Function y = cos ² Pie x-sin ² Period of pie x

Function y = cos ² Pie x-sin ² Period of pie x

y=cos ² πx-sin ² πx
=cos(2πx)
So t = 2 π / 2 π = 1

Function f (x) = cos ² x+2sinxcosx—sin ² What is the minimum positive period of X?

∵2sinxcosx=sin2x ,cos ² x—sin ² x=cos2x
∴f（x)=cos ² x+2sinxcosx—sin ² x
=sin2x+cos2x
=√2(√2/2sin2x+√2/2cos2x)
=√2sin(2x+π/4)
The minimum positive period of F (x) t = 2 π / 2 = π
I hope it will help you

Function y = Cos2 (x - π) 12)+sin2(x+π 12) The minimum positive period of - 1 is __

y=1
2[1+cos2（x-π
12]+1
2[1-cos2（x+π
12]-1=1
2[cos（2x-π
6）-cos（2x+π
6）]=sinπ
6•sinx=1
2sinx．T=π．
So the answer is: π

Known function f (x) = √ 3 (SIN) ² x-cos ² x)-2sinxcosx ① Find the minimum positive period of F (x) ② let x ∈ [- π / 3, π / 3] find the value range and monotonic increasing interval of F (x)

f(x)=-√3cos2x-sin2x=-2[(√3/2)cos2x+(1/2)sin2x]=-2cos(2x-π/6)
Minimum t = 2 π / 2 = π
x=π/12,f(π/12)=-2
x=π/3,f(π/3)=0
-2≤f(x)≤0
-π / 3 ≤ x ≤ π / 12 monotonically decreasing
π / 12 ≤ x ≤ π / 3 monotonically increases

Known function f (x) = √ 3 (SIN) ² x－cos ² x) - 2sinxcosx, (1) find the minimum positive period of F (x) (2) x ∈ {- π / 3, π / 3}, find the value range and monotonically increasing interval of F (x),

f(x)=√3(sin^2x-cos^2x)-2sinxcosx
=-√3cos2x-sin2x
=-2sin(2x+π/3)
1. Find the minimum positive period T = π
2. Let x ∈ [- π / 3, π / 3] find the value range and monotone interval of the function
-π/2

Known function FX = cos ² x-2sinxcosx-sin ² x Find the maximum and minimum values of the function on the interval [- Pai / 2,0]

f(x)=cos ² x-2sinxcosx-sin ² X = cos2x sin2x = root 2Sin (2x + 3 / 4 * π) x belongs to [- π / 2,0] 2x + 3 / 4 * π∈ [- 1 / 4 π, 3 / 4 * π], so the maximum value is taken as root 2 when 2x + 3 / 4 * π = 1 / 2 * π, and the minimum value is taken as -