What to do with the value range of the function Y sin2X cos2X

Solution: First convert it to a simpler form, and then evaluate the range.

y＝sin2x+cos2x

2)[(2)/2*sin2x+)[2)/2*cos2x](√2)(sin2xcos45°+cos2xsin45°)(2)sin(2x+45°)

1≤sin(2x+45°)≤1

The range of the original function of 2 ( 2)sin(2x+45°) 2 is [- 2, 2].

Use derivatives. y=sin2x+cos2x

y'=2cos2x-2sin2x

When y'=0.

Lingy'=2cos2x-2sin2x=0, solve the equation cos2x=sin2x, look at the image of the functions sin2x and cos2x, in a period [0, ], the solution of the equation is x1 = 8, x2 = 5 8

When x = 8, y = 2 under the root number

When x=5 8, y=- 2

Give the money.

sin2x＋cos2x

2(√2/2sin2x+√2/2cos2x)√2(sin2xcos45+cos2xsin45)√2sin(2x+45)

There is no range of x involved in the problem, so the range of sin(2x+45) is (-1,1), so the range of y sin2x+cos2x is.

You can multiply both sides by 2 2 at the same time, and then simplify it to 2 2*y=sin(2x+pi 4), and the value range is [- 2, 2].

y=√2（√2/2sin2x+√2/2cos2x)√2sin(2x+π/4）

sin(2x+π/4）∈【1,1】

So 2sin(2x+ 4) [2, 2] uses an auxiliary angle.

Simplify y= 2sin(2x+pi 4) and then evaluate the range.

After extracting 2, change the root number 2 to sin45cos45 and use the addition formula.

Summary. For the function y=cos(x-2), x is a real number, and the range of the cos function is [-1,1], so the range of cos(x-2) is also [-1,1].

cos(x-2).

For the function y=cos(x-2), x is the real number, and the range of the cos function is [-1,1], so the range of cos(x-2) is also [-1,1]. Brigade resistance.

Can you add, I don't quite understand it.

When x-2=0, cos(x-2)=cos(0)=1, the maximum value is 1. When x-2= is used, cos(x-2) = cos( )1, and the imaginary pose is taken to the minimum value of -1. Therefore, the range of cos(x-2) is [-1,1].

y=cos2x-sinx+1 =1-2(sinx) 2-sinx+1 =-2(sinx) 2-sinx+2 =-2[(sinx) 2+sinx]+2 =-2[(sinx)+1 2] 2+3 2 -1 mold cavity liquid sinx 1 When sinx=-1 2, y is the largest round denier, which is 3 2;When sinx=1, y is the smallest, which is -1 -1 y 3 2, that is, the value range is [-1,3 2].

Solution: 1. Xuntan sinx+cosx= 2(sinx* 2 2+cosx*2).

Because cosx = 2 2 and sinx = 2 2 then sinx+cosx= 2(sinxcos 4+cosxsin 4) = 2sin(x+4).

2、sinx＋cosx

2 ( 2 2 * sinx + 2 2 * cosx) 2 (sinxcos45 degrees + cosxsin45 degrees) 2sin (x+45 degrees).

sinx ten cosx universal formula:

sin = [2tan( 2)] 1+【tan( 2)】noisy state }cos = [1-tan( 2) ] 1+[tan( 2)] tan = [2tan( 2)] 1-[tan( 2)].

The method is as follows, please comma circle for reference:

If there is help from the landslide, please celebrate.

=1-2sins squared x +2sinx

2 (sinx-1 2) square + 3 2

3/2≤sinx-1/2≤1/2

9 2 -2 (sinx-1 2) square 0-3 -2 (sinx-1 2) square +3 2 3 2 2 2 value range [-3, 3 2].

=1-2sin²x-sinx

2(sin²+1/2sinx+1/16)+1/8+1=-2(sinx+1/4)²+9/8

Because -3 4 sinx+1 4 5 4

So 0 (sinx+1 4) 25 16 so-25 8 -2 (sinx+1 4) 0 so-2 -2 (sinx+1 4) +9 8 9 8 so the range of the function y=cos2x-sinx is [-2,9 8].

y=cos2x-sinx=y=(1-2(sinx)^2)-sinx

Let t=sinx,-1<=t<=1 delete.

y=-2*t 2+t+1 into a range of unary quadratic functions.

y=2sinx -cos2x = 2sinx -(1-2(sinx)^2)

Let t = sinx, then -1<=t <=1

Bring it in and get. y=2t -1 +2t 2 = 2t 2 +2t -1 The symmetry axis of this function is t=-1 2, and the minimum value obtained here is y= 2*1 4 -2 *1 2 -1 = 3 2

The maximum value is obtained at t=1 and y=3

So the range is [-3 2, 3].

cos2x+cosx+1=2(cosx)^2+cosx=2(cosx+1/4)^2-1/8

The range of cosx is [-1,1].

When cosx=-1 4 is activated, the minimum value is -1 8, and when cosx=1, the maximum value is 3

So the range is [[-1, 8,3].

Method 1: Separation Constant Method.

y=(cosx+2-2) (cosx+2)=1-2 (cosx+2) Because -1<=cosx<=1, 1<=cosx+2<=32 3<=2 (cosx+2)<=2

2<=-2/(cos+2)<=-2/3

Therefore, -1<=y<=1 3

That is, the range is: [-1,1 3].

Method 2: Take advantage of the boundedness of trigonometric functions.

y(cosx+2)=cosx

cosx=2y/(1-y)

Since -1<-cosx<=1, so.

1<=2y/(1-y)<=1

Solve the left inequality to get (y+1) (1-y)>=0,-1<=y<1 solve the right inequality to get (3y-1) (1-y)<=0,y<=1 3 or y>1 to get -1<=y<=1 3

That is, the range is: [-1,1 3].

Solution: y=cosx (cosx+2)=(cosx+2-2) (cosx+2)=1-2 (cosx+2).

Since cosx belongs to [-1,1], cosx+2 belongs to [1,3], so y=1-2 (cosx+2) belongs to [-1,1 3], i.e., the y=cosx (cosx+2) range is [-1,1 3].

Method 2: y(cosx+2)=cosx

1-y)cosx=2y

cosx=2y/(1-y)

Due to |cosx|<=1, so there is |2y/(1-y)|<=1 solution: -1<=y<=1 3

y=cosx/(cosx+2)=(cosx+2-2)/(cosx+2)=1-2/(cosx+2)

1<=cosx<=1

The value range is [1 3, 1 2].

Alternative: y=1 (1+2 cosx).

When cosx=-1, max=12

When cosx=1, min=13