Cos α + cos β = 1 / 3 to find the maximum and minimum value of cos α - Sin ^ 2 β

Cos α + cos β = 1 / 3 to find the maximum and minimum value of cos α - Sin ^ 2 β

Cos α - Sin ^ 2 β = 1 / 3-cos β - Sin ^ 2 β = cos ^ 2 β - cos β - 2 / 3cos α ∈ [- 1,1], so cos β∈ [- 2 / 3,1] is x ∈ [- 2 / 3,1], find the maximum and minimum value of x ^ 2-x-2 / 3. According to the image of quadratic function, the minimum value is the vertex of the image, that is, the lowest point, and the maximum value of - 11 / 12 is when x = - 2 /

What is the maximum value of a × sin α + cos α What about sin α + a × cos α

sqrt(a^2+1)

The maximum value of sin (sin + COS)

Let t = SiNx + cosx
Then y = the maximum value of Sint is required
And T = √ 2Sin (x + π / 4)
We get: | t|

Find the minimum sum of positive periods of the function y = sin ^ 4x + 2 √ 3sinxcosx cos ^ 4

The title should be: y = sin ^ 4x + 2 √ 3sinxcosx cos ^ 4x, y = sin ^ 4x + 2 √ 3sinxcosx cos ^ 4x = sin ^ 4x cos ^ 4x + √ 3sin2x = (sin? X + cos? X) (sin? X-cos? X) + √ 3sin2x = sin? X-cos? X + √ 3sin2x = - cos2x +

Find the period of the function y = root 3cos (3x / 2 + 2x) + cos ^ 2 x - Sin ^ 2 X. when x is the value, y takes the maximum and minimum values

[the title should be: y = radical 3 cos (3 π / 2 + 2x) + cos ^ 2 x - Sin ^ 2 x]
Y = radical 3 cos (3 π / 2 + 2x) + cos ^ 2 x - Sin ^ 2 x
=Radical 3 cos {2 π - (π / 2-2x)} + {cos ^ 2 x - Sin ^ 2 x}
=Radical 3 cos (π / 2-2x) + cos2x
=Radical 3 sin2x + cos2x
= 2 ( sin2xcosπ/6 + cos2xsinπ/6)
= 2 sin(2x+π/6)
Period T = 2 π / 2 = π
When 2x + π / 6 = 2K π + π / 2, that is, x = k π + π / 6, where k ∈ Z, there is a maximum value ymax = 2
When 2x + π / 6 = 2K π - π / 2, that is, x = k π - π / 3, where k ∈ Z, there is a minimum value of Ymin = - 2

The function f (x) = cos2x - sin2x 2，g(x)=1 2sin2x-1 Four (I) how can the image of function f (x) be obtained from the image of function g (x)? (II) find the minimum value of the function H (x) = f (x) - G (x), and find the set of X whose minimum value is obtained by using H (x)

(I) f (x) = 12cos2x = 12sin (2x + π 2) = 12sin2 (x + π 4), so to get the image of F (x), we only need to shift the image of G (x) to the left by π 4 unit length, and then move the resulting image upward by 14 unit length. (II) H (x) = f (x) - G (x) = 12cos2x-12sin2x + 14 = 22cos (2)

Find the function f (x) = sin4x + cos4x + sin2xcos2x The minimum positive period, maximum value and minimum value of 2-sin2x

f(x)=(sin2x+cos2x)2-sin2xcos2x
2-2sinxcosx
=1-sin2xcos2x
2(1-sinxcosx)
=1
2(1+sinxcosx)
=1
4sin2x+1
Two
So the minimum positive period of the function f (x) is π and the maximum value is 3
4, the minimum value is 1
4．

Find the period, monotone decreasing interval and maximum value of the function y = cos ^ 2x sin ^ 2 - √ 3cos (3 π / 2 + 2x) + 1

y=cos^2x-sin^2-√3cos(3π/2+2x)+1
=cos2x-√3sin2x+1
=2cos(2x+π/3)+1
When 2K π + π / 2 ≤ 2x + π / 3 ≤ 2K π + 3 π / 2, the function decreases monotonically
2kπ+π/2≤2x+π/3≤2kπ+3π/2
2kπ+π/6≤2x≤2kπ+7π/6
kπ+π/12≤x≤kπ+7π/12
So the monotone decreasing interval: [K π + π / 12, K π + 7 π / 12], K ∈ Z
Max: ymax = 3
Minimum value: Ymin = - 1

SiNx cosx = 1 / 2 find the value of sin * 3x-cos * 3x?

Sin ^ 3x cos ^ 3x = (SiNx cosx) (sin ^ 2x + sinxcosx + cos ^ 2x) = 1 / 2 * (1 + 3 / 8) 11 / 16

F (x) = cos cube x + sin squared X - the maximum value on cosx is equal to?

F (x) = cos cube x + sin squared x-cosx
=Cos cube x + 1-cos square x-cosx
Cosx = t belongs to [- 1,1]
f(X)=t^3-t^2-t+1
Derivation f '= 3T ^ 2-2t-1 = (3T + 1) (t-1)
[- 1, - 1 / 3] increases [- 1 / 3,1] decreases
T = - 1 / 3 at the maximum value
Substitution
f(X)max=32/27