y = sec
⎛ x⎞
⎝ 2⎠
The asymptotes for 1 period of!
The asymptotes for 1 period of
y = sec ( x ) are x =
−π π 3π
, ,
!
2 2 2
The min max occur at x = 0 , π !
−π π 3π
, ,
2 2 2
The min max occur at x = 0 , π
Set these values!
Set these asymptotes!
x
equal to
and solve for x
2
y = sec ( x ) are x =
equal to 3x and solve for x!
−π
π
3π
⎛ x⎞
= 2•
, 2• , 2•
⎝ 2⎠
2
2
2
1
1 −π 1 π 1 3π
• ( 3x ) = •
, • , •
3
3 2
3 2 3 2
−π π 3π
The asy. for y = sec ( 3x ) are
, ,
6 6 6
!
1
1
1
min - max: • 3x • 0 , • π
3
3
3
π
The min - max for y = sec ( 3x ) are 0 ,
3
asy: 2 •
Graph y = sec (3x ) using these values!
Graph y = sec
asy:
The asy. for y = sec
⎛ x⎞
are − π , π , 3π
⎝ 2⎠
x
min - max: 2 •
= 2• 0 , 2•π
2
⎛ x⎞
The min - max for y = sec
are 0 , 2π
⎝ 2⎠
!
⎛ x⎞
using these values
⎝ 2⎠
y
y
π
3
1
x
2π
1
–1
x
–1
−π
6
π
6
3π
6
−π
!
π
3π
The period for y = sec (x) is 2π
The period for y = sec
The period for y = sec (x) is 2π
The period for y = sec ( 3x ) is
3x = 2π
1
1
• 3x = • 2π
3
3
2π
→ 1 period =
3
Math 370 Section 1 – 5B !
⎛ x⎞
is
⎝ 2⎠
x
= 2π
2
x
2 • = 2 • 2π
2
→ 1 period = 4π
!
!
Page 5 of 7!
© 2016 Eitel
Example 9!
y = tan
Example 10
⎛ 2x ⎞
!
⎝ 3⎠
y = cot
The asymptotes for y = tan ( x ) are
−π π
, !
2 2
The root for 1 period of y = tan ( x ) is 0 !
Set these values equal to
2x
!
3
The asymptotes for y = cot ( x ) are 0,π
The root for 1 period of y = cot ( x ) is
Set these values equal to
and solve for x!
asy:
The asy. for y = tan
asy:
−3π 3π
⎛ 2x ⎞
are
,
⎝ 3⎠
4 4
5x
2
2 ⎛ 5x ⎞ 2
2
•
= • 0 , •π
5 ⎝ 2⎠ 5
5
The asy. for y = cot
3x
3
= •0= 0
2
2
⎛ 2x ⎞
The root for y = tan
is 0
⎝ 3⎠
!
!
⎛ 2x ⎞
Graph y = tan
using these values!
⎝ 3⎠
root:
2π
⎛ 5x ⎞
are 0 ,
⎝ 2⎠
5
2 5x
2 π π
•
= • =
5 2
5 2 5
π
⎛ 5x ⎞
The root for y = cot
is
⎝ 2⎠
5
root:
Graph y = cot
y
⎛ 5x ⎞
using these values
⎝ 2⎠
y
x
3π
4
−3π
4
The period for y = tan
x
π
5
2π
5
!
The period for y = tan ( x ) is π
→ 1 period =
x
2
and solve for x
3 ⎛ 2x ⎞ 3 −π 3 π
•
= •
, •
2 ⎝ 3⎠ 2 2
2 2
2x
=π
3
3 2x 3
•
= •π
2 3 2
⎛ 5x ⎞
⎝ 2⎠
The period for y = cot ( x ) is π
⎛ 2 x⎞
is
⎝ 3⎠
3π
2
Math 370 Section 1 – 5B !
The period for y = cot
⎛ 5 x⎞
is
⎝ 2⎠
5x
=π
2
2 5x 2
•
= •π
5 2
5
2π
→ 1 period =
5
!
!
Page 6 of 7!
© 2016 Eitel
Example 11!
Example 12
y = csc
y = csc ( 4 x ) !
The asymptotes for y = csc ( x ) are 0 , π , 2π
The asymptotes are 0 , π , 2π !
x 3x
,
!
2
2
The min max occur at x =
⎛ x⎞
⎝ 3⎠
x 3x
,
2
2
x
Set these values equal to
3
and solve for x
⎛ x⎞
asy: 3•
= 3• 0 , 3• π , 3• 2π
⎝ 3⎠
The min max occur at x =
Set these values equal to 4 x !
and solve for x!
1
1
1
1
• ( 4 x ) = • 0 , • π , • 2π
4
4
4
4
π π
The asy. for y = csc ( 4 x ) are 0 , ,
4 2
1
1 π 1 3π
min - max:
• 4x = • , •
4
4 2 4 2
π 3π
The min - max for y = csc ( 4 x ) are
,
8
8 !
!
asy:
Graph y = csc (4 x ) using these values!
The asy. for y = sec
⎛ x⎞
are 0 , 3π , 6π
⎝ 3⎠
x
π
3π
= 3• , 3•
2
2
2
3π 9π
⎛ x⎞
The min - max for y = sec
are
,
⎝ 3⎠
2
2
min - max: 3•
Graph y = csc
⎛ x⎞
using these values
⎝ 3⎠
y
y
3π
8
1
π
8
–1
0
x
9π
2
1
π
4
π
2
0
!
x
3π
2
–1
3π
6π
The period for y = csc (x) is 2π
The period for y = csc ( 4 x ) is
4 x = 2π
1
1
• 4 x = • 2π
4
4
π
→ 1 period =
2
The period for y = csc (x) is 2π
The period for y = csc
x
= 2π
3
x
3• = 3• 2π
3
→ 1 period = 6π
!
Math 370 Section 1 – 5B !
⎛ x⎞
is
⎝ 3⎠
!
Page 7 of 7!
© 2016 Eitel