2. The definition domain of the function y = arccosx is [- 1, 1], and the value range is [0, π].

3. The function y = arctgx has a domain of r and a range of.

4. The definition domain of the function y = arcctgx is r, and its value range is (0, π).

5. Arcsine (-) =; arccos(-)=； arctg(- 1)=； arcctg(-)=

6 . sin(arccos)=； ctg[arcsin(-)]=； TG(arctg)=； cos(arcctg)=。

7. If cosx =-, x ∈ (,π), then x =.

8. If sinx =-, x ∈ (-, 0), then x =.

9. If 3ctgx+ 1 = 0, x ∈ (0, π), then x =.

Two. Basic requirements:

1. Understand the definition of inverse trigonometric function correctly and master the inverse function relationship between trigonometric function and inverse trigonometric function;

2. Grasp the definition and value range of inverse trigonometric function, y = arcsinx, x ∈ [- 1], y ∈ [-,], y = arccosx, x ∈ [- 1, 65438.

3. The symbol arcsinx can be understood as an angle or arc on [-,] or as a real number on the interval [-,]; The same symbol arccosx can be understood as an angle or arc on [0, π] or as a real number on the interval [0, π];

4.Y = Arcsinx is equivalent to Siny = X, Y ∈ [-] and Y = Arccosx is equivalent to Cosy = X, X ∈ [0, π], which is the main basis for solving the inverse trigonometric function problem;

5. Pay attention to the identity sin (Arcsinx) = x, x ∈ [- 1], cos (Arccosx) = x, x ∈ [-1], Arcsin (.

6. Master the judgment of the parity and increase and decrease of the inverse trigonometric function, and in most cases, it can be understood and applied by combining the images and properties of the corresponding trigonometric function;

7. Pay attention to the application of the equations Arcsinx+ArcCosx =, ArcTGX+ArcCTGX =.

Example 1. In the following categories, (c) is correct.

(A)arcctg(- 1)=-(B)arccos(-)=-

sin[arcsin(-)]=-(D)arctg(TGπ)=π

Solution: In (a) and (b), there are problems in the range, that is, ARCTTG (- 1) ∈ (0, π), ARCCOS (-) ∈ [0, π],

In (d), the arc TG (TG π) ∈ [-,], while π [-,] and ∴ (a) (b) (d) are incorrect.

Example 2. In the following functions, (d) has an inverse function.

(A)y=sinx，x∈[-π，0] (B)y=sinx，x∈[，]

(C)y=sinx，x∈[，] (D)y=sinx，x∈[，]

Solution: This problem is to determine the interval in which the function y = sinx is a monotone function. Since y = sinx is a monotonically decreasing function in the interval [,], D is selected.

Example 3. The arcsine (SIN 10) is equal to (c).

(A)2π- 10(B) 10-2π(C)3π- 10(D) 10-3π

Solution: This question is to judge which angle's sine value is equal to sin 10 and the angle is on [-,].

Since sin (3 π-10) = sin (π-10) = sin10, 3 π- 10 ∈ [-], C is selected.

Example 4. Find the inverse function of the following function, and find its domain and value domain.

( 1)f (x)=2sin2x，x∈[，]； (2)f (x)=+arccos2x。

Solution: (1) x ∈ [,], 2x ∈ [,], 2x-π ∈ [-,], -2 ≤ y ≤ 2.

From y = 2sin2x, sin2x =, sin (2x-π) =-sin2x =-,∴ 2x-π = arcsin (-),

∴x =- inverse symplectic, ∴f- 1(x)=- inverse symplectic, -2≤x≤2, y∈[,].

(2) f (x)=+arccos2x，x∈[-，]，y∈[，]，

∴ arccos2x=y-，2x=cos(y-)，x=cos(y-)=siny

∴f - 1(x)=sinx，x∈[，]，y∈[-，]。

Example 5. Find the domain and value of the following function:

( 1)y = arccos； (2)y = arcsine (-x2+x); (3) y=arcctg(2x- 1)，

Solution: (1) y = arccos, 0

(2)y = arcsine (-x2+x),-1≤-x2+x≤ 1, ∴ ≤x≤,

Because-x2+1=-(x-) 2+,∴-1≤-x2+x ≤, ∴-≤ y ≤ arcsine.

