The minimum positive period of the function y = cos ^ 2 (2x + π / 3) - Sin ^ 2 (2x - π / 3) is

The minimum positive period of the function y = cos ^ 2 (2x + π / 3) - Sin ^ 2 (2x - π / 3) is

It's lazy enough
cos^2(2x+π/3)=[cos(4x+2π/3)-1]/2
sin^2(2x-π/3)=[1-cos(4x-2π/3)]/2
Take Huajian, don't be so lazy!

The minimum value of the function y = √ 3sinx-cosx-1 is Hurry!

y=√3sinx-cosx-1
=2（√3/2sinx-1/2cosx）-1
=2（cosπ/6sinx-sinπ/6cosx）-1
=2sin（x-π/6）-1
Because - 1 ≤ sin (x - π / 6) ≤ 1
So the maximum value of Y is 1 and the minimum value is - 3

Simplify the following functions and find the minimum positive period and the maximum minimum (1) f (x) = √ 3sinx / 2 + cosx / 2 （2）f（x）=√3cos2x-sin2x

(1) F (x) = √ 3sinx / 2 + cosx / 2 = 2 × (√ 3 / 2sinx / 2 + 1 / 2cosx / 2) = 2Sin (x / 2 + π / 6); minimum positive period T = 2 π × (1 / 2) = 4 π; maximum value = 2; minimum value = - 2; (2) f (x) = √ 3cos2x-sin2x = 2 × (√ 3 / 2cos2x-1 / 2sin2x) = 2cos (2x + π / 3); minimum positive period T = 2 π ×

Find the maximum and minimum value of the function y = root 3sinx + cosx

y=2(sinx*√3/2+cosx*1/2)
=2(sinxcosπ/6+cosxsinπ/6)
=2sin(x+π/6)
sin(x+π/6)∈[-1,1]
So max = 2, min = - 2

Given the function y = 1 / 2 cosx (cosx + Radix 3 SiNx) + 1, and the image crossing point P (PAI / 6,7 / 4) of the function: when y takes the maximum value, find the set of independent variable x

F (Π / 6) = 7 / 4, i.e. 1 / 2 * a * cos Π / 6 (COS Π / 6 + √ 3 sin Π / 6) + 1 = 7 / 4, and a = 1 is obtained
Therefore, after dissolving, f (x) = 1 / 2 sin (2x + Π / 6) + 5 / 4
If y is the maximum value (7 / 4), then sin (2x + Π / 6) = 1
That is, 2x + Π / 6 = 2K Π + Π / 2, x = k Π + Π / 6 (K ∈ z) is obtained

The minimum positive period of the function y = | cosx |

The period of y = cosx is 2 π, but due to the symmetry of absolute value, the period becomes π!

Given the vector a = (2sinx / 2, 3 + 1 under the root), vector b = (cosx / 2-3sinx under the root sign, f (x) = vector a, vector B + M 1. Finding the monotone interval of F (x) on [0,2 school] 2. When x belongs to [0, PAI-2], the minimum value of F (x) is 2. Find the value set of X if f (x) is greater than or equal to 2 3. If there are real numbers a, B, C, such that a [f (x) - M] + B [f (x-C) - M] = 1, for any x, it is r-invariant. Then we find the value of bcosc

Vector a = (2sinx / 2, 3 + 1 under root), B = (cosx / 2-3sinx / 2,1) f (x) = a ● B + M = 2sinx / 2 (cosx / 2 - √ 3sinx / 2) + 3 + 3 + 1 + M = 2sinx / 2cox / 2-2-2 √ 3sinx / 2 + (3 + 2 + 3 + 1 + 3 + 1 + 3 + 1 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 1 + 1 + M = 2 (1 / 2 SiNx + 3 / 2cosx) + 1 + 1 + 1 + 1 + 1 + 1 + 1 + M = 2 (1 / 2 SiNx + 3 / 2cosx) + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + M = 2Sin (x + π /...)

The known function f (x) = sin (x + π) 6)−cos(x+π 3)+cosx， (I) find the minimum positive period of function f (x) and write out all monotone decreasing intervals; (II) if x ∈ [− π) 2，π 2] Find the maximum value m and minimum value m of function f (x)

(I) f (x) = sin (x + π 6) - cos (x + π 3) + cosx = 32sinx + 12cosx - (12cosx-32sinx) + cosx = 3sinx + cosx = 2Sin (x + π 6), ∵ ω = 1, ∵ t = 2 π, let 2K π + π 2 ≤ x + π 6 ≤ 2K π + 3 π 2, then 2K π + π 3 ≤ x ≤ 2K π + 4 π 3, then the monotone decreasing interval of the function is as follows

The known function f (x) = (2 √ 2) cosx / [cos (x / 2) - sin (x / 2)] Finding monotone interval of function How is the first step drawn out? I don't understand~

Therefore, f (x) = 2 √ 2 [cos (x / 2) - sin (x / 2)] so f (x) = 2 √ 2 [cos (x / 2) - sin (x / 2)] / [cos (x / 2) - sin (x / 2)] = 2 √ 2 [cos (x / 2) + sin (x / 2)] = 2 √ 2 [cos (x / 2) + sin (x / 2)] = 2 √ 2 [2 [cos (x / 2) + sin (x / 2 + π / 4)] = 4sin (x / 2 + 2 + π / 4) SiNx increasing interval is (2k π - π / 2,2k π + π + π + π π + π + π + π π π / 2,2k π + π + π/ 2) minus interval

The known function FX = [cosx + cos (π / 2-x)] [cosx + sin (π + x)] Finding the minimum positive period of function FX If 0

f(x_ =So the angle of α - sin2 / S is α - sin2 / S = α - sin2 / S = α - sin2 / S = α - sin2 / S = α - sin2 / S = α - sin2 / S = α - sin2 / S = α - sin2 / S = α - sin2 / S = α - sin2 = α - sin2 / S = α - SiNx = α - sin2 / S = α - sin2 / S = α - sin2 = α - sin2 / S = α - sin2 / S = α - SiNx = α = sin2 / S = α = sin2 / S = α = sin2 / S = α = SiNx = 2