If Cotx / (2cosx + 1) = 1, the value of cos2x / (1 + sin2x) is

If Cotx / (2cosx + 1) = 1, the value of cos2x / (1 + sin2x) is

Cotx / (2cosx + 1) = 1cosx / (2sinxcosx + SiNx) = 1cosx / (sin2x + SiNx) = 1cosx = sin2x + sinxcosx SiNx = sin2x (cosx SiNx) ^ 2 = (sin2x) ^ 21-sin2x = (sin2x) ^ 2sin2x = (- 1 + √ 5) / 2 calculate the value of cos2x, and it will be good for us

Is sin2x equal to 2sinx? Is cos2x equal to 2cosx?

sin2x=2sinx*cosx;
cos2x=cosx*cosx-sinx*sinx

Let f (x) = sin (2x + φ), where φ is a real number, if f (x) ≤| f (π / 6) | is r constant for X, and f (π / 2) > F (π), Then the monotone interval of FX is? Tell me the first step, because f (x) ≤| f (π / 6) | is r constant for X, so | f (π / 6) | is 1. Why

The known function f (x) = sin (2x + φ), the known function f (x) = sin (2x + φ), where φ is a real number, if f (x) ≤| f (f (π / 6) | | for X ∈ r holds, and f (π / 2) > F (π (π / 2) > F (π), then the monotone increasing interval of F (x) is analytic: ∵ function f (x) = sin (2x + φ), f (x) function f (x) = sin (2x + φ), f (x) ≤ | f (π / 6) | for X ∈ R | f (x) x (x) take the maximum value of x = π / 6) | f (f (f (π / it's a good idea

(2011 · Anhui) known function f (x) = sin (2x + φ), where φ is a real number, if f (x) ≤| f (π) 6) | holds for X ∈ R, and f (π) 2) If f (π), then the monotone increasing interval of F (x) is () A. [kπ-π 3，kπ+π 6]（k∈Z） B. [kπ，kπ+π 2]（k∈Z） C. [kπ+π 6，kπ+2π 3]（k∈Z） D. [kπ-π 2，kπ]（k∈Z）

If f (x) ≤| f (π 6) | for X ∈ r holds, then f (π 6) is equal to the maximum or minimum value of the function, that is 2 × π 6 + φ = k π + π 2, K ∈ Z, then φ = k π + π 6, K ∈ Z and f (π 2) > F (π 2) > F (π) is sin | 0, let k = - 1, at this time, φ = - 5 π 6, satisfy the conditions that 2x π6 ∈ [2K π - π 2, 2K π + π + π + π + π + π + π + π 2, 2K π + π + π + π, π, 2], K ∈

Let f (x) = sin (2x + φ), where φ is a real number if f (x) ≤| f (π) 6） | is constant for X ∈ R, and f (π 2) If f (π), then the value of F (0) is () A. −1 Two B. 1 Two C. Three Two D. − Three Two

If f (x) ≤| f (π)
6) | holds for X ∈ R,
Then f (π)
6) Equal to the maximum or minimum value of the function,
That is 2 × π
6+φ=kπ+π
2，k∈Z，
Then φ = k π + π
6，k∈Z，
F (π)
2) F (π), i.e. sin φ < 0,
Let k = - 1, and then φ = - 5 π
6, satisfying the condition of sin ﹤ 0
Then f (0) = sin (− 5 π)
6)=-1
2．
Therefore, a

Let f (x) = sin (2x + φ), where φ is a real number, and f (x) ≤ f (2 π / 9), all x ∈ r hold. Let P = f (2 π / 3) q = f (5 π / 6), If r = f (7 π / 6), then the relation between the large and small of P, Q, R is

Given the function f (x) = sin (2x + φ), where φ is a real number, and f (x) ≤ f (2 π / 9), all x ∈ r holds. Let P = f (2 π / 3), q = f (5 π / 6), R = f (7 π / 6), then the relationship between the large and small of P, Q, R is
Analysis: the ∵ function f (x) = sin (2x + φ), where φ is a real number and f (x) ≤ f (2 π / 9) all x ∈ r hold,
∴f(2π/9)=sin(4π/9+φ)=1==>4π/9+φ=π/2==>φ=π/18
∴f(x)=sin(2x+π/18)
F (x) takes the maximum value at x = 4 π / 18, the minimum value at x = 13 π / 18, and the maximum value at x = 22 π / 18
∴P=f(2π/3)

The known function f (x) = 3sinx•cosx+sin2x． (I) find the minimum positive period and monotone increasing interval of function f (x); (II) how can the image of function f (x) be obtained from the image of function y = sin2x?

(1) ∵ function f (x) =
3sinx•cosx+sin2x=
Three
2sin2x+1−cos2x
2=sin(2x−π
6)+1
Two
The minimum positive period of the function f (x) is π (5 points)
By 2K π - π
2≤2x−π
6≤2kπ+π
2(k∈Z)⇒kπ−π
6≤x≤kπ+π
3(k∈Z)，
The monotonic increasing interval of F (x) is [K π − π
6，kπ+π
3]，（k∈Z）．　… (8 points)
（Ⅱ）∵f(x)＝sin(2x−π
6)+1
2＝sin2(x−π
12)+1
2，
ν first shift π from the image of the function y = sin2x to the right
12 units, and then move the image up 1
The graph of function f (x) can be obtained by 2 units (12 points)

Given the function f (x) = 1 / 2Sin ^ 2x + √ 3sinx × cosx-1 / 2cos ^ 2x, find the minimum positive period of F (x) and the symmetry axis of function image

First, we use the formula of double angle and auxiliary angle
y=sin(2x-Pi/6)
Period T = Pi
Axis of symmetry: x = pi / 3 + K pi / 2

Given the function f (x) = 2Sin (x - π / 6) cosx + 2cos ^ 2x, find the monotone increasing interval of F (x)

2 (cosx) ^ 2 = 2 (sinxcospi / 6-cosxsinpi / 6) cosx + 2 (cosx) ^ 2 = 2 (sinxcospi / 6-cosxsinpi / 6) cosx + 2 (cosx) ^ 2 = √ 3sinxcos x + (cosx) ^ 2 = 2sinxcos x * √ 3 / 2 + (1 + cos2x / 2) / 2 = 3 / 2 * sin2x + 1 / 2 * cos2x + 1 / 2 = sin (2x + pi / 6) + 1 / 2, so a = 1, w = 2, φ = pi / 6, H = H = 1 = 1, w = 2, φ = Pi / 6, H = H = 1 = 1, w = 1, w = 2, φ = pi / 6, H = 1 = 1 / 2, so a = 1 1 / 2 T = 2pi / w = Pi, so - pi / 2 + 2kpi

Calculation: root 12 * root 3 - root 3 * root 75

=√36-√225
=6-15
=-9