Let a = (cosx / 2, Tan (x / 2 + π / 4)), B = (√ 2Sin (x / 2 + π / 4), Tan (x / 2 - π / 4)), Let f (x) = ab Find the maximum value and minimum positive period of function f (x), and write the monotone interval of F (x) on [0, π]

Let a = (cosx / 2, Tan (x / 2 + π / 4)), B = (√ 2Sin (x / 2 + π / 4), Tan (x / 2 - π / 4)), Let f (x) = ab Find the maximum value and minimum positive period of function f (x), and write the monotone interval of F (x) on [0, π]

(x) = 2cosx / 2 × ((x / 2 + π / 4) + Tan (x / 2 + π / 4) + Tan (x / 2 + π / 4) × Tan (x / 2 - π / 4)) = √ 2 [sin (x + π / 4) + sin (π / 4)) + [1 + Tan (x / 2)] / [1-tan (x / 2)] / [1-tan (x / 2)] / [Tan (x / 2)] / [1 + Tan (x / 2)] = √ 2Sin (x + π / 4) maximum value = √ 2, minimum positive cycle = 2 π, 2 π = 2 π = 2 π 2 π = 2 π = 2 π = 2 π = 2 π = 2 π = 2 π = 2 π = 2 π the increasing region of SiNx

Given vector a = (radical 3sin ω x, cos ω x) B = (COS ω x, - cos ω x) (ω > 0), function f (x) = vector a dot product vector B, and the minimum positive period of function f (x) is π (1) When x ∈ [0,2 π], find the monotone increasing interval of function f (x) (2) In the triangle ABC, the opposite sides of angles a, B and C are a, B, C respectively. If the square of B = AC, find the value range of F (b)

(2) 1 / 2 * (2 √ 3 sinwx-2cos (2) Wx + 1-1) = 1 / 2 * 1 / 2 * (√ 3sin2wx-cos2wx) - 1 / 2 = (3 / 2 * sin2wx-cos2wx) - 1 / 2 = √ 3 / 2 * sin2wx-1 / 2 * cos2wx-1 / 2 = sin (2wx - π / 6) - 1 / 2 = sin (2wx - π / 6) - 1 / 2, the minimum positive period is π (2 π / (2WW) = π w = 1 / w = π w = 1 / w = 1 / F (x) = sin (2x - 2x - 2x - 2x - 2WW) = π w = π w = π w = 1, f (x) = 1 / F (x) = sin2xπ / 6) - 1 / 2 Ling -

Given that ω > 0, vector M = (√ 3sin ω x, cos ω x), vector n = (COS ω x, - cos ω x), and f (x) = m · n + 1 / 2 And the minimum positive period π of F (x) = m · n + 1 / 2 (1) Find the analytic expression of F (x); (2) It is known that a, B, C are the edges of △ ABC inner angles a, B, C respectively, and a = √ 19, C = 3, and cosa is exactly the minimum value of F (x) on [π / 12,2 π / 3], and find the areas of B and △ ABC

⑴f(x)=m•n+1/2=√3sinωxcosωx-cos²ωx+1/2=√3/2•sin2ωx-1/2•cos2ωx-1/2+1/2=sin(2ωx-π/6),∵ω>0,∴T=π=2π/2ω => ω=1,∴f(x)=sin(2x-π/6)；----------------------------------...

Given vector M = (2 √ 3sin (x / 4), 2), vector n = (COS (x / 4), cos ^ 2 (x / 4)), function f (x) = vector m × vector n 1. Finding the minimum positive period of function f (x) 2. If f (a) = 2, find the value of COS (a + π / 3)

The minimum positive period of the minimum positive period of the minimum positive period of the 1 f (x) 3 sin (x / 2) + cos (x / 2) + 1 = 2Sin (x / 2) + cos (x / 2) + 1 = 2Sin (x / 2 + π / 6) + 1, f (x) t = 2 π / w = 4 π, 2, f (a) = 2 sin (A / 2 + π / 6) + 1 = 1 = 2, find a = 4K π or 4 / 3 π + 4 / 3 π + 4K π (k belongs to z) cos (a + π / π + 4K π (k belongs to Z) cos (a + π / π / 1 + 4 K belongs to Z) cos (a + π / π / π / A + π / π / 4 3) =

We know two nonzero vectors M = (√ 3sin ω x, cos ω x), vector n = (COS ω x, cos ω x), ω > 0 (1) When ω = 2, X belongs to (0, π), the vector m is collinear with N, and the value of X is calculated (2) If the distance between the image of the function f (x) = vector m multiplied by vector N and any two adjacent intersections of the line y = 1 / 2 is π / 2; 1. When f (A / 2 + π / 24) = 1 / 2 + √ 2 / 6, a ∈ (0, π), find the value of cos2a 2. Let g (x) = SiNx multiply cosx / F (x / 2 + π /) - 1 / 2, X ∈ [0, π / 2], try to find the value range of function g (x)

We know two nonzero vectors M = (√ 3sin ω x, cos ω x), vector n = (√ 3sin ω x, cos ω x), ω > 0. (1) when ω = 2, X belongs to (0, π), vector m and N are collinear, find the value of x (2) if you function f (x) = vector m, the distance between the image multiplied by vector N and any two adjacent intersection points of the line y = 1 / 2 is π / 2

The square of an axis of symmetry of the image of the function y = sin (3x + π / 3) cos (x - π / 6) + cos (3x + π / 3) cos (x + π / 3)

cos(x+π/3)
=sin[π/2-(x+π/3)]
=sin(π/6-x)
=-sin(x-π/6)
So y = sin (3x + π / 3) cos (x - π / 6) - cos (3x + π / 3) sin (x - π / 6)
=sin[(3x+π/3)-(x-π/6)]
=sin(2x+π/2)
=cos2x
The axis of symmetry of cosx is the place where x = k π
So here is 2x = k π
x=kπ/2

What is the equation of an axis of symmetry of the image of the function y = sin (3x + π / 3) cos (x - π / 6) - cos (3x + π / 3) sin (x - π / 6)?

The solution consists of y = sin (3x + π / 3) cos (x - π / 6) - cos (3x + π / 3) sin (x - π / 6)
=sin[(3x+π/3)-(x-π/6)]
=sin(2x+π/2)
=cos2x
Let 2x = k π, K belong to Z
Therefore, the symmetric axis equation x = k π / 2, K belongs to Z
Therefore, a symmetric axis equation of the function is x = 0

A symmetric axis equation of the image of the function y = sin (3x - π / 2) - 1 is a.x = π / 6 b.x = π / 3 c.x = π / 2 D.X = 3 π / 2

y=sin(3x-π/2)-1
3x-π/2=2kπ+π/2
3x=2kπ+π
x=2kπ+π/3
When k = 0, x = π / 3 is one of the symmetry axes
So choose B

The symmetric axis equation of the function y = sin (3x + π / 4) is?

The axis of symmetry of sine function is the maximum or minimum value of X
That is, y = ± 1
Then 3x + π / 4 = k π + π / 2
3x=kπ+π/4
X = k π / 3 + π / 12, K is an integer

A symmetric axis equation of the image of the function y = sin (x / 2) + √ 3 · cos (x / 2) is () a.x = 11 / 3 π b.x = 5 π / 3 c.x = - 5 π / 3 A.x=11/3π B.x=5π/3 C.x=-5π/3 D.x=-π/3

C.
First, y = 2Sin (x / 2 + π / 3)
Let (x / 2 + π / 3) = k π + π / 2
Where x = - 5 π / 3 satisfies the equation