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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) (2) from (1) we know that f (x) = - sin (2x - π 6) + 12 2 12 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\3
Given the vector a = 2 (COS α x, cos α x), vector b = (COS α x, Radix 3sin α x) (0 < α < = 1), the function f (x) = vector a * vector B If the line x = π / 6 is a symmetric axis of the image of function f (x) (1) Try to find the value of α (2) write how f (x) is transformed from y = SiNx. (3) find the minimum value of function f (x) in [0, π / 2] and the corresponding value of X Online, etc
(1) F (x) = 2cos ^ 2aX + 2 radical 3sinax cosax = 1 + cos2ax + 3sin2ax = 1 + 2Sin (2aX + π / 6)
If the line x = π / 6 is a symmetric axis of the image of function f (x), then there is sin (a π / 3 + π / 6) = 1 or - 1
Because 0 (2) f (x) = 1 + 2Sin (2x + π / 6)
Y = SiNx shifts one π / 6 units to the left, the abscissa becomes 1 / 2 of the original, the ordinate remains unchanged, the abscissa remains unchanged, the ordinate becomes twice the original, and then moves up one unit
（3）x∈[0,π/2],2x+π/6∈[π/6,7π/6],2sin(2x+π/6)∈[-1,2]
F (x) max = 3, where x = π / 6
F (x) min = 0, where x = π / 2
Vector M = (- 1, coswx + Radix 3sinwx), vector n = (f (x), coswx), where w > 0, vector m, vertical vector n, f (x) the distance between any two adjacent symmetric axes Vector M = (- 1, coswx + Radix 3sinwx), vector n = (f (x), coswx), where w > 0, vector m, vertical vector n, f (x) the distance between any two adjacent symmetry axes is 3 / 2, (1) find the value of W (2) let a be the first quadrant angle, and f (3a / 2 + Pai / 2) = 23 / 26, find the value of [sin (a + Pai / 4)] / [cos (4 Pai + 2a)] I calculated w = 2 / 3,
Creepy, let's work out w = 1 / 3
m⊥n,
∴0=m*n=-f(x)+coswx[coswx+√3sinwx],
∴f(x)=(coswx)^2+√3sinwxcoswx
=(1/2)[1+cos2wx+√3sin2wx]
=1/2+sin(2wx+π/6),
(1) The distance between any two adjacent symmetry axes of F (x) image is 3 π / 2,
∴2π/（2w)=3π,w=1/3.
(2)f(x)=1/2+sin(2x/3+π/6),
f(3a/2+π/2)=1/2+sin(a+π/2)=23/26,
cosa=5/13,
A is the angle of the first quadrant,
∴sina=12/13,
sin(α+π/4)/cos(4π+2α)
=(sina+cosa)/[√2cos2a]
=1/[√2（cosa-sina)]
=(-13√2）/14.
Given the vector M = (√ 3sin ω x, 0), n = (COS ω x, - sin ω x) (ω > 0), in the image of function f (x) = m * (M + n) + T, the minimum distance between the symmetry center and the symmetry axis is π / 4; and when x ∈ [0, π / 3], the maximum value of F (x) is 3 / 2 (1) Find the analytic expression of F (x); (2) Find the monotone increasing interval of F (x)
1) m+n=(√3sinωx+cosωx,-sinωx)
m*(m+n)=3sin^2(ωx)+√3sin2ωx/2
=3*(1-cos2ωx)/2+)+√3sin2ωx/2
=√3sin2ωx/2-3cos2ωx/2 +3
=√3sin(2ωx-π/3)+3
f(x)=√3sin(2ωx-π/3)+3+t
The minimum distance from the center of symmetry to the axis of symmetry is π / 4, indicating that the period is π
ω=1
f(x)=√3sin(2x-π/3)+3+t
When x ∈ [0, π / 3]
-π/3
The known function f (x) = √ 3sin (ω x + φ) - cos (ω x + φ) (0
It is known that the function f (x) = (√ 3) sin (ω x + φ) - cos (ω x + φ) (o0) is an even function and the distance between two adjacent symmetry axes of the image of function y = f (x) is π / 2
Let f (x) = √ 3sin (ω x + φ) - cos (ω x + φ) (o0) be even functions To find the value of F (Π / 8), another question is: after the image of function y = f (x) is shifted to the right by Π / 6 units, the image of function y = g (x) is obtained, and the monotone decreasing interval of G (x) is obtained,
f(x)=√3sin(ωx+φ)-cos(ωx+φ)=2sin(ωx+φ-∏/6).
