Graph of Tangent Function
x
y = tan x
y=sin x / cos x
-π/2
=-1/0=UNDEFINED
-π/3
-1.732
-π/4
-1.000
-π/6
-0.577
0
0.000
π/6
0.577
π/4
1.000
π/3
1.732
π/2
(−
(−
)
3 2 (− 1 2) = − 3
2 2
)(
( 3 2) = -1
(−1 2)
)
2 2 = −1
3 =− 3 3
= 0/1 = 0
) 3= 3
( 2 2 ) ( 2 2 )= 1
( 3 2 ) (1 2 ) = 3
(1 2 )
(
3 2 =1
=1/0= UNDEFINED
2π/3
-1.732
3π/4
-1.000
5π/6
-0.577
π
0.000
7π/6
0.577
5π/4
1.000
4π/3
1.732
3π/2
=0/-1 = 0
= -1/0 = UNDEFINED
5π/3
-1.732
7π/4
-1.000
11π/6
-0.577
2π
0.000
13π/6
0.577
9π/4
1.000
7π/3
1.732
= 0/1 = 0
3
Recall that the domain of the tangent
function is all real numbers except for
odd multiples of π/2 (i.e. π/2, 3π/2, 5π/2,
etc.), because at those values the tangent
function is undefined. The vertical lines
through these x-values are the vertical
asymptotes of the graph. We should also
note that as x approaches values close to
an odd multiple of π/2, the absolute value
of y = tan x gets very large. Notice that
π/2 ≈ 1.57. If we let x= 1.56, y= tan 1.56
≈ 92.62. We also know that the period
(or cycle) of the tangent function is π,
which means the values of the tangent
function repeat themselves after the
length of π on the x-axis.
We can see from the graph that the
tangent function is symmetric with
respect to the origin, so the tangent
function is odd.
This means tan(-x) = - tan x.
Vertical Asymptotes at - π/2, π/2,3π/2, 5 π/2, etc.
2
5π/2
1
Where are the xintercepts?
(9π/4, 1)
0
π/2
-π/2
Where are the yintercepts?
(5π/4, 1)
(π/4, 1)
(-π/4, -1)
-1
-2
(3π/4, -1)
π
3π/2
2π
(7π/4, -1)
5π/2
Transformations of Tan x
Example 1 Graphing Variations of y=tan x using tranformations.
Graph y = 2tan x
2 is our amplitude, A, which means it is the factor that will vertically stretch
or shrink our graph. Since A=2, our graph will vertically stretch by a factor
of 2.
For instance, when x =π/4, tan x = sin π/4/cos π/4 = 1, so 2tan π/4 = 2(1) = 2
3
2
(5π/4, 2)
(π/4, 2)
(9π/4, 2)
1
0
π/2
-π/2
π
2π
3π/2
5π/2
-1
(-π/4, -2)
(3π/4, -2)
-2
(7π/4, -2)
-3
π
Example 2 Graph y = − tan⎛⎜ x + ⎞⎟
⎝
4⎠
This graph will have 2 changes. First, since it is negative, it be reflected on the x-axis
(turned upside-down). Also, as in our y = Asin(wx+h) + v, h is the horizontal
displacement. Since h >0, the graph is shifted to the left.
y = -tan x
y = -tan (x+ pi/4)
2
2
1
1
0
π/2
-π/2
0
π
3π/2
2π
5π/2
-π/2
-1
-1
-2
-2
π/2
π
3π/2
2π
5π/2
Vertical Asymptotes now at π/4, 5π/4 ,9π/4, etc.
Graph of Cot Function
Recall that the domain of the cotangent function is all real numbers
except for all integral multiples of π (i.e. 0, π, 2π, 3π, etc.), because
at those values the tangent function is undefined (at those values
cos x = ±1 and sin x = 0, and cot x = cos x/ sin x).
Vertical Asymptotes
2
-π/2
(-π/4, -1)
1
(π/4, 1)
0
π/2
-1
-2
(9π/4, 1)
(5π/4, 1)
π
(3π/4, -1)
3π/2
2π
(7π/4, -1)
5π/2
2.8 Phase Shifts and Sinusoidal Curve Fitting
Previously, when talking about transformations, I used the following equation for a
sinusoidal graph:
y = Asin(ωx + h) + v
This is not the standard form. The standard form is:
y = Asin(ωx - φ) + B
Notice in this form we use – φ, instead of + h. So this means
if φ > 0, the graph is shifted to the RIGHT, and
if φ < 0, the graph is shifted to the LEFT.
