Given that vector a = (SiNx, 3 / 2), vector b = (cosx, - 1). Find the value range of F (x) = (vector a + vector b) * vector B on [- π / 2,0]

Given that vector a = (SiNx, 3 / 2), vector b = (cosx, - 1). Find the value range of F (x) = (vector a + vector b) * vector B on [- π / 2,0]

By F (x) = (vector a + vector b) * vector B
=(sinx+cosx,3/2-1)*(cosx,-1)
=sinxcosx+(cosx)^2-1/2
=(1/2)sin2x+(1/2)cos2x+1/2-1/2
=(√2/2)sin(2x+π/4)
When x ∈ [- π / 2,0], 2x + π / 4 ∈ [- 3 π / 4, π / 4]
In other words, sin (2x + π / 4) ∈ [- 1, √ 2 / 2]
That is, the range is [- √ 2 / 2,1 / 2]

Given the vector a = (SiNx, 2 / 3), B = (cosx, - 1), find the value range of F (x) = (a + b) * B on [- π / 2,0]

If I'm not wrong, it should be [root of minus two two plus five sixths, Four Thirds]
The final function expression I worked out is: F (x) = root two * sin (2x + quartile) + 5 / 6

Given the vector a = (cosx + SiNx, 2sinx), B = (cosx SiNx, - cosx) f (x) = AB, find the minimum positive period of F (x)

f(x)=ab=(cosx+sinx,2sinx)(cosx-sinx,-cosx)
=cos²x-sin²x-sin2x
=cos2x-sin2x
=-√2sin(2x-π/4)
Minimum positive period = 2 π / 2 = π

Given vector M = (cosx, 2sinx), vector n = (2cosx, - SiNx), f (x) = vector m * vector n (1) Find the value of F (- 3009 / 3 π) (2) When the sum of [2x, 2x] = (2x, 2x) = (2x, 2x) = (2x) = (2x) = (2) = (2) = (1) = (2) = (2) = (2) = (2) = (2) = (2) = (2) = (2) = (2) = (2) Please give me the detailed process

∵ vector M = (cosx, 2sinx), vector n = (2cosx, - SiNx),
/ / F (x) = vector m * vector n
＝2cos^2x-2sin^2x
=2cos2x
(1)f(-3009/3π)=2cos（-2006π）=2cos2006π=2
(2）g(x)=1/2f(x)+sin2x
=cos2x+sin2x
=√2sin(2x+π/4）
∵0≤x≤π/2
∴π/4≤2x+π/4≤5π/4
∴√2/2≤sin(2x+π/4）≤1
∴1≤√2sin(2x+π/4）≤√2
So the maximum value of G (x) = 1 / 2F (x) + sin2x is √ 2
The minimum value is 1

Given the vector a = (2cosx, SiNx ^ 2), vector b = (2sinx, cosx ^ 2), find the maximum value of the function f (x) = / A / - / B /

Known vector a = (2cosx, SiNx), vector b = (2sinx, cos, x), vector b = (2sinx, cos, x), the function f (x) = the maximum value of the maximum value of ∣ a ∣ a ∣ a ∣ B ∣ B ∣ f (x) = ∣ a ∣ a ∣ a ∣ B ∣ B ∣ B ∣ B ∣ B ∣ we have known vector a = (2cosx, SiNx), vector b = (2sinx, cos ͆ x, vector b = (2sinx, cos, cos ∣ x) ∣ f (x) s

The vector m (SiNx, - cosx) n = (COSA, - Sina) where 0

F (x) = sinxcosa + cosxsina = sin (x + a), f (π) = sin (π + a) = - Sina = - 1, because 0f (c) = sin (c + π / 2) = cos (c) = 1 / 2, so C = π / 3, and SINB = 2sina = 2Sin (2 π / 3-B) = (root 3) CoSb + SINB, so CoSb = 0, so B = π / 2, so a = π / 6
Conclusion

For any x ∈ R, the quadratic function f (x) holds f (1-x) = f (1 + x). Let a = (SiNx, 2), B = (2sinx, 1 / 2), C = (cos2x, 1), d = (1,2) When x belongs to [0, π], find the solution set of inequality f (vector a times vector b) > F (vector c times vector D) What I want to ask is that f (1-x) = f (1 + x) can deduce that f (2-x) = f (x) is not? Then f (a * b) = f (2-cos2x) = f (cos2x)? In this way, f (cos2x) > F (cos2x + 2) Because of cos2x

a*b=2(sinx)²+1.

Given that the quadratic function f (x) = x ^ 2 + MX + n for any x belongs to R, f (x) = f (2 + x) holds. Let vector a = (SiNx, 2) vector b = (2sinx, 1 / 2), vector C = (cos2x, 1), vector D = (1,2) (1) find the monotone interval of function f (x). (2) when x belongs to [0, π], find the solution set of inequality f (vector a, vector b) > F (vector C, vector D) F (- x) = f (2 + x); not f (x) = f (2 + x)

Axis of symmetry x = 1
a.b=2-cos2x,c.d=2+cos2x
The absolute value of 2-cos2x-1 is greater than the absolute value of 2 + cos2x-1
Again - 1

Let f (x) = f (2-x) hold for any ∈ R, let a = (SiNx, 2), B = (2sinx, 1 / 2) C = (sin? X, 3), d = (- 2,1). Find the solution set of the inequality f (a · b) > F (C · d)

Because of a / b = 2 * (SiNx) ^ 2 + 1 = 2-cos2xc · d = - 2 * (SiNx) ^ 2 + 3 = 2 + cos2x, therefore, f (a · b) = f (2-cos2x) = f (cos2x) f (cos2x) f (C · d) = f (2 + cos2x) = f (- cos2x) if f (x) = PX ^ 2 + QX + R, then f (cos2x) > F (- cos2x) = > cos2x > - cos2x, that is, cos2x > cos2x > cos2x > cos2x > 0 = = > - π / 2 / 2 = cos2x > 0 = = = > - π / 2 / 2 > 0 = = = > - π / 2 > 0 = > 0 = > 0 = > 0 = > + 2K π

Given the vector M = (cosx, - SiNx), n = (cosx, sinx-2 √ 3cosx), Let f (x) = m * n, X belong to R 1) If f (x) = 24 / 13 and X belongs to [π / 4, π / 2], find the value of sin2x

The minimum positive period of the function f (x) of the function f (x) f (x) = (cosx) ^ 2 - (SiNx) ^ 2 + 2 + 2 √ 3 SiNx cosx = cos2x + 2 + (6) 1) 1) function f (x) is the smallest positive period of the function f (x) is π 2) (x) = 24 / 13, and X belongs to [π / 4, π / 2] f (x) = 2Sin (2x + π / 6) = 24 / 13, sin (2x + π / 6) = 12 / 132x + π / 6 = arcsin12 / 13 = arcsin12 / 13 = arcsin12 / 13 = > 2x = arcsin12 = arcsin12 / 13 = = > 2x = arcsin12 = arcsin12 / 13 = > 2x = arcsin12