If the function y = 2cos (2x + φ) is odd and is an increasing function on (0, π / 4), then the real number φ may be () A. - π / 2 b.0 C. π / 2 d. π

If the function y = 2cos (2x + φ) is odd and is an increasing function on (0, π / 4), then the real number φ may be () A. - π / 2 b.0 C. π / 2 d. π

Y = 2cos (2x + φ) is an odd function, so y (0) = 0, φ = k π + π / 2, K belongs to Z odd function, and it is an increasing function on (0, π / 4), so it is an increasing function on (- π / 4,0); so it is an increasing function on (- π / 4,) and an increasing interval of cosine function on (- π / 4,) ((2k-1) π, 2K π), so (- π / 4) x2 + φ > = (2k-1) π

It is known that there exists a real number W, Fai (where W is not equal to 0, belonging to Z) such that f (x) = 2cos (Wx + FAI) is an odd function and an increasing function on (0, π / 4)
1) Guess the values of W and Fai of the two groups and verify that they are consistent with the meaning of the question
2) Find out all the W and Fai that fit the meaning of the question

F (x) = 2cos (Wx + FAI) is an odd function, f (x) = 2cos (Wx + FAI) = - f (- x) = - 2cos (- Wx + FAI) so cos (Wx) cos (FAI) = 0so Fai = k π + π / 2F (x) = 2cos (Wx + K π + π / 2) is an increasing function on (0, π / 4), f (0) = 0, so K is even, that is, f (x) = - 2Sin (Wx) from 0 to right is not

Trigonometric function, f (x) = 2cos (Wx + φ) + B, for any real number xdouyou f (x + π, 4) = f (- x)
Process, thank you
wrote it wrong
Trigonometric function, given that f (x) = 2cos (Wx + φ) + B, for any real number x, f (x + π / 4) = f (- x) holds, and (π / 8) = - 1, then the value of real number B is?

From F (x + π / 4) = f (- x), 2cos (Wx + W π / 4 + φ) + B = 2cos (- Wx + φ) + B
That is, Wx + W π / 4 + φ = - Wx + φ or Wx + W π / 4 + φ = Wx - φ
That is, 2wx = - w π / 4 (rounding off) or W π / 4 = - 2 φ
F (π / 8) = - 1 = 2cos (w π / 8 + φ) + B
Substituting 2cos0 + B = - 1
The solution is b = - 3

It is known that the function f (x) = sin2xcosfai-2cos ^ 2xsin (x-fai) - cos (π / 2 + FAI) (- π / 2 < Fai < π / 2) has the maximum value when x = π / 6
Finding the value of FAI
The abscissa of each point on the image of function y = f (x) is expanded to 2 times of the original, and the ordinate remains unchanged. The image of function y = g (x) is obtained. If G (α) = 1 / 3, α∈ (- π / 2,0), the value of cos α is obtained

Sin (π - φ) - otherwise it can't be calculated
f(x)=sin2xcosφ-2cos^2xsin(π-φ)-cos(π/2+φ)（-π/2＜φ＜π/2）
=sin2xcosφ-sinφ(2cos^2x-1)
=sin2xcosφ-cos2xsinφ
=sin(2x-φ)
When x = π / 6, the maximum value is obtained (- π / 2 ＜φ＜π / 2)
π / 3 - φ = π / 2 + 2K π, K is an integer
φ=-π/6
g(x)=sin(x+π/6)
cos(a)=sin(a+π/6)=1/3
α∈（-π/2,0）
α+π/6∈（-π/3,π/6)
cos(α+π/6)=√(1-1/9)=2√2/3
cosα=cos((α+π/6)-α)
=cos(α+π/6)cosπ/6+sin(α+π/6)sinπ/6
=(2√6+1)/6