(3) y = arctg (2x- 1) because 2x-1>; - 1，∴0 & lt； arcctg(2x- 1)& lt； ，∴ x∈R，y∈(0，)。

Example 6. Find the scope of the following functions:

( 1) y=arccos(sinx)，x∈(-，)； (2) y=arcsinx+arctgx。

Solutions: (1) ∵ x ∈ (-), ∴ sinx ∈ (-, 1), ∴ y ∈ [0,].

(2)∫y = arcs inx+arctgx。 , x ∈ [- 1, 1], and Arcsinx and Arctgx are both increasing function.

∴ -≤arcsinx≤, -≤arctgx≤, ∴ y∈[-，]。

Example 7. Determine the parity of the following functions:

( 1)f(x)= xarcsin(sinx)； (2) f (x)=-arcctgx。

Solution: the domain of (1) f (x) is r, f (-x) = (-x) arcsin [sin (-x)] = xarcsin (sinx) = f (x),

F (x) is an even function;

(2) the domain of f (x) is r,

f(-x)=-arc ctg(-x)=-(π-arc ctgx)= arc ctgx-=-f(-x)，

F (x) is odd function.

Example 8. Make an image with the function y = arcsin (sinx), x ∈ [-π, π].

Solution: y = arcsin (sinx), x ∈ [-π, π], the image is abbreviated.

Example 9. Compare the sizes of Arcsin, ArcTG and Arccos (-).

Solution: Arcsin

Let Arcsin = α, sinα= =, ArcTG = β, Tgβ =, ∴ Sinβ =

arctg & ltarcsin & ltarccos(-)。

Example 10. Solving inequality: (1) Arcsinx <

Solution: (1) x ∈ [- 1, 1], when x =, arcsinx=arccosx, arcsinx is increasing function and arccosx is a decreasing function,

∴ When x ∈ [- 1,), Arcsinx

(2) ∵ arccosx =-arcsinx, ∴ The original formula is simplified to 4arcsinx & gt, arcsinx & gt=arcsin,

Arcsinx is increasing function, ∴ < x≤ 1.

Three. Basic skills training issues:

1. The following relationship always holds (b).

(A)π-arc cosx & gt； 0(B)π-arcctgx & gt； 0 (C)arcsinx-≥0 (D)arctgx->0

2. The subtraction function defined on (-∞, ∞) is (d).

(A)y = arcs inx(B)y = arc cosx(C)y = arct GX(D)y = arcctgx

3. The solution set of inequality arcsinx & gt is. 4. inequality arccosx >; The solution set of is.

4. Selected questions:

(1) Multiple choice questions:

The value of 1.cos (arccos) is (d).

(A) (B) (C)cos (D) does not exist.

2.arcsinx & gt 1, then the range of x is (c).

(A)sin 1 & lt； x & lt(B)sinx & lt； x ≤( C)sin 1 & lt； x≤ 1 (D)

3. Given y = arcsinx arctg | x | (-1≤ x ≤1), then this function (a).

(a) is an odd function; (b) is an even function; (c) It is both odd function and even function; (d) is an odd or even function.

4. if a = arcsin (-), b = arctg (-) and c = arccos (-), then the relationship between a, b and c is (b).

(A)A & lt； b & ltc(B)a & lt； c & ltb(C)C & lt； a & ltb(D)c & lt； b & lta

5. Given tgx =-, x ∈ (,π), then x = (c).

(A)+arctg(-)(B)π-arctg(-)(C)π+arctg(-)(D)

6. The inverse function of the function f (x) = 2 arccos (x-2) is (d).

(A)y =(cosx-2)(0≤x≤π)(B)y = cos(x-2)(0≤x≤2π)

(C)y = cos(+2)(0≤x≤π)(D)y = cos+2(0≤x≤2π)

7. If arccosx≥ 1, the value range of x is (d).

(A)[- 1， 1] (B)[- 1，0] (C)[0， 1] (D)[- 1，arccos 1]

