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Known a＝(− 3sinωx，cosωx)， B = (COS ω x, cos ω x) (ω > 0), Let f (x) = a• b. The minimum positive period of F (x) is π (1) Find the value of ω; (2) Find the monotone interval of F (x)
（1）f（x）=−
3sinωxcosωx+cos2ωx=-
Three
2sin2ωx+1
2cos2ωx+1
2=-sin（2ωx-π
6）+1
2．
∵ω＞0，∴T=2π
2ω=π，
∴ω=1．
(2) From (1), we know that f (x) = - sin (2x - π)
6）+1
2．
∵2kπ-π
2≤2x-π
6≤2kπ+π
2，k∈Z，
K π - π
3≤x≤kπ+2π
3. The K ∈ Z function is a decreasing function
From 2K π + π
2≤2x-π
6≤2kπ+3π
2，k∈Z，
K π + 2 π
3≤x≤kπ+5π
3. K ∈ Z function is increasing function
So the monotone decreasing interval of the function is [K π - π
3，kπ+2π
3]，k∈Z．
The monotone increasing interval of the function is [K π + 2 π]
3，kπ+5π
3]，k∈Z．
Vector M = (- 1, cos ω x + Radix 3sin ω x), n = (f (x), cos ω x), where ω > 0, and the distance between any two adjacent symmetry axes of M ⊥ n, f (x) is 3 π / 2 (1) Find the value of ω (2) Let α be the first quadrant and f (3 / 2 α + π / 2) = 23 / 26, and find the value of sin (α + π / 4) / cos (4 π + 2 α)
⊥ n, ⊥ n, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\/ 2 + sin (2x / 3
The radical of F X is known as 3sin (Wx + φ) - cos (Wx + φ) (W > 0,0)
(x) = (x) = (3) sin (ω x + φ) - cos (ω x + φ) = 2 {[(3 / 3) / 2] sin (ω x + φ) - (1 / 2) cos (ω x + φ)} = 2 [sin (π / 3) sin (ω x + φ) - cos (π / 3) cos (ω x + φ) - cos (π / 3) cos (ω x + φ + π / 3) f (- x) = - 2cos (ω x + φ + π / 3) because f (- x) = f (x), so cos (COS (COS) (because f (- x) = f (x), so cos (COS) cos (COS (COS) because f (- x) = f (x) = f (x) = f (x), so cos (COS) cos (COS) because f (- x) = f φ + π / 3 - ω x) = Co
The known function f (x) = √ 3sin (Wx + φ) - cos (Wx + φ) (0
F (x) = (x) = (x) = (x) = (x) = (2) [(3 / 2Sin (Wx + φ) - 1 / 2cos (Wx + φ)] = 2Sin (Wx + φ - π / 6) symmetry axis: let Wx + φ - π / 6 = k π + π / 2 (K ∈ z) solution: x = (K π + 2 π + π / 2 (K ∈ z) solution: x = (K π + 2 π / 3 - φ) / W, so the distance between two adjacent symmetry axes is π / | w | w | w | w | w | w | w | w | w | w | W F (x) is even function
If the vector a = (root 3coswx, sinwx) B (sinwx, 0), where w > 0, Let f (x) = (vector a + vector b) * vector B-1 / 2 if the image of function f (x) is the same as The straight line y = m is tangent, and the abscissa of the tangent point is an isochromatic sequence with tolerance π Find the expression of F (x) and the value of M When x ∈ (π / 2,7 π / 4), the abscissa of the intersection point of G (x) = cos α forms an equal proportion sequence, and the value of obtuse angle α is calculated
(1) F (x) = (a + b) * B-1 / 2 = a * B + b * B-1 / 2 = a * B + b * B ^ 2-1 / 2 = 2 = √ 3sin ω x * cos ω x + 0 + (sin ω x) ^ 2-1 / 2 / 2 = √ 3 / 2 * sin (2 ω x) - 1 / 2 * cos (2 ω x) = sin (2 ω X - π / 6), from known, known, the maximum value of function is | M = 1, and the period T = 2 π / (2 ω) = π, therefore, ω = 1, and M = ± 1, so, so, function function function function function function function, so, therefore, function function function function function function function function function function function function function function function function function function function function function function f (x) =
The positive period of (cosinx) = (Wx) is the minimum of (Wx), which is a positive vector of (Wx) 1, find the value of W, 2, shift the image of function y = f (x) to the left by π / 12 units, and then extend the left and horizontal sides of the point on the obtained image to the original 4, and the ordinate remains unchanged, so as to obtain the image of function y = g (x), and find the monotone decreasing interval of function y = g (x)
