The method of taking five points is: let x = ω x+φ, take 0, π, π from X, find the corresponding value of X and the corresponding value of Y, and then plot the points.
3. image transformation: function y = asin (ω x+φ) (A > 0, ω > 0) can be obtained by transforming the image with the function y = sinx as follows:
(1) phase transition: y = asinx→ y = asin (x+φ), and all points on the image with y = asinx are moved to the left (φ >: 0) or to the right (φ < 0) by |φ| units in parallel;
(2) periodic transformation: y = sin (x+φ) → y = sin (ω x+φ),
Extend y = sin (x+φ) the abscissa (0; 1) to the original number of times (the ordinate is unchanged).
(3) amplitude transformation: y = sin (Ω x+φ) → y = asin (Ω x+φ), which lengthens (a >: 1) or shortens (00, Ω >) the ordinate of each point on the image of y = sin (Ω x+φ); 0, x (-∞, +∞)) represents a vibration quantity, so a is called amplitude, t = is called period, f = is called frequency, ω x+φ is called phase, and φ is called initial phase.
The period of the function y = acos (ω x+φ) is.
The period of the function y = atan (ω x+φ) is.
5. The symmetry axis of sine curve y = sinx is x =+kπ (kz) and the center of symmetry is (kπ, 0) (kz).
The symmetry axis of cosine curve y = cosx is x = kπ (kz) and the center of symmetry is (kz).
The symmetry center of the image with function y = tanx is (kz).
Test center sparring 1. The definition domain of the function y = sinx is [a, b], and the value range is [- 1,], so the sum of the maximum and minimum values of b-a is ().
A.B2π
C.D4π
Analysis: This topic aims to examine the images and properties of trigonometric functions, as well as the ability to collect and process image information.
Find the maximum and minimum of B-A, that is, find the maximum and minimum length of the independent variable value interval corresponding to the value range of function y = sinx [- 1,]. Draw the image of function y = sinx in the range of [0,3 π], and the observed image can be solved intuitively.
Analysis: As shown in the figure below, it is easy to get (b-a) max =-=,
(b-a)min=-=。
(b-a)max+(b-a)min=2π。
Comments: The key to solve this problem is to observe the intervals of independent variables with the corresponding function value range of [- 1,] on the image, and then compare the lengths of these intervals, and take the maximum length interval and the minimum length interval as the solution. Such problems lie in observation, calculation and comparison.
2. Shift the image of function y = f(x) sinx to the right by one unit, and then get the image of y = 1-2 sin2x through the symmetrical transformation about the x axis, then f(x) can be ().
A.cosx B.2cosx
C.sinx D.2sinx
3. Use the function y = sinx (xr) to move all points on the image to the left by one unit length, and then shorten the abscissa of all points on the obtained image to the original multiple (the ordinate is unchanged), and the function represented by the obtained image is ().
A.y=sin，xR
B.y=sin，xR
C.y=sin，xR
D.y=sin，xR
Analysis: Move all points on the image with Y = sinx to the left by one unit length to get Y = sin. Then the abscissa of all points on the image is shortened to the original abscissa to get Y = sin.
4. The image of function y = f(x) is translated by vector a =, and the image of function y = sin+2 is obtained, so the analytical expression of function f(x) is ().
A.f(x)=sinx B.f(x)=cosx
C.f(x)=sinx+2 D.f(x)=cosx+4
Analysis: When F (x) = SIN, after translation by vector A =, there is exactly Y = SIN+2. So I chose B.
The five-point method in formula 1 is y = asin (ω x+φ) (A > 0, ω > 0) Preparation for sketch problem solving: Determining "five points" is the key to five-point drawing, so that z = ω x+φ and z take 0, π, π, π respectively, thus determining the coordinates of five points.
Typical example 1 Let the function f (x) = sin (2x+φ) (-π < φ < 0), and Y = f (x) One symmetry axis of the image is a straight line x =.
[Resolution] (1) x = is the image symmetry axis of the function y = f (x).
sin(2×+φ)= 1，+φ=kπ+，kZ。
∵-π 0) or y = acos (ω x+φ) (A > 0, ω >; 0) Form; Find the period t =；; Find the amplitude a; List five special points in a period of time. When drawing an image within a specified interval, list the special points within the interval.
