"End of Chapter Integration" Trigonometric Functions PPT

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"End of Chapter Integration" Trigonometric Functions PPT

Part One: Topic 1: Graphics of trigonometric functions and their transformations

Example 1 Function f(x)=Asin(ωx+φ) A, ω, φ are constants, A>0, ω>0, |φ|<π/2. The partial image is shown in the figure. , then the value of f(0) is _________.

Analysis: From the picture in question, we can see that A=√2, T/4=7π/12-π/3=π/4, so T=π, ω=2π/T=2. And the function graph passes through the point (π/ 3 "," 0),

Therefore, 2×π/3+φ=π, then φ=π/3, so the analytical formula of the function is f(x)=√2sin(2x+π/3), so f(0)=√2sinπ/3= √6/2.

Answer:√6/2

To summarize, when finding the analytical formula of the function y=Asin(ωx+φ)(A>0,ω>0) from the known function graph, the commonly used problem-solving method is the undetermined coefficient method. From the maximum or minimum value in the graph A is determined by the value, ω is determined by the period, and φ is determined by the coordinates of the point that fits the analytical formula. However, the analytical formula of y=Asin(ωx+φ)(A>0,ω>0) obtained from the image is generally not Uniquely, only by limiting the value range of φ can we get a unique solution. Otherwise, the value of φ is uncertain and the analytical formula will not be unique.

Variation training 1 It is known that the lowest point on the graph of the function y=f(x)=Asin(ωx+φ) A>0,ω>0,0<φ<π/2 is M(2π/3 " ,-" 2),The period is π.

(1) Find the analytical formula of f(x);

(2) Stretch the abscissa coordinates of all points on the image of y=f(x) to twice the original value (the ordinate remains unchanged), and then translate the resulting image to the right along the x-axis by units of length , get the image of the function y=g(x), and write the analytical formula of the function y=g(x);

(3) When x∈[0"," π/12], find the maximum and minimum values ​​of the function f(x).

Analysis: (1) First determine ω from the period of the function graph as π, and then determine A, φ from a lowest point of the graph as M (2π/3 ",-" 2). (2) Through image transformation The relationship with the analytical expression determines g(x). (3) The range of ωx+φ is determined by x∈[0"," π/12], thereby determining the optimal value.

Integrate PPT at the end of the chapter, the second part of the content: Topic 2 Evaluation of trigonometric functions

Example 2: Try to find the value of tan 10°+4sin 10°.

Analysis: The formula you are looking for contains tangent functions and chord functions. You should first divide the tangent chords into universal components, and then solve the problem based on the relationship between the angles.

Solution: Original formula = (√3 sin10"°" +4sin10"°" cos10"°" )/cos10"°"

=(√3 sin10"°" +2sin20"°" )/cos10"°" =(√3 sin"(" 30"°-" 20"°)" +2sin20"°" )/cos10"°"

=(√3 sin30"°" cos20"°-" √3 cos30"°" sin20"°" +2sin20"°" )/cos10"°"

=(√3/2 cos20"°" +1/2 sin20"°" )/cos10"°"

=(sin"(" 60"°" +20"°)" )/cos10"°"

=sin80"°" /cos10"°" =1.

PPT is integrated at the end of the chapter, and the third part is: Topic 3: Simplification and proof of trigonometric functions

Example 4 Simplification: ("(" 1+sinα+cosα")" (sin α/2 "-" cos α/2))/√(2+2cosα)(π<α<2π).

Analysis: Observe that it contains the angle α/2 and its double angle α, and the denominator contains the root sign. Consider using the power-raising formula to convert cos α into the formula related to cos2α/2, and remove the root sign.

Solution: original formula

=(2cos^2 α/2+2sin α/2 cos α/2)(sin α/2 "-" cos α/2)/√(4cos^2 α/2)

=(2cos α/2 (cos α/2+sin α/2)(sin α/2 "-" cos α/2))/2|cos α/2|

=(cos α/2 (sin^2 α/2 "-" cos^2 α/2))/|cos α/2|

=-(cos α/2 cosα)/|cos α/2| .

∵π<α<2π,

∴π/2<α/2<π.

∴cosα/2<0.

∴Original formula=cos α.

PPT is integrated at the end of the chapter, and the fourth part is: Topic 4: Comprehensive application of properties of trigonometric functions and transformation formulas

Example 6 When x=π/4, the function y=f(x)=Asin(x+φ)(A>0) obtains the minimum value, then the function y=f(3π/4 "-" x) is ()

A. An odd function obtains the maximum value when x=π/2

B. Even function and the graph is symmetric about the point (π,0)

C. An odd function obtains the minimum value when x=π/2

D. Even function and the graph is symmetric about the point (π/2 "," 0)

Analysis:∵f(π/4)=-A,

∴sin(π/4+φ)=-1,

∴φ=5π/4+2kπ,k∈Z,∴y=f(3π/4 "-" x)=Asin(-x)=-Asin x,

∴y=f(3π/4 "-" x) is an odd function and obtains the minimum value when x=π/2.

Answer:C

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Update Time: 2024-10-06

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