1.cos(wx+π/2)=-sinwx
Then (cwsinx) = (3wsinx) = (- cwsinx) = (- cwsinx) = (- cwsinx) = (- cwsinx) = (- cwsinx) = (- cwsinx) = (- cwsinx) = (- cwsinx) = (- cwsinx) = (- cwsinx) = (- cwsinx) = (- cwsinx) = (- cwsinx
=(1-cos2wx)/2 + (√3/2)sin2wx
=(√3/2)sin2wx - (1/2)cos2wx + 1/2
=sin(2wx - π/6) + 1/2
F (x) minimum positive period = 2 π / 2W
The minimum positive period of ∵ f (x) is π
∴2π/2w=π
W=1
2. Shift π / 12 to the left in units: F (x) = sin (2x - π / 6) = sin [2 (x - π / 6)] = sin [2 (x - π / 6 + π / 12)] = sin [2 (x - π / 12)] = sin (2x - π / 6)
The horizontal left side of the point on the image is 4 times longer than the original: F (x) = sin [2 (x / 4) - π / 6] = sin (x / 2 - π / 6)
The decreasing interval of the function y = SiNx is [2K π + π / 2,2k π + 3 π / 2] (K ∈ z)
ν let X / 2 - π / 6 ∈ [2K π + π / 2,2k π + 3 π / 2] (K ∈ z)
x/2∈[2kπ+2π/3,2kπ+5π/3] （k∈Z）
x∈[4kπ+4π/3,4kπ+10π/3] （k∈Z）
The monotone decreasing interval of G (x) is: X ∈ [4K π + 4 π / 3,4k π + 10 π / 3] (K ∈ z)
Vector a = (coswx sinwx, sinwx), vector b = (- coswx sinwx, 3coswx) f (x) = vector a + vector B + y (question too long, supplement below) X ∈ R, f (x) symmetric about x = P, w ∈ (1 / 2,1) 1: Find the minimum positive period 2: Find the value of F (x) on (0,3p / 5) by (P / 4,0) 6p / 5.2: [- 1-radical 2,2-radical 2] Sorry, it should be F (x) = vector a * vector B + y Two questions are the range of values (- P / 4,0)
Please check the question: F (x) = vector a * vector B?
Let vector a = (2coswx, root 3), vector b = (sinwx, cos? Wx sin? Wx) (W > 0) function f (x) = vector a The distance between a symmetry center of F (x) image and its adjacent symmetry axis is π / 4 (1) Find the analytic formula of F (x) (2) In the acute triangle ABC, and f (a) = 0, B = π / 4, a = 2, find the angle C
Vector a = (2coswx, root 3), vector b = (sinwx, cos? Wx sin? Wx) (W > 0)
Function f (x) = vector a ● B
=2sinwxcoswx+√3(cos²wx-sin²wx）
=sin2wx+√3cos2wx
=2(1/2sin2wx+√3/2cos2wx)
=2sin(2wx+π/3)
The distance between a symmetry center of F (x) image and its adjacent symmetry axis is π / 4
Then t / 4 = π / 4, t = π
From 2 π / (2W) = π, w = 1
∴f(x)=2sin(2x+π/3)
(2)
∵f(A)=2sin(2A+π/3)=0
A is the inner angle of the triangle
Then 2A + π / 3 = π, a = π / 3
B = π / 4, a = 2,
∴C=π-A-B=π-π/3-π/4=5π/12
It is known that f (x) = sin 2wx + (radical 3 / 2) sin2wx - (1 / 2) (x ∈ R, w > 0) if the minimum positive period of F (x) is 2 π (1), find the expression of F (x) and monotonically increasing interval of f (x) (2) find the maximum and minimum values of F (x) in the interval [- (π / 6), (5 π / 6)]
(1) The minimum positive period of the minimum positive period of F (x) is 2 π, 2 π / (2W) = 2 π, w = w = w = 1 / 2 / 2, f (x) = 1 / 2 (1-cos2wx) + and √ 3 / 2sin2wx-1 / 2 / 2 sin2wx-1 / 2cos2wx = sin (2wx - π / 6) ∵ f (x) the minimum positive period of F (x) is 2 π, 2 π / (2W) = 2 π, w = w = 1 / 2 ɀ f (x) = sin (x - π / 6) from 2K π - π / 2 / 2 / 2, from 2K π - π / 2 / 2 / 2 / 2 / 2) from 2K π - π / 2 / 2 / 2 it is found that the ratio of X - π / 6 ≤ 2K π + π
The known function f (x) = Three 2sinωx−sin2ωx 2+1 The minimum positive period of 2 (ω > 0) is π (I) find the value of ω and the monotone increasing interval of function f (x); (II) when x ∈ [0, π 2] Find the value range of function f (x)
（Ⅰ）f(x)＝32sinωx−1−cosωx2+12=32sinωx+12cosωx=sin(ωx+π6)．… Because the minimum positive period of F (x) is π, so ω = 2 From 2K π - π 2 ≤ 2x + π 6 ≤ 2K π + π 2, K ∈ Z, K π - π 3 ≤ x ≤ K π is obtained
RELATED INFORMATIONS
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