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The known function f (x) = 2Sin (π - x) cosx (1) Find the minimum positive period of F (x); (2) Find the monotone increasing interval of F (x) and its symmetry axis
∵f（x）=2sin（π-x）cosx=2sinxcosx=sin2x
(1) According to the periodic formula, t = π can be obtained
(2) Order − 1
2π+2kπ≤2x≤1
2 π + 2K π, − π
4+kπ≤x≤π
4+kπ，k∈Z
Let 2x = k π + 1
2 π, x = π
4+kπ
2，k∈Z
The monotonic increasing range of F (x) is [− π
4+kπ，π
4 + K π], its axis of symmetry x = π
4+kπ
2，k∈Z
Let f (x) = 2Sin (π - x) cosx. (1) find the minimum period of F (x). (2) find f (x) in the interval [- π - 6, π - 2] Maximum and minimum on The period is the minimum positive period
⑴f(x)＝2sin(π－x)cosx＝2sinxcosx＝sin2x
Minimum positive period T = 2 π / 2 = π
⑵∵x∈[－π/6,π/2]
∴2x∈[－π/3,π]
∴f(x)max＝sin(π/2)＝1,f(x)min＝sin(－π/3)＝－1/2.
Given the function f (x) = 2Sin (Pi-X) cosx, find the minimum positive period of F (x)?
f(x)=2sin(π-x)cosx=2sinxcosx=sin2x
The minimum positive period T = 2 π / w = 2 π / 2 = π
Given the function f (x) = 2Sin (Pi-X) cosx, find the minimum positive period of F (x) and find f (x) in the interval [- 6 / PI, 2 The known function f (x) = 2Sin (Pi-X) cosx Finding the minimum positive period of F (x) Find the maximum and minimum of F (x) on the interval [- 6 / Pai, 2 / PAI]
Finding the minimum positive period of F (x) by known function f (x) = 2Sin (Pi-X) cosx
The minimum positive period of the function f (x) = 2Sin (Pi-X) cosx = 2sinxcosx = sin2xf (x) t = 2 π / 2 = π
The maximum and minimum of F (x) on the interval [- 6 / Pai, 2 / PAI]
X belongs to [- π / 6, π / 2]
2X belongs to [- π / 3, π]
X = - π / 6 min = - √ 3 / 2
X = π / 4 max = 1
The known function f (x) = 2Sin (Pi-X) cosx (1) Find the minimum positive period of F (x) (2) find the maximum and minimum of F (x) on the interval of [- Pai / 6, Pai / 2]
F (x) = 2Sin (pai-x) cosx
=2sinxcosx=sin2x
Minimum positive period = 2pi / 2 = pi (PI is pie)
F (- pi / 6) = sin (- pi / 3) = - (radical 3) / 2
f(pi/2)=sin(pi)=0
f(pi/4)=sin(pi/2)=1
Max = 1
Minimum = - (radical 3) / 2
Given the function f (x) = 2Sin (π - x) cosx, find the minimum positive period of F (x)
f(x)=2sin(π-x)cosx
=2sinxcosx
=sin2x
T=2π/2=π
Answer: the minimum positive period is π
It is known that the function f (x) = 2Sin ω x (ω > 0) is in the interval [- π 3，π 4] If the minimum value on is - 2, then the minimum value of ω is equal to () A. 2 Three B. 3 Two C. 2 D. 3
The function f (x) = 2Sin ω x (ω > 0) is in the interval [- π
3，π
4] If the minimum value on is - 2, then the value range of ω x is [− ω π
3，ωπ
4]，
∴−ωπ
3≤−π
2 or ω π
4≥3π
2，
The minimum value of ω is equal to 3
2，
Therefore, B
It is known that the function f (x) = 2Sin ω x (ω > 0) is in the interval [- π 3，π 4] If the minimum value on is - 2, then the minimum value of ω is equal to () A. 2 Three B. 3 Two C. 2 D. 3
The function f (x) = 2Sin ω x (ω > 0) is in the interval [- π
3，π
4] If the minimum value on is - 2, then the value range of ω x is [− ω π
3，ωπ
4]，
∴−ωπ
3≤−π
2 or ω π
4≥3π
2，
The minimum value of ω is equal to 3
2，
Therefore, B
Find the function y = 1 / 2 SiNx + radical 3 / 2 cosx, find the maximum and minimum value x ∈ (- 90,90) Now I need to... Do a little favor
y=1/2 sinx+√3/2 cosx
=Cos60 degree SiNx + sin60 degree cosx
=Sin (x + 60 degrees)
X = 30 degrees, maximum = 1, x = - 90 degrees, minimum = - 0.5
If f (x) = 2 roots, 3 * SiNx / 3 * cosx / 3-2sin square X / 3 1. X belongs to [0, π]. Find the range of F (x) and the coordinate of symmetric center 2. In △ ABC, the opposite sides of a, B and C are a, B and C respectively. If f (c) = 1 and B squared = AC, find Sina
F (x) = 2 radicals 3 * SiNx / 3 * cosx / 3-2sin squared X / 3 = 4sinx / 3 (√ 3cosx / 3-sinx / 3) / 2
=4sinx/3sin﹙π﹣x﹚/3=﹣2[cosπ/3－cos﹙2x-π﹚/3]=2cos﹙2x-π﹚/3－1
The range of F (x) = - 1 + 2cos (2x - π) / 3 is - 3 ≤ f (x) ≤ 1
∵ ﹙2x-π﹚/3=kπ+π/2 ＝＝＞x=﹙3kπ+5π/2﹚／2
ν symmetric center coordinate (3K π + 5 π / 2) / 2, - 1) (k is an integer)
2.f（C）=1 ＝＝＞cos﹙2x-π﹚/3=1 ∴﹙2x-π﹚/3=0 x=π/2
The △ ABC is a right triangle, and bsquare = AC, C? = a? + B? Sina = A / C > 0
∴a²/c²+a/c﹣1 =0
∴sinA=a/c=﹙√5-1﹚/2
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