8. Function y = y = arccos (sinx) (-<

(A)(，)(B)[0，] (C)(，)(D)[，]

9. Given x ∈ [- 1, 0], the following equation holds (b).

(A)arcsin = arc cosx(B)arcsin =π-arc cosx

(C)arccos = arcsinx(D)arccos =π-arcsinx

10. The inclination of the line 2x+y+3 = 0 is equal to (c).

(A)arctg 2(B)arctg(-2)(C)π-arctg 2(D)π-arctg(-2)

(2) Fill in the blanks:

1 1. If cos α =- (

12. The minimum value of the function y = (Arcsinx) 2+2 Arcsinx- 1 is -2.

13. The inverse function of the function y = 2sin2x (x ∈ [-,]) is.

14. the definition domain of the function y = arcsin is x≤ 1 or x≥3, and the range of values is

15. arctangent means that the inclination of the straight line AX-Y+A = 0 (a ≠ 0) is α =

(3) Answer questions:

16. Find the inverse of the following function:

( 1) y=3cos2x，x∈[-，0]； (2)y =π+arccos x2(0 & lt； x≤ 1)。

Solution: (1) x ∈ [-0], ∴ 2x ∈ [-π, 0], and the function y = 3cos2x is a single-valued function on the domain.

And -3 ≤ y ≤ 3. ∴ π+2x ∈ [0，π]，y = 3cos2x =-3cos (π+2x)，cos (π+2x) =-，

∴ π+2x=arccos，∴x=arccos-，

∴ y = 3cos2x, the inverse function of x ∈ [-, 0] is y = arccos-, -3 ≤ x ≤ 3.

(2)∵0 & lt； x≤ 1，π≤y & lt； ，∴ arccosx2=y-π，x2=cos(y-π)，x=，

The inverse function of the original function is y =, π≤ X.

17. Find the maximum value of the function y = (arccosx) 2-3 arccosx and the corresponding value of x.

Solution: function y = (arccosx) 2-3 arccosx, x ∈ [- 1, 1], arccosx ∈ [0, π]

Let arccosx = t, 0 ≤ t ≤ π, ∴ y = t2-3t = (t-) 2-,

When t =, that is, x = cos, the function takes the minimum value,

When t = π, that is, x =- 1, the function obtains the maximum value π 2-3 π.

18. If f (arccosx) = x2+4x, find the maximum value of f (x) and the corresponding value of x. ..

Solution: let arccosx = t, t ∈ [0, π], x = cost, and substitute it to get f (t) = cos2t+4cost.

∴ f (x)=cos2x+4cosx，x∈[0，π]，cosx∈[- 1， 1]，f (x)=(cosx+2)2-4

When cosx =- 1, that is, x = π, the function gets the minimum value of -3.

When cosx = 1, that is, x = 0, the function gets the maximum value of 5.

19.( 1) Find the monotone decreasing interval of function y = arccos (x2-2x); (2) Find the monotonic increasing interval of the function ARCTG (x2-2x).

Solution: (1) function y = arccosu, u ∈ [- 1, 1] is a decreasing function,

∴- 1 ≤ x2-2x ≤ 1， 1-≤ x ≤ 1+，x2-2x = (x- 1) 2+ 1，

∴ 1 ≤ x ≤ 1+, u = x2-2x is a increasing function, and according to the concept of composite function, the original function is a subtraction function.

(2) the function y = arctgu increasing function, u∈R, and x2-2x = (x- 1) 2+ 1,

When x≥ 1, the original function is increasing function.

20. find a point on the curve y = y=5sin(arccos) to make it farthest from the straight line x+y- 10 = 0 and find the farthest distance.

Solution: let arccos = α, -3 ≤ x ≤ 3, cos α =,

y=5sinα=5，

Properties and images of trigonometric functions

[Key]: Properties and images of compound trigonometric functions.

[Difficulties]: Image transformation of compound trigonometric function

[Example]

Example 1. Find the domain of the function: f(x)= 1

Solution:

( 1):2kπ≤x ≤( 2k+ 1)π(k∈Z)

(2):-4 & lt； X< iv

The domain is.

