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A small problem in trigonometric function of senior one mathematics (for finding value range) Known function f (x) = [6 × The fourth power of cosx + 5 × Square - 4] / cos2x of SiNx, find its range Talk about your ideas, (you'd better attach the answer) sin ² We haven't learned the conversion of x = (1-cos2x) / 2. Can you tell me why it is so converted?
sin ² x=(1-cos2x)/2
cos ² x=(cos2x+1)/2
(cosx)^4=(cos ² x) ²= (cos2x+1) ²/ four
So f (x)
=[6(cosx)^4＋5sin ² x－4]／cos2x
=(3(cos2x+1) ²/ 2+(5(1-cos2x)/2-4)/cos2x
=[(3/2)(cos ² 2x+2cos2x+1)-(5/2)cos2x-(3/2)]/cos2x
=[(3/2)cos ² 2x+(1/2)cos2x]/cos2x
=(3/2)cos2x+(1/2)
-1=
Definition domain and value domain of inverse trigonometric function Especially the range of values. Why are they different in different questions Why is the domain of SiNx (- Pai / 2, Pai / 2), not r
It can be inferred from the definition of inverse trigonometric function:
1) Let SiNx = a, X ∈ [- Pai / 2, Pai / 2], a ∈ [- 1,1], then x = arcsin a
Therefore, the definition field of y = arcsinx: [- 1,1], value field: [- Pai / 2, Pai / 2]
2) Similarly, the inverse cosine range is: [0, Pai], and the arctangent range: (- Pai / 2, Pai / 2)
Then answer: only monotone functions can have inverse functions. To be exact, only one-to-one mapping can have inverse mapping
If x ∈ R, then when a = 0, arcsin a = 0, pie, or
At this time, y = arcsin x, for the same value of X, there are multiple y corresponding to it, which does not meet the function definition
Finding the definition domain and value domain of inverse trigonometric function f（x）=arc cos（3x+5）
f（x）=arc cos（3x+5） -1
Kneel down and find the respective definition fields and value fields of trigonometric function and inverse trigonometric function. Don't tell me what's wrong,
The definition domain of trigonometric function sin (x) and COS (x) is r, and the value domain is [- 1,1]. The definition domain of Tan (x) is x not equal to π / 2 + K π (K ∈ z), and the definition domain of r.cot (x) is x not equal to K π (K ∈ z), and the value domain is R. inverse trigonometric function y = arcsin (x), definition domain [- 1,1], value domain [- π / 2, π / 2] y = arccos (x
The value domain or definition domain of trigonometric function or inverse trigonometric function when double integral changes the integration order in polar coordinates An example: D:-π/2≤ θ ≤π/2; 0≤ ρ ≤acos θ It's right first ρ Integral, back pair θ integral After changing the integration sequence θ Integral, back pair ρ integral Change to D: 0 ≤ ρ ≤a;- arccos( ρ/ a)≤ θ ≤arccos( ρ/ a) What I want to ask is By ACOs θ=ρ obtain θ= arccos( ρ/ a), So use ρ express θ Integral time limit -arccos( ρ/ a) How did you get it?
How to determine the integral limit: 1. If ρ Integral, back pair θ Integral, then θ Take it as a constant and observe it ρ From the conditions you give, the integral field is a ρ= acos θ A is the diameter of the circle θ When fixed, ρ The variation range of is 0 ≤ ρ ≤acos θ, then θ My fan
Is the definition domain of inverse trigonometric function [- 1,1] or [- π / 2, π / 2]
The definition field is - 1 to 1, and the value field is the latter. In fact, the definition field is replaced with the original value field
RELATED INFORMATIONS
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