Function f (x) = cosx, (cosx is greater than or equal to SiNx) and SiNx, cos is less than SiNx, then the value range of F (x) is?

Function f (x) = cosx, (cosx is greater than or equal to SiNx) and SiNx, cos is less than SiNx, then the value range of F (x) is?

Draw an image, [(root 2) / 2,1]

Given the vectors a = (cos3 / 2x, SIN3 / 2x), B = (cosx / 2, - SiNx / 2), and X ∈ [0, π / 2], find the modules of (1) a · B and a · B; (2) If f (x) = a · B-2 λ| The minimum value of a + B | is - 3 / 2. Find the real number λ A and B in the question are vectors (1) The module questions of a · B and a + B are wrong

You have the wrong title. Is it the model of a + B?

Given vector a = (cos3 \ 2x, SIN3 \ 2x), vector b = (cosx \ 2, - SiNx \ 2), and X belongs to [0, π \ 2], find 1. | vector a + vector B| Find 1. | vector a + vector B| 2. Find the minimum value of function f (x) = vector A. vector B-4 | vector a + vector B | I hope you can help me understand the following

/a+b/=√a^2+b^2+2ab,
a^2+b^2=2,
ab=cos3/2x*cosx/2-sin3/2x*sinx/2=cos2x,
/a+b/=2cosx,
f(x)=cos2x-8cosx=2cos^2x-8cosx-1,
0

Known vector a=(cos3 2x，sin3 2x)， b=(cosx 2，-sinx 2) And X ∈ [0, π 2] , find: （1） a• b； (2) If f (x)= a• b-2 λ| a+ The minimum value of B | is - 3 2. Seek λ The value of the

（1）
a•
b=
(
a+
b)2=
2+2cos2x=2cosx（x∈[0，π
2]）
(2) From (1): F (x) = cos2x-4 λ cosx=2cos2x-4 λ cosx-1=2（cosx- λ） 2-2 λ 2-1
∵x∈[0，π
2]
∴cosx∈[0，1]，
When λ ∈ [0,1], f (x) min = - 2 λ 2-1, and f (x) min = - 3
2，
So - 2 λ 2-1=-3
2， λ= one
2，
When λ＜ At 0, f (x) min = f (π)
2)=2 λ 2-2 λ 2-1=-1，
And f (x) min = - 3
2. It does not meet the meaning of the question
When λ＞ 1, f (x) min = f (0) = 2-4 λ- 1=-4 λ+ 1, and f (x) min = - 3
two
So - 4 λ+ 1=-3
2， λ= five
8 this and λ＞ 1 contradiction
Comprehensive above λ The value of is 1
2．

Known vectors a = (cos3 / 2x, SIN3 / 2x), B = (cosx / 2, - SiNx / 2), where x belongs to [0, π / 2] If the minimum value of F (x) = a.b-2 / A + B / is - 3 / 2, calculate the input value

A * b = cos3 / 2x * cosx / 2-sin3 / 2xsinx / 2 = cos2x | a + B | 2 = a ^ 2 + B ^ 2 + 2Ab = 2 + 2cos2x use m to represent the letter. It's not easy to type F (x) = cos2x-2m root sign (2 + 2cos2x) = cos2x-4m ︱ cosx ︱ = 2cosx ^ 2-4m ︱ cosx - 1 = 2 (︱ - M) ^ 2-2m ^ 2-1. The minimum value is - 3 / 2. Then in | cosx |

Known vector a =（cos three two x，sin three two x）， b =（cos x two ，-sin x two ）And X ∈ [0, π two ]， (1) Beg a • b And| a + b |； (2) If f (x)= a • b -2 λ| a + b |The minimum value of is- three two , find the real number λ The value of the

(1) According to the meaning of the question, a • B = cos32xcosx2-sin32xsinx2 = cos2x, a + B = (cos32x + cosx2, sin32x-sinx2), (a + B = (sin3x2 + cosx2) 2 + (sin3x2 − sinx2) 2 = 2 + 2cos2x = 2|cosx|. ∵ x ∈ [0, π 2], | 1 ≥ cosx ≥ 0, | a + B | = 2cosx. (2) from (