In the previous form, I said that h was the horizontal displacement. However, the actual
amount of the horizontal displacement, or phase shift, is φ/ω. How do we get this?
Recall that the period of the sine function is 2π.
sin(ωx - φ) = sin((ωx – φ) + 2π )
The period will begin with ωx – φ = 0 Æ x = φ/ω = starting point of one cycle. We see
from this that this shows that the starting point is shifted from 0 to φ/ω.
And end with ωx – φ = 2π Æ x = 2π/ω + φ/ω = ending point of one cycle.
Also remember that the new period is T = 2π/ω
Example 1
Find amplitude, period, phase shift of y = 3sin(2x –π)
A = 3, ω = 2, φ = π
Amplitude = A = 3
This that the y values range from -3 to 3.
Period T = 2π/ω = 2π/2 = π
This means that the whole sine cycle is squeezed into a period of π.
Phase shift = φ/ω = π/2.
x = φ/ω = starting point of one cycle = π/2.
x = 2π/ω + φ/ω = ending point of one cycle = 2π/2 + π /2 = 3π/2
Since φ > 0, the graph is shifted to the right, so start the cycle at x = φ/ω = π/2.
4.000
3.000
2.000
1.000
0.000
-1.000
-2.000
-3.000
-4.000
π/4
π/2
3π/4
π
5π/4
5π/4 3π/2
2π
Example 3 Finding a Sinusoidal Functions from Temperature Data
Average
Monhly
Temp
Month
1
29.7
2
33.4
3
39
4
48.2
5
57.2
6
66.9
7
73.5
8
71.4
9
62.3
10
51.4
11
39
12
31
Average Montly Temperature for Jan-Dec.
80
70
60
50
40
Series1
30
20
10
0
0
1
2
3
4
5
6
7
8
9
10
11
12
We will fit this to the STANDARD sinusoidal function:
y = Asin(ωx - φ) + B
Step 1: To find the amplitude A, we compute
A = largest data value - smallest data value
=
2
73.5 − 29.7
= 21.9
2
Step 2: Determine the vertical shift, B, by finding the average of the highest
and lowest value.
B = 73.5 + 29.7
2
= 51.6
Step 3: Find frequency, ω. Cycle will (hopefully) repeat itself every year,
so period = 12 months.
T = 12. Use the fact that T = 2π/ ω to solve for ω.
ω = 2π/T = 2π/12 = π/6
Step 4: Using A=21.9, ω = π/6, and B = 51.6,determine horizontal shift by
choosing an arbitrary data point (x,y) from the given table and solving the
equation for φ.
⎛π
⎞
y = 21.9 sin ⎜ x − ϕ ⎟ + 51.6
⎝6
⎠
Let’s choose the first data point, x = 1 (Jan.), y = 29.7
⎛π
⎞
29 .7 = 21 .9 sin ⎜ (1) − ϕ ⎟ + 51 .6
⎝6
⎠
⎛π
⎞
− 21 .9 = 21 .9 sin ⎜ − ϕ ⎟
⎝6
⎠
⎛π
⎞
− 1 = sin ⎜ − ϕ ⎟
⎝6
⎠
When does sin = - 1? For consistenc y, we will choose an angle between
π
-π
and .
2
2
⎛ -π ⎞
sin ⎜
⎟ = −1
⎝ 2 ⎠
π
6
π
6
−ϕ = −
+
π
2
π
2
=ϕ =
Solve for ϕ by getting all other term s on the other side.
π
6
+
3π
4π
2π
=
=
6
6
3
So.. A sine function that fits the data is
2π
⎛π
y = 21.9 sin ⎜ x −
3
⎝6
⎞
⎟ + 51.6
⎠
Avg Monthly Temp
y=21.9*sin((pi/6)*x - 2pi/3) + 51.6
80
70
60
50
40
30
20
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Homework
[(From last lecture) Ch. 2.6 p. 166-169
#11* , 23*,25,27, 33*,35, 39*,43, 53, 63*, 67*, 69, 81 ]
Ch. 2.7 p.176 #11,13,19
Ch. 2.8 p.189 #5,11,13,19,21