Find the minimum positive period and parity of the function y 2cos 2 x 45 degrees 1

Since cos(2a)=cos2(a)-sin 2(a)=2cos2(a)-1, then 2cos2(a)=cos(2a)+1

So y=2cos2(x-45)-1=cos(2x-90)=sin(2x).

t=2k +2 2 (k is an integer) then k=0 is the minimum positive period.

Minimum positive period:

Let f(x)=sin(2x) then f(-x)=-sin(-2x)=-f(x).

So the function is odd.

Parity: Odd function (when judging parity, it is necessary to ensure that the defined domain is symmetrical with respect to the origin of the coordinate axis, and x is r in this question).

Solution: y=2cos 2(x-45°) 1

cos2(x-45°)

cos(2x-90°)

sin2x, so the minimum positive period t=2 2= , it is easy to know that y=sin2x is an odd function, so the original function is an odd function.

y=2cos2(x-45 degrees)-1

cos(2(x-45))

cos(2x-90)

sin2x minimum positive period:

Parity: Odd functions.

The answer can be found by using 2cos 2(a)-1=cos2a. The period is , odd function.

f(x)=y=2sin^2x+sin2x

1-cos2x+sin2x

1 + root number 2sin (2x-4).

The minimum positive period t=2 2=

The definition field is r, but f(-x) does not =f(x) nor =-f(x).

2 plus a quarter of non-odd and non-even.

By the double angle of the god file type.

y=cos)2x-π/2)

So y=sin2x

So t=2 2=

f(x)=sin2x

f(-x)=sin(-2x)=-sin2x=-f(x) defines the domain as r, and the symmetry of the original model is noisy.

So it's an odd function.

Summary. The function y=2cos( 3-x 2), find the minimum positive period of y.

Hello, this minimum positive period is 4 factions, because w is equal to one-half, see the figure below for specific operations

f(x)=2cos(2x+ 2)=-2sin(2x)period: t=2 blindness2=

Parity: f(-x)=-2sin(-2x)=2sin(2x)=-f(x).

Therefore, it is not known as a meganaqi function.

Single minus interval: because - 2+2k

y=2cos (x- 4)-1=cos(2x- laughing2)=cos( 2-2x)=sin2x

The function key grinding number y=2cos (x-4)-1.

Minimum positive period:

Be. Odd functions.

f(x)=4sin(6+x)=1-2x[1-cos(3+2x)]=1-2+2cos(3+2x).

2cos(π/3+2x)-1

f(x)=4sin ( 6+x) minimum positive period This type of question is used.

cos2x=cos²x-sin²x=2cos²x-1=1-2sin²x

y=2cos²(x-π/4)

cos[2(x-π/4)]+1-1

cos(2x-π/2)

cos( No difference 2

2x)sin(2x)

sin(-2x)=-sin(2x)

The number of piths is a function of odd milling.

Answer: Your input should be y=2cos (x- 4)-1, then y=2cos (x- 4)-1

cos(2x-π/2)

sin2x period t=2 2=

is an odd function.

The x-factor is 2 so the minimum period t=2 2=

It is defined as an even function according to the even function.