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Graph of Tangent Function
5 π/
7 π/3 1.
9 π/4 1.
13 π/6 0.
2 π 0.000 = 0/1 = 0
11 π/6 -0.
7 π/4 -1.
5 π/3 -1.
3 π/2 = -1/0 = UNDEFINED
4 π/3 1.
5 π/4 1.
7 π/6 0.
π 0.000 =0/-1 = 0
5 π/6 -0.
3 π/4 -1.
2 π/3 -1.
π/2 =1/0= UNDEFINED
π/3 1.
π/4 1.
π/6 0.
0 0.000 = 0/1 = 0
- π/6 -0.
- π/4 -1.
- π/3 -1.
- π/2 =-1/0=UNDEFINED
x y = tan x y=sin x / cos x
(− 12 ) ( 32 ) =- 1 3 =− 33
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.
≈ 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.
( − 3 2 ) (− 12 ) =− 3
( 1 2 ) ( 3 2 ) = 1 3 = 33
(− 2 2 ) ( 2 2 ) =− 1
( 2 2 ) ( 2 2 ) = 1 ( 3 2 ) ( 1 2 ) = 3
0
1
2
(π/4, 1)
(5π/4, 1) (^) (9π/4, 1)
- π/2^ π/2^ π^3 π/2^2 π^5 π/
(-π/4, -1) (3π/4, -1)^ (7π/4, -1)
Vertical Asymptotes at - π/2, π/2,3π/2, 5 π/2, etc.
Where are the x- intercepts?
Where are the y- intercepts?
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
⎟ ⎠
⎜ ⎞ ⎝
= − ⎛^ + 4
y tan x^ π
**-
-**
0
1
2
3
(π/4, 2) (5π/4, 2)^ (9π/4, 2)
- π/2^ π/2^ π^3 π/2 2 π (^5) π/
(-π/4, -2) (3π/4, -2)^ (7π/4, -2)
Example 2 Graph
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.
- -
0
1
2
y = -tan x
- -
0
1
2
y = -tan (x+ pi/4)
- π/2 π/2^ π (^3) π/2 2 π (^5) π/2 -π/2 π/2^ π (^3) π/2 2 π (^5) π/
Vertical Asymptotes now at π/4, 5π/4 ,9π/4, etc.
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) + 2 π ) 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 = 2 π Æ x = 2 π/ ω + φ/ω = ending point of one cycle. Also remember that the new period is T = 2 π/ 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 = 2 π/ 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 = 2 π/ ω + φ/ω = ending point of one cycle = 2 π/ 2 + π /2 = 3 π/ 2 Since φ > 0, the graph is shifted to the right, so start the cycle at x = φ/ω = π/2.
-4.
-3.
-2.
-1.
π/4 π/2^3 π/4^ π^5 π/4^5 π/4^3 π/2^2 π
Example 3 Finding a Sinusoidal Functions from Temperature Data
12 31
11 39
10 51.
9 62.
8 71.
7 73.
6 66.
5 57.
4 48.
3 39
2 33.
1 29.
Average Monhly Month Temp
Average Montly Temperature for Jan-Dec.
0
10
20
30
40
50
60
70
80
0 1 2 3 4 5 6 7 8 9 10 11 12
Series
We will fit this to the STANDARD sinusoidal function: y = Asin(ωx - φ) + B
Step 1: To find the amplitude A, we compute A =
Step 2: Determine the vertical shift, B, by finding the average of the highest and lowest value. B =
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/
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.
largestdatavalue-smallestdatavalue
21. 9 sin ⎟+
y = ⎛^ π^ x −ϕ
Homework
[(From last lecture) Ch. 2.6 p. 166-
#11* , 23,25,27, 33,35, 39,43, 53, 63, 67*, 69, 81 ]
Ch. 2.7 p.176 #11,13,
Ch. 2.8 p.189 #5,11,13,19,