If the distance between two adjacent symmetry axes is Π / 2, then the half period of the function is Π / 2, and the period is Π, then ω = 2
Function is even function, and o
Vector M = (sinwx + coswx, root 3coswx) (W > 0), n = (coswx sinwx, 2sinwx). Function f (x) = m * n + T If the distance between two adjacent symmetry axes on the image is 3 π / 2, and when x ∈ [0, π], the minimum value of F (x) is 0. (1) find the expression of function f (x), (2) in △ ABC, if f (c) = 1 and 2Sin? B = CoSb + cos (A-C), find the value of sina
(1) In the process of simplification, we only need to pay attention to two points: one is the application of double angle formula, the other is the application of trigonometric sum formula. Finally, t, X are determined according to the minimum value of F and symmetry axis
(2) First substitute f to find C, then according to the given formula and a = π - B-C, we can solve B, and then find a
Vector a = (√ 3coswx, sinwx), B = (sinwx, 0), where w belongs to (- 1 / 2,5 / 2), the function f (x) = (a + b) * B-1 / 2, and f (x) is about the straight line x = π / 3 pairs 1. Find the analytic formula of function f (x) What should I do after I have calculated sin (2wx - π / 6)
Sin (2wx - π / 6) from here we can see that the image of F (x) is a sine curve. If you think about the image, there are countless symmetry axes of any sine curve. If you take any one of them, you will find that this line passes through the peak or trough of the sine curve
Given that vector a = (root 3, coswx), vector b = (sinwx, 1), function f (x) = vector a * vector B, and the minimum positive period is 4 π (2) Let a, B belong to [π / 2, π], f (2A - π / 3) = 6 / 5, f (2B + 2, π / 3) = - 24 / 13, find the value of sin (a + b)? (3) if x belongs to [- π, π], find the value range of function f (x)?
F (x) = (radical 3, coswx) * (sinwx, 1) = radical 3 * sinwx + coswx * 1 = 2 [((radical 3) / 2) * sinwx + (1 / 2) coswx] = 2Sin (Wx + pi / 6) (1) 2 * pi / W = 4 * PI w = 1 / 2 F (x) = 2Sin (1 / 2x + pi / 6) (2) introduce x = 2A - π / 3 into f (x) with sina = 3 / 5; X = 2B + 2 π / 3, with introgression, with cos
Vector M = (sinwx + coswx, root 3coswx), vector n = (coswx sinwx, 2sinwx), w > 0, Let f (x) = m ` n, f (x) The distance between two adjacent axes of symmetry is equal to pi / 2, 1. Find the analytic formula of function f (x) 2. In the triangle ABC, a, B, C are the opposite sides of a, B and C respectively, B + C = 4, f (a) = 1, find the maximum area of the triangle Be sure to be detailed thank you
f(x)=(cos²wx-sin²x)+2√3coswxsinwx
=cos(2wx)+√3sin(2wx)
=2sin(2wx+π/6)
1, the distance between two adjacent symmetry axes is π / 2,
It is shown that the minimum positive period of F (x) t = 2 × (π / 2) = π
And T = 2 π / (2W) = π / W, so w = 1
Then f (x) = 2Sin (2x + π / 6)
2,f(A)=2sin(2A+π/6)=1
So sin (2a + π / 6) = 1 / 2
And 0
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