Note: the unit of independent variable x in sinx is "radian", x ∈ r.

Example 2. Find the increasing interval of y=cos( -2x).

Analysis (1): This function is a composite function of y=cosu and u= -2x.

∵ u= -2x is a decreasing function, and you need the increasing interval of y=cos( -2x), and you only need to find the decreasing interval of y=cosu.

Method (1):∵y = COSU, the decreasing interval is 2kπ≤u≤π+2kπ (k∈Z).

∴ Let 2kπ≤ -2x≤π+2kπ, -kπ≤ x≤-kπ (k ∈ z).

∫-k is equivalent to k, and the ∴ increasing interval is [- +kπ, +kπ] (k∈Z).

Analysis (2): ∵ COSU is an even function, ∴ y=cos(2x-)

Let y = cost and t=2x,

∵ t=2x- is the increasing function, and the increment interval of y=cos(2x-) is needed, and only the increment interval of y=cost is needed.

Method (2): the increasing interval of ∵ y = cost is π+2kπ≤t≤2π+2kπ (k∈Z).

∴ Let π+2kπ≤2x- ≤2π+2kπ, +kπ≤x≤ +kπ (k∈Z).

∴ The increasing interval is +kπ≤x≤ +kπ (k∈Z).

Note: The results obtained by the two methods are different on the surface, but the range represented by the two forms is exactly the same from the figure.

Example 3. Find the period and range of the function y = sin2x+sinx sin (x+).

Analysis: To find the period, range and monotone interval of a function, the common method of trigonometric function is to convert it into an angle trigonometric function.

Solution: y=

=

=

=

∴ T= =π, the range is [].

Example 4. Find the maximum value of the function y = sinx cosx+sinx+cosx.

Analysis: sinx+cosx and sinxcosx are mutually transformed. If sinx+cosx is regarded as a whole and set as a new element, then the function can be transformed into a function of Singapore dollar. Pay attention to the range of Singapore dollar.

Solution: let sinx+cosx=t, t∈[-,].

Then (sinx+cosx)2=t2, that is, 1+2sinxcosx=t2, sinxcosx=,

y = t+=(T2+2t)-=(t+ 1)2- 1，

When t= ymax=+.

Example 5. Judge the parity of the following functions

( 1)y=sin(x+ )- cos(x+)

(2)y=

Analysis: the domain is r, which is symmetrical about the origin. After equivalent deformation, it is transformed into a trigonometric function of an angle as much as possible, and then its parity is judged.

Solution: (1)y=2[ sin(x+ )- cos(x+ )]

=2sin[(x+ )- ]

=2sinx

This function is an odd number function.

(2) The domain x≡π+2kπ and (k∈Z) can be obtained from the denominator, and the domain is asymmetric about the origin in the rectangular coordinate system.

∴ Function is odd and even.

Example 6. Write the analytical expressions of the following function images.

(1) Shift all points on the image with function y=sinx to the left by one unit, and then expand the abscissa of each point on the obtained image to twice the original one, thus obtaining the image of the required function.

(2) Reduce the abscissa of all points on the image of function y=cosx to half of the original value, keep the ordinate unchanged, and then move the image to the left by one unit to get the image of the required function.

(1) Analysis: According to the sequence of image transformation, the variation of independent variable X is:+; Time magazine.

The analytical formula of the image is: y=sinx→y=sin(x+ )→y=sin ().

Solution: The analytic expression of the function image is y=sin (), which can also be written as y=sin (x+).

(2) Analysis: According to the sequence of image transformation, the change of independent variable X is: 2 times; + 。

The analytical formula of the image is: y=cosx→y=cos2x→y=cos2(x+).

Solution: The analytic expression of the function image is y=cos2(x+), which can also be written as y=cos(2x+).

Example 7. Known function y=sin(3x+)

(1) Judge the parity of the function;

(2) Judging the symmetry of the function.

Analysis: There are both connections and differences between the parity of function and the symmetry of function. By definition, alternative methods.

