The function f (x) = 3sin (2x - π) 6) In the interval [0, π 2] The value range on is______ .

The function f (x) = 3sin (2x - π) 6) In the interval [0, π 2] The value range on is______ .

∵x∈[0，π
2]，
∴2x∈[0，π]，
∴2x-π
6∈[-π
6，5π
6]，
∴sin（2x-π
6）∈[-1
2，1]，
∴f（x）=3sin（2x-π
6）∈[-3
2，3]；
That is, f (x) is in the interval [0, π]
2] The value range on is [- 3
2，3]．
So the answer is: [-3
2，3]．

The function y = 3sin ^ 2x-4cosx + 1, X belongs to the range of [π / 3,2 π / 3]

y=3sin^2x-4cosx+1
y=3(1-cos^2x)-4cosx+1
=3-3cos^2x-4cosx+1
=-3cos^2x-4cosx+4
=-3(cosx-2/3)^2+16/3.
Because x belongs to [π / 3,2 π / 3],
So, - 1 / 2

Given the function f (x) = cos ^ 2 (x - π / 6) - 1 / 2cos2x, find the minimum positive period of function f (x) Given the function f (x) = cos ^ 2 (x-pai / 6) - 1 / 2cos2x, find the minimum positive period of function f (x) and the equation of symmetry axis of image (2) Find the value range of function f (x) on the interval [- Pai / 12, Pai / 2]

(1) f(x)=cos^2(x-π/6)-1/2cos2x
=[cos(2x-π/3)+1]/2-1/2cos2x
=1/2[1/2cos2x-√3/2sin2x]-1/2cos2x
=-1/2(1/2cos2x+√3/2sin2x)
=-1/2sin(2x+π/6)
Therefore, the minimum positive period is: T = π;
Because the symmetry axis of sin2x is: x = π / 4 + K π / 2
So the equation of symmetry axis of F (x) image is: x = (π / 4 - π / 6) + K π / 2 = π / 12 + K π / 2
(2) Because - π / 12 ≤ x ≤ π / 12
So - π / 6 ≤ 2x ≤ π / 6
So 0 ≤ 2x + π / 6 ≤ π / 3
According to the image of sine function, the range of F (x) is: [0, √ 3 / 2]

Given the function f (x) = √ 2cos (x - π / 12), X ∈ R is used to find the value of F (3 / π). If cos θ = 3 / 5, θ {(3 / 2 π, 2 π), find f (θ - π / 6) Cos θ = 3 / 5, θ (3 / 2 π, 2 π) how do we find sin θ?

I'm very happy for you 1. F (π / 3) = √ 2cos (x - π / 12) = √ 2cos (π / 3 - π / 12) = √ 2cos (π / 4) = 12. Because cos θ = 3 / 5, θ ﹤ 2 π, 2 π), sin θ = - 4 / 5 (according to (COS θ) ^ 2 + (sin θ) ^ 2 = 1, so sin θ has two values 4 / 5 and - 4 / 5, but according to the value range of θ

The function y = 2sinx+ 2cos（x+π 4) The maximum value of is () A. Six B. Two C. 2+ Two D. Ten

From the meaning of the title, y = 2sinx+
2cos（x+π
4）
=2sinx+
2（cosxcosπ
4-sinxsinπ
4）
=2sinx+cosx-sinx=sinx+cosx=
2sin(x+π
4)
When sin (x + π)
4) When = 1, the maximum value of the function y is 0
2，
Therefore, B

If the function f (x) = 3-2sinx-2cos? X, then the minimum value and maximum value of F (x) are

f(x)=3-2sinx-2cos²x
=3-2sinx-2(1-(sinx)^2)
=2(sinx)^2-2sinx+1
=2(sinx-(1/2))^2+(1/2)
And: - 1

Finding the minimum positive period of the function y = 1 / 2cos? X

y=1/2cos²x
Cos2x = 2cos? X-1 is known from the angle doubling formula
Then cos? X = (1 + cos2x) / 2
y=(1+cos2x)/4
So the minimum positive period is π

The period of function f (x) =2cos 2 (x- π /4) -1 is

According to cos2a = 2cos? A-1
f(x)=2cos²(x-π/4)-1
=cos2(x-π/4)
=sin(2x-π/2)
=-cos2x
∴ T=2π/2=π

The function y = 2cos2 (x − π) 4) The minimum positive period of − 1 is______ , parity is______ Function

f（x）=2cos2（x-π
4）-1=cos（2x-π
2) If = sin2x, then the function is odd with period T = π,
So the answer is: π; odd

The minimum positive period for the function y equal to 2cos squared minus 1 is

Y = 2 (COSA) ^ 2-1 = 2 (COSA) ^ 2 - [(COSA) ^ 2 + (Sina) ^ 2] = (COSA) ^ 2 - (Sina) ^ 2 = cos2a, so the minimum positive period T = 2 π △ 2 = π