Function y = cosx-sin2x-cos2x + 7 The maximum value of 4 is () A. 4 Seven B. 2 C. 11 Four D. 15 Four

Function y = cosx-sin2x-cos2x + 7 The maximum value of 4 is () A. 4 Seven B. 2 C. 11 Four D. 15 Four

y=cosx-sin2x-cos2x+7
4=-cos2x+cosx+7
4=-(cosx-1
2)2+2．
Because - 1 ≤ cosx ≤ 1,
So when cosx = 1
2, the function gets the maximum value of 2
Therefore, B

Function y = cosx-sin2x-cos2x + 7 The maximum value of 4 is () A. 4 Seven B. 2 C. 11 Four D. 15 Four

y=cosx-sin2x-cos2x+7
4=-cos2x+cosx+7
4=-(cosx-1
2)2+2．
Because - 1 ≤ cosx ≤ 1,
So when cosx = 1
2, the function gets the maximum value of 2
Therefore, B

0 < x < π / 4, find the minimum value of the function f (x) = (cos2x + 1) / (sinxcosx sin ^ 2x)!

In the first step, f (x) is simplified to f (x) = 2cos ^ 2x / (sinxcosx sin ^ 2x). In the second step, we know that cosx is not 0, f (x) is divided by the square of cosx to get f (x) = 2 / (TaNx Tan ^ 2x). In addition, G (x) = TaNx Tan ^ 2x. The range of TaNx is (0,1 / 4). When G (x) reaches the maximum value, f (x)

It is known that the function y = cos2x + sin squared x-cosx is in the range of [0360], the set of X when the function obtains the maximum and minimum values Answer me as soon as you know,

Cos2x = cosx ^ 2-sinx ^ 2 y = cosx ^ 2-cosx = (cosx-0.5) ^ 2-0.25 Max 2 min -0.25

The known function y = cos2x + sin x-cosx (1) The maximum and minimum of Y; (2) In the range of [0,2 π], the set of X when the function obtains the maximum and minimum value

y = cos2x + sin²x - cosx= cos²x - cosx = (cosx - 1/2)² - 1/4x = 2kπ + π ,max(y) = 2x = 2kπ ± π/3 ,min(y) = -1/4x∈[0,2π]x ∈ {π} ,max(y) = 2x ∈ {π/3 ,8π/3} ,min(y) = -1/4

Find the maximum and minimum values of the function y = cosx - 1 / 2 cos2x + 1

y=cosx - 1/2 cos2x +1
=cosx-1/2(2cos²x-1)+1
=-cos²x+cosx+3/2
=-(cosx-1/2)²+7/4
This is a quadratic function about cosx
cosx∈[-1,1]
When cosx = 1 / 2, the maximum value of Y is 7 / 4
When cosx = - 1, y gets the minimum value - 1 / 2

Y = 2sinx-3cosx 2. Y = cos? X-cos quartic x 3. Y = cos quartic x-sin quartic X

1、y=√13sin（x+B）,tanB=3/22、y=cos²x（1-cos²x）=cos²x*sin²x=1/4sin²2x3、y=（cos²x+sin²x）（cos²x-sin²x）=cos²x-sin²x=cos2x

Find the value range of function y = cos2x + 2sinx-2

∵y=cos2x+2sinx-2
=1-sin2x+2sinx-2
=-（sinx-1）2，
∵-1≤sin≤1，
∴-2≤sin-1≤0，
∴（sinx-1）2∈[0，4]，-（sinx-1）2∈[-4，0]．
The value range of the function y = cos2x + 2sinx-2 is [- 4, 0]

Let f (x) = sin2x + cos (2x + π / 6), where x ∈ R ① Finding the minimum positive period of function f (x)

f(X)=sin2x+cos(2x+π/6)
=Sin2x + (radical 3 / 2) * cos2x - (1 / 2) * sin2x
=(1 / 2) * sin2x + (radical 3 / 2) * cos2x
=sin(2x+π/3）
Therefore, the minimum positive period is 2 π / 2 = π

Known vector a=(2cosx，-2)， b=(cosx，1 2)，f(x)= a• b. If x ∈ R, then f (x) is () A. Even functions with minimum positive period π B. Odd functions with minimum positive period π C. The minimum positive period is π Even function of 2 D. The minimum positive period is π Odd function of 2

∵f(x)=
a•
b=2cos2x-1=cos2x，∴f（-x）=cos（-2x）=cos2x=f（x）
The function f (x) is the minimum positive period of 2 π
Even function of 2 = π
So choose a