Solution: (1) is defined as r and f(x)=sin(3x+).

f(-x)=sin[3(-x)+ ]=-sin(3x-)

∫sin[3(-x)+]≠sin(3x+)

sin[3(-x)+]≦- sin(3x+)

The function y=sin(3x+) is neither a odd function nor an even function.

(2) The image with function y=sin(3x+) is an axisymmetric figure, and the equation of symmetry axis is 3x+ =kπ+.

That is, x= (k∈Z)

The image with function y=sin(3x+) is also a central symmetric figure, and the coordinate of the symmetric center of the image with function y = sinu is (kπ, 0).

Let 3x+ =kπ and x= (k∈Z).

∴ y=sin(3x+) The coordinate of the symmetry center of the image is (,0) (k∈Z).

test

Multiple choice

1 domain. Y = is (below k ∈ z) ().

(A)[2k ] (B)[2k ]

(C)[2k ] (D)(-∞，+∞)

2.f (x) = cos (3x-θ)-sin (3x-θ) is odd function, then θ = () (hereinafter referred to as k∈Z).

(A)kπ (B)kπ+ (C)kπ- (D)kπ+

3. The same function as the image with function y=cos(x-π) on [] is ().

(A)y =(B)y =(C)y = cos(x-)(D)y = cos(-x-4π)

4. Shift the image with function y=sin(2x-) to the right by one unit, and the corresponding function of the obtained image is ().

(a) Non-odd and non-even functions (b) are both odd function and even functions.

(c) odd function (d) even function

5. Transform the image with function y=sin () as follows to get the image with function y=sin x ().

(a) Translation to the right (b) Translation to the left (c) Translation to the right (d) Translation to the left.

6. The function f (x) = sin (ω x+θ) cos (ω x+θ) (ω >; 0) Take 2 as the minimum positive period and get the maximum value when x=2, then one value of θ is ().

(A)- π (B)- π (C) π (D)

7.ω is a positive real number, and the function is increasing in number, then ().

(A) (B) (C) (D)

8. The monotonic increasing interval of 8.y = cos (+2x) sin (-2x-2x) is (below k∈Z) ().

(A)[ ] (B)[ ]

(C)[ ] (D)[ ]

9. The function y = 3sin (the maximum value of x+is ().

Article 4, paragraph 2, paragraph 3, paragraph 7, paragraph 4, item 8

10. When x∑ (), f(x)=|sin(3kx+ )| has a complete period, then the smallest positive integer value that k can take is ().

12 (B) 13 (C)25 (D)26

Answer and analysis

Answer: 1, D 2, C 3, A 4, D 5, C 6, A 7, A 8, A 9, D 10, b.

Analysis:

1. For x∈R,-1≤sinx≤ 1, cos(sinx)>0 is a constant, so x ∈ r.

2. If f(x)=2sin(+θ-3x) is sorted out, it can be verified by substituting f(0)=0.

Note: A property of odd function: If there is 0 in the domain of odd function's f(x), then f(0)=0 (otherwise it may not be true).

3. First, y=cos(x-π)=-cosx,

y = = | cosx | =-cosx(∵x∑[]，cosx & lt0)

Y = (when x = is meaningless, it is obviously not the answer)

y=cos(x- π)=-sinx，

y=cos(-x-4π)=cosx .

4 . y = sin(2x-)y = sin(2(x-))=-cos2x .

Note: For the translation of function images, master the law of adding left and subtracting right (adding a number to X when translating left and subtracting a number from X when translating right). Note, just change (x).

5.y=sin x=sin[ (x- )+ ]，y=sin( x+ )→y=sin[ (x- )+ ]

That is, x becomes x-, so shift the unit to the right.

6.f(x)= sin(2ωx+2θ). When t = 2, ω = and x=2, take the maximum value of f(x) and substitute it into the options for verification.

7. Let ωx=t, because f(x)=2sint is the increasing function on [-,].

So -≤t≤, that is, -≤ωx≤,-≤x≤,

It is known that f(x) increases on [-,], so we can solve 0.

