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Given that the minimum positive period of the function y = 1 / 2Sin [(x + π) / a] (a > 0) is 3 π, then a =, the value range of the function is:
∵ y = 1 / 2Sin [(x + π) / a] (a > 0) the minimum positive period is 3 π
∴T=2π/(1/A)=3π
A = 3 / 2
The original formula is reduced to
y=1/2sin【3/2X+3/2π】
When 3 / 2x + 3 / 2 π = π / 2 + 2K π (K ∈ n), that is, x = 4 / 3K π - 2 / 3 π (K ∈ n), the maximum value y = 1 / 2 is obtained
When 3 / 2x + 3 / 2 π = - π / 2 + 2K π (K ∈ n), i.e. x = 4 / 3K π - 2 π (K ∈ n), the minimum value y = - 1 / 2 is obtained
The value range of the function is y ∈ [- 1 / 2,1 / 2]
Let (2x) be the even (2x) of (2x) + the (2x) of (2x) + the (2x) of (2x) + θ
If f (x) = sin (2x + θ) + 2 √ 3 [1 + cos (2x + θ)] / 2 - √ 3 = sin (2x + θ) + 3cos (2x + θ) = 2Sin (2x + θ + π / 3) is the even function, f (- x) = f (x) 2Sin (- 2x + θ + π / 3) = 2Sin (2x + θ + π / 3) = 2Sin (2x + θ + π / 3) so - 2x + θ + π / 3 = 2K π + 2x + θ + π / 3 or 3 or 3 or 3 or 3 or 3 or 3 or 3 or 3 or 3 or 3 or 3 or 3, or 3 = 2K π + 2x + 2x + 2x + 2x + 2x + π- 2x + θ + π / 3 = 2K π + π - (2x + θ + π / 3) -
Given the function f (x) = 2Sin (x + α / 2) cos (x + α / 2) + 2 √ 3cos ^ 2 (x + α / 2) -√ 3, α is a constant Finding the minimum positive period of function f (x)
f(x)=2sin(x+α/2)cos(x+α/2)+2√3cos^2(x+α/2)-√3
f(x)=sin(2x+α)+2√3[cos(2x+α)+1]/2-√3
f(x)=sin(2x+α)+√3[cos(2x+α)+1]-√3
=sin(2x+α)+√3cos(2x+α)
=2×[1/2sin(2x+α)+√3/2cos(2x+α)]
=2[sin(2x+α)·cosπ/3+cos(2x+α)·sinπ/3]
=2sin(2x+α+π/3) ∴ ω = 2
T=2π/|ω|=π
The known function f (x) = [2 √ 3cos (x / 2) + 2Sin (x / 2)] cos (x / 2) (1) Find f (17 π / 12) (2) In ⊿ ABC, the sides of angles a, B and C are a, B and C respectively. If f (c) = √ 3 + 1 and B ^ 2 = AC, find the value of sina
(x / 2) = 2 √ 3 [cos (x / 2) + 2Sin (x / 2)] cos (x / 2) = 2 √ 3 [cos (x / 2)] ^ 2 + SiNx = 3 {2 [cos (x / 2)] ^ 2 + SiNx = 3 {2 [cos (x / 2)] ^ 2-1} + 3 + SiNx = 3 + SiNx = 3 = 2 (SiNx * 1 / 2 + 3 / 2 * cosx) + 3 = 3 = 2Sin (x + π / 3 + 3) + 3 (1) for f (17 π / 12) = 2F (17 π / 12) = 2F (17 π / 12) = 2F (17 π / 12) = 2F (17 π / 12) = 2F (17 π / 12) = 2 (1) (1) for f (1 sin (7 π / 4) + √ 3 = √ 3 - √ 2 (2) f (c) = √ 3
Given the function y = 2Sin ω xcos ω x + 2 √ 3cos2 ω X - √ 3 (ω >), the straight lines X = X1 and x = x2 are any two symmetric axes of y = f (x), The minimum value of absolute value x1-x2 is π / 2 (1) Find the value of ω (2) if f (a) = 2 / 3, find the value of sin (5 π / 6-4a) ω＞0
The original function can be reduced to y = radical 13sin (2wx + U) - radical 3
1, the minimum value of absolute value X1 -- X2 is π / 2, and W > 0, so t / 2 = π / 2 = 2 π / 2W, w = 1
2, y = radical 13sin (2a + U) - radical 3 = 2 / 3
Well, the second question has not been answered for the time being
Find the value range of the function y = 2Sin xcos x + 2Sin x + 2cos x + 4
t=sinx+cosx=√2 sin(x+π/4)
-√2 =
The known function y = 2Sin (π of x-3) 1. Find its definition domain and minimum positive period 2. Use "five point method" to draw the image of this function in a period (to be listed)
1. The definition domain is the RX coefficient is 1, so t = 2 π / 1 = 2 π 2, 2, 5-point method, that is, sin takes 0, π / 2, π, 3 π / 2, π, π then x-π / 3 = 0, x = π / 3, sin (x-π / 3) = 0x - π / 3 = π / 2, x = 5 π / 6, sin (x-π / 3) = 1 x-π / 3 = π, x = 4 π / 3, sin (x-π / 3) = 0x - π / 3 = 3 = 3 π / 2, 3, sin (x - π / 3) = 0x - π / 3 = 3 = 3 π / 2, 2, 3, 3, 3, 3 π / 2, 3 = 3 = 3 x = 11 π / 6, sin (x - π / 3) =
The known function f (x) = 2Sin ω xcos ω x (ω > O, X belongs to R) (1) Find the range of F (x) (2) If the minimum positive period of F (x) is 4 π, find the value of ω
F (x) = sin (2wx), so the range is [- 1,1]
T = 2 π / 2W = 4 π, so w = 1 / 4
F (x) = 3sin ω xcos ω x + √ 3cos2 ω X - √ 3 / 2 + 1 find the value range of function y = f (x) F (x) = 3sin ω xcos ω x + √ 3cos ^ 2 ω X - √ 3 / 2 + 1
f(x)=3sinωxcosωx+√3cos^2ωx-√3/2 +1
=3/2sin2ωx+√3/2cos2ωx+1
=√3sin(2ωx+π/6)-√3/2+1
Therefore, the range is: [- 3 √ 3 / 2 + 1, √ 3 / 2 + 1]
It is known that the function f (x) is equal to the square of 2Sin (4 / 4 row plus x) minus the root sign 3cos2x minus 1. X belongs to R. find the minimum positive period of function f (x)
f(x)=2sin^2 (x+π/4)-√3cos2x-1
=-cos(2x+π/2)-√3cos2x
=sin2x-√3cos2x
=2sin(2x-π/3)
When x belongs to R, the minimum positive period of function f (x) is t = 2 π / 2 = π
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