Find the maximum value of the function y = sin2x * sin (π / 3-2x)

Find the maximum value of the function y = sin2x * sin (π / 3-2x)

The main purpose of this paper is to investigate the formula of sum difference product and product sum difference, because: cos (a + b) = cosacosb sinasinb; (1) cos (a-b) = cosacosb + sinasinb; (2) from (1) and (2), we can get: sinasinb = (COS (a-b) - cos (a + b)) / 2; (3) from (3), we can get: y = (COS (4x pi / 3) - cos (PI / 3)) / 2

The function y = sin (π / 3-2x) + sin 2x is transformed into a sine function, and the maximum value and minimum positive period of the function are obtained

Sin (π / 3-2x) = - cos2x, so y = sin2x-cos2x = √ 2 [sin2x / √ 2-cos2x / √ 2] = √ 2Sin (2x - π / 4),
The maximum value of this function is √ 2 and the minimum positive period is 2 π / 2 = π

The function y = sin (Π / 3-2x) + sin 2x is transformed into a sine function, and the maximum value and minimum period of the function are obtained
Do you use any formula? Do you add sin2x?

Using the formula: Sina + SINB = 2Sin [(a + b) / 2] cos [(a-b) / 2] y = sin (Π / 3-2x) + sin2x = 2Sin [(Π / 3-2x + 2x) / 2] cos [(Π / 3-2x-2x) / 2] = 2Sin Π / 6cos (Π / 6-2x) = cos (Π / 6-2x) = sin [Π / 2 - (Π / 6-2x)] = sin (2x + Π / 6) ymax = 1; Tmin = 2 Π / 2 = Π

Given the function f (x) = sin2x-2sin2x (I), find the minimum positive period of function f (x); (II) find the minimum value of function f (x) and the set of X when f (x) takes the minimum value

(I) because f (x) = sin2x - (1-cos2x) = 2Sin (2x + π 4) - 1, the minimum positive period of function f (x) is t = 2 π 2 = π (II) from (I), when 2x + π 4 = 2K π − π 2, that is, x = k π − π 8 (K ∈ z), the minimum value of F (x) is − 2 − 1; therefore, when f (x) takes the minimum value, the set of X is: {x | x = k π - π 8, K ∈ Z}

If cos2x = - 3 / 4, then sin ^ 4 + cos ^ 4=

cos2x=(cosx)^2 - (sinx)^2 = -3/4
also
(cosx)^2 + (sinx)^2 = 1
be
sin^4+cos^4
=[ ((cosx)^2 + (sinx)^2 )^2 + ((cosx)^2 - (sinx)^2 )^2 ] /2
=(1 + 9/16)/2
=25/32

The plane vector a = (COS θ, sin θ) is known,
The vector a = (COS θ, sin θ), B = (cosx, SiNx), C (sin θ, - cos θ) where 0 〈θ 〈π, and the function f (x) = (a · b) cosx + (B · C) SiNx image passing point (π / 6,1)
(1) I find f (x) = cos (2x - θ), θ = 60 degrees, but I don't know how to find the value of θ
(2) If the abscissa of each point on the image of function y = f (x) is changed to twice of the original, and the ordinate remains unchanged, the image of function y = g (x) will be obtained, and the maximum and minimum values of y = g (x) on [0, π / 2] will be calculated

(1) If f (x) = cos (2x - θ) is right, then the point (π / 6,1) in F (x) = cos (2x - θ) is: 1 = cos (2 * π / 6 - θ). That is, cos (π / 3 - θ) = cos0 °, π / 3 - θ = 0. θ = π / 3.. θ = 60 ° (2)..., which changes the abscissa of function f (x) = cos (2x - θ) to twice the original

The range of function y = 2Sin (x - π / 6) (x ∈ [0, π])
Request details

Because - Π / 6=

The function y = 2Sin (x + π / 6), X ∈ [0, π / 2], and the range is

x∈【0,π/2】
x+π/6∈【π/6,2π/3】
sin(x+π/6)∈【1/2,1】
2sin(x+π/6)∈【1,2】
Range = [1,2]

1. Let f (x) = x ^ 2 + 1, G (x) = f [f (x)], f (x) = g (x) - AF (x), ask whether there is a real number a, so that f (x) is a decreasing function in the interval (- 1,0) and an increasing function in the interval (- 1,0)?

f(x)=x^2+1,
g(x)=f(x^2+1)=(x^2+1)^2+1=x^4+2x^2+2,
F(x)=(x^4+2x^2+2)-a(x^2+1)=x^4+(2-a)x^2+(2-a)
=[x^2+(1-a/2)]^2+(1-a^2/4).
x

1. If f (x) = x ^ 2 + (LGA + 2) + LGB, and f (- 1) = - 2, and f (x) ≥ 2x holds for all real numbers, then a=_ ,b=_.
2. Calculation: LG (radical 3 + radical 5 + radical 3 - radical 5)
3. Given AB > 0, a ^ 2-2ab-9b ^ 2 = 0, find the value of LG (a ^ 2 + ab-6b ^ 2) - LG (a ^ 2 + 4AB + 15b ^ 2)

1.
Because f (- 1) = - 2
So - 2 = 1-lga-2 + LGB
Then: LGA LGB = 1
A / b = 10
Also: for all real numbers x, f (x) > = 2x,
So f (x) - 2x
=x2+(lga)x+lgb
>=0 established
Therefore, the discriminant is as follows
(lga)^2-4lgb