8. Simplify y=-sin4x=- sin4x+. The original problem is to find the decreasing interval of sin4x.

2kπ+ ≤4x≤2kπ+ π ≤x≤ π.

9. Pay attention to simplify the formula y=8cos(x-).

10. The function f(x) = is solved according to the meaning t, that is, the period t of k≥4π.

Note: The period of the function f(x)=|sinωx| is T=.

Trigonometric function problem with parameters

The question with parameters has been a hot topic in the college entrance examination in previous years because it can well examine the mathematical thought of classified discussion and in-depth study of mathematical ability. However, due to the difficulty, it has cooled down in the past two years. Problems with parameters often appear in inequalities and functions, and are sometimes involved in trigonometric functions. However, because trigonometric functions are mostly low and intermediate questions in the college entrance examination, this part is more difficult.

The so-called parameters are related to variables. Therefore, to deal with this kind of problems, we should have the idea of variables, that is, we should regard parameters as a moving and changing quantity. When this parameter changes to different values, it may have different effects on the problem-solving process and needs to be discussed in categories. The following examples are all about the combination of images and properties of parametric variables and trigonometric functions.

Example 1. If cos2x=acos2x+bcosx+c holds for all real numbers x, then a2+B2+C2 = _ _ _ _ _.

Analysis: When the variable x changes, the value of cosx also changes, but this change cannot affect the value of the whole formula.

Solution: The original formula is arranged as: (a-2)cos2x+bcosx+c+ 1=0, that is, this formula holds no matter what value X takes.

Then a-2=0, b=0, c+ 1=0 must be established at the same time. Solve a=2, b=0, c=- 1, then a2+b2+c2=5.

Note: only a=0 is necessary to make acosx unaffected by the change of x value.

Example 2. It is known that α, β ∈ [-,], sinα= 1-a, sinβ= 1-a2, α+β.

Analysis: In order to get the range of the variable A, we must find an inequality containing A according to the known conditions, and pay attention to the boundedness of the sine function in this question.

Solution: Because α+β

According to y=sinx is the increasing function on [-,], we get sinα.

So there is, solve 1

Note: This topic mainly examines the range of trigonometric functions and the flexible use of monotonicity.

Example 3. The image of the function y=sin2x+acos2x is symmetric about the straight line x=-, so what is the value of a?

Analysis: If the image of the function f(x) is symmetrical about the straight line x=a, then there is f(a+x)=f(a-x).

Solution: Let f(x)=sin2x+acos2x, and F (-+x) = f (-x) hold for any X according to the meaning of the question.

That is, sin (-+2x)+acos (-+2x) = sin (-2x)+acos (-2x).

sin(-+2x)+sin(+2x)= a[cos(+2x)-cos(-+2x)]

( 1+a)sin2x=0

For the above formula to hold (not affected by the value of x), it must be 1+a=0, that is, a=- 1.

Note: 1, is there anything similar to the example 1

2. For multiple-choice questions, you can substitute two points about x=- symmetry for verification, for example.

Example 4. It is known that the equation 2sin2x-cos2x+2sinx+m=0 has a solution, and the range of seeking truth from the number m.

Analysis: put the variable m aside and examine the range of values on the other side.

Solution: m=-3sin2x-2sinx+ 1 is obtained from the original formula.

Let y=-3sin2x-2sinx+ 1, then y has a maximum value and a minimum value. As long as m is within this range, the original equation has a solution.

Let t=sinx, then-1≤t≤ 1, and find the range of y=-3t2-2t+ 1 According to the image with quadratic function -4≤y≤,

That is, when -4≤m≤, the original equation has a solution.

Note: It is also a skill to separate variables and put them aside. The following Example 5 was also used.

Example 5. Given 0≤θ≤ find cos2θ+2msinθ-2m-2: (sinθ-1< 0)

Let y=, then y is a variable, and let 2m >；; Y holds, as long as the maximum value of 2m>y is enough.

The maximum value of y (0 ≤ sinθ

y=

=sinθ+

=sinθ+ 1+

=-[( 1-sinθ)+ ]+2

∫( 1-sinθ)+ When 1-sinθ= 1, that is, θ=0, the minimum value is 3.