Preparation of type ⅱ trigonometric function image transformation: When trigonometric function image transformation, pay attention to the amount and direction of translation and expansion. (1) Translation transformation: translation along the X axis, according to the rule of "left plus right minus"; Translate along the y axis according to the law of "addition and subtraction"; (2) Telescopic transformation: when stretching along the X axis, the abscissa X is elongated (0; 1) is the original multiple (the ordinate y remains unchanged); When stretching along the y axis, the ordinate y is elongated (a >；; 1) or shortened by (00) units. When panning to the left, X is replaced by X+A, and when panning to the right, X is replaced by X-A, and any other values and symbols remain unchanged. If the abscissa of each point on the graph is extended to the original ω times (ω >); 1), you just need to change X. If you shorten the abscissa of each point on the image to the original (ω >; 1), you just need to change x into ω X.
Equation 3 is known as y = asin (ω x+φ) (a >; 0，φ& gt； 0) Preparation of image and analytical formula: The difficulty of image analytical formula y = asin (ω x+φ)+b lies in the determination of φ, which is essentially an undetermined coefficient method. The basic method is the "five-point method", which is determined by one of the "five points". Image transformation method, that is, the image whose transformed function is known, can usually determine φ from the zero point or the maximum point.
Example 3 the function f (x) = asin (x+φ) (a >; 0，0 & lt； φ& lt； π), the maximum value of xR is 1, and its image passes through point m.
(1) Find the analytical formula of f(x);
(2) Given α and β, and F (α) =, F (β) =, find the value of F (α-β).
[Analysis] (1) The maximum value of the function f(x) is 1, and a = 1.
The image of f(x) passes through point m,
sin=。
0 & ltφ& lt； π，& lt+φ& lt； π,
+φ=π,φ=.
So f (x) = sin = cosx.
(2) ∵ f (α) = cos α =, f (β) = cos β =, and α, β,
sinα=，sinβ=，
f(α-β)= cos(α-β)= cosαcosβ+sinαsinβ
=×+×=.
Preparation for solving the fourth kind of symmetry axis (symmetry center) problem: the symmetry problem of the image with function y = asin (ω x+φ).
(1) The image of the function y = asin(ωX+φ) is symmetrical about the straight line x = xk (where ωxk+φ= kπ+, kZ), that is, the straight line passing through the peak or trough and perpendicular to the X axis is its symmetry axis.
(2) The image of the function y = asin(ωX+φ) is centrosymmetric about the point (xj, 0) (where ωXJ+φ= kπ, kZ), that is to say, the intersection of the function image and the X axis (translation position point) is its symmetry center.
Example 4 shows that the minimum positive period of the function f(x) = asinω x+bcos ω x (where a, b, ω are real constants and ω > 0) is 2, and when x =, f(x) takes the maximum value of 2.
(1) Find the expression of function f(x);
(2) Is there a symmetry axis of f(x) in the closed interval [,]? If it exists, the equation of its symmetry axis is obtained; If it does not exist, explain why.
[Analysis] (1) f (x) = sin (ω x+φ) (φ is the auxiliary angle).
Suppose = 2, t = 2, ω = = π,
f(x)=2sin(πx+φ)，
When x =, f(x) takes the maximum value of 2,
2 = 2=2sin, that is, sin = 1
φ = 2kπ+from+φ = 2kπ+，kZ。
∴f(x)=2sin.
(2) by π x+= kπ+(kz),
X = k+, which is the symmetry axis of this function,
Let ≤ k+≤, get ≤k≤(kZ), k = 5,
So the symmetry axis of f(x) exists on [,], and its equation is x =.
[Comment] When the image of y = asin(ωx+φ) or some characteristics of its image are known to determine its analytical formula, a should be determined by the maximum value of the function, ω should be determined by the period t, and φ is more difficult to determine. Generally, φ should be determined by the key point of the image, but it should be noted that this key point is the first point in the "five-point method" drawing, and the corresponding value of ωx+φ is.
The image symmetry axis of y = asin (ω x+φ) can be solved by ω x+φ = kπ+(kz). There are countless kinds of symmetry axes, and sometimes we can use the test method to determine whether a straight line is its symmetry axis, or whether there is a symmetry axis in a certain range.
Inquiry: (1) Find the symmetry axis equation and the symmetry center coordinates of the function y = 2sin;
(2) Find the symmetrical center coordinates of the function y = 3 tan.
Analysis: (1) Observe the image of y = sinx, where x = kπ+(kz) is its symmetry axis and the coordinate (kπ, 0)(kZ) is its symmetry center.
Then 3x+= kπ+(kz) and x =+(kz) are called symmetry axes.
X =-(kZ) From 3x+= kπ, we know that (kZ) is the center of symmetry.
(2) Function y = y = tanx symmetry center of the tanx image is (kZ), then x =-(kz) is obtained from 2x+= (kz).
The symmetry center of the function y = 3 tan is (kz).
Skills If the function y = sin2x20 12, the mathematics of the college entrance examination always reviews a round of "No.1" courseware. Lecture 20