∴ y max =- 1，2m & gt- 1，m & gt- ,

Therefore, when θ = m takes any real number, the original formula holds.

When 0 ≤ θ

Note: 1, this topic is comprehensive, which is a difficult topic. There is more knowledge to investigate, but we must understand the idea of variables.

2. find the function y = x+(a >；; 0), we can observe the image at (0, +∞) according to the image, as shown in the figure (odd function).

Summary: Examples 1, 3, 4 and 5 all reflect the idea of variables, so we should pay attention to experience. Example 5 deeply examines the idea of classified discussion. In addition, the problem of parameters is often associated with the range of values, which is doomed to be associated with inequality.

High-quality college entrance examination questions

1. Among the following four functions, the function with π as the minimum positive period and decreasing in the interval is ().

a、y=cos2x B、y=2|sinx| C、D、y=-cotx

Solution: y=cos2x, the period is π, and the interval is increasing function.

Y=2|sinx|, the period is π, and it is a decreasing function in the interval.

At least, it can be judged that it is not a decreasing function in the interval.

Y=-cotx, increasing function in the interval, ∴ B

2. The approximate image of the function y=x+sin|x|, x∈[-π, π] is ().

Solution: delete a, b, d from the parity (non-odd and non-even) of the function and the coordinates of special points. Angle

3. Let the function f(x)=sin2x. If f(x+t) is an even function, one possible value of t is _ _.

Solution: Draw a sketch with f(x)=sin2x. It is not difficult to see that if you move the image horizontally to the left, you can get an image that is symmetrical about the Y axis.

It should be filled in.

4. Part of the image of the function y=-xcosx is ().

Solution: ∫f(x)=-xcosx, ∴ f(-x)=-(-x)cos(-x)=xcosx=-f(x)

Then f(x) is odd function (x∈R), which can be selected from b and d,

Let a point on the image be below the x axis.

∴

5. The known function f (x) = x2+2x tan θ- 1, where.

(1) When, find the maximum and minimum of the function f(x);

(2) Find the range of θ so that y=f(x) is a monotonic function in the interval.

Solution: (1) When,

The minimum value of f(x) is,

When x=- 1, the maximum value of f(x) is.

(2) Function f(x)=(x+tanθ)2- 1-tan2θ The symmetry axis of the image is x=-tanθ,

Y = f (x) is a monotone function in the interval [- 1,].

∴-Tan θ≤ 1 or,

That is, tanθ≥ 1 or tanθ≤0,

Therefore, the range of θ is.

Comments: This question is a synthesis of the basic knowledge of quadratic function and trigonometric function. In solving the problem (1), after getting the analytical formula of quadratic function, we should pay attention to correctly comparing the function value at the end of the interval with the maximum value of this function and make a choice.

In the problem (2), it is assumed that f(x) is a monotone function on the interval, which should be considered by classification. If it is monotonically increasing, then -tanθ≤ 1, if it is monotonically decreasing, then this step is the focus and difficulty of solving the problem.

6. The known function x ∈ r.

(i) When the function y takes the maximum value, find the set of independent variables x;

(II) What kind of translation and scaling transformation can be obtained from the image of y=sinx(x∈R)?

Solution: (1)

It is necessary and only necessary for y to get the maximum value

That's k ∈ z.

So when the function y takes the maximum value, the set of independent variables x is.

(II) Transform the function y=sinx as follows:

(i) shifting the image of the function y=sinx to the left to obtain the image of the function;

(ii) shortening the abscissa of each point on the obtained image to the original time (the ordinate is unchanged) to obtain the image of the function;

(iii) shortening the ordinate of each point on the obtained image to the original time (the abscissa is unchanged) to obtain the image of the function;

(IV) shifting the obtained image upward by a unit length to obtain an image of the function;

To sum up, the image of the function is obtained.

Comments: By using trigonometric formula, the known function can be transformed into a simple analytic function of an angle, and its maximum value can be discussed. The solution to this problem is explained in detail with the corresponding image transformation, which should be understood and mastered.