Known vector M = (root 3sin (x / 4), 1), vector n = (COS (x / 4), cos ^ 2 (x / 4)) f (x) = M.N (2) Note that f (x) = M.N. in △ ABC, the opposite side of ABC is ABC, which satisfies (2a-c) CoSb = bcosc, and find the value range of F (a) I hope to give a detailed process. The answer is (1,3 / 2) Very urgent

Known vector M = (root 3sin (x / 4), 1), vector n = (COS (x / 4), cos ^ 2 (x / 4)) f (x) = M.N (2) Note that f (x) = M.N. in △ ABC, the opposite side of ABC is ABC, which satisfies (2a-c) CoSb = bcosc, and find the value range of F (a) I hope to give a detailed process. The answer is (1,3 / 2) Very urgent

m={√3sin(x/4),1},n={cos(x/4),cos^(x/4)}m*n=√3sin(x/4)*cos(x/4)+1*cos^(x/4)=(√3/2)*sin[2*(x/4)] + {1+cos[2*(x/4)]}/2=(√3/2)*sin(x/2) + (1/2)*cos(x/2) + (1/2)=sin(x/2)*cos(π/6)+cos(x/2)*sin(π/6) + (1/2)=sin(x/2 + π/6) + (1/2) f(x)=sin(x/2 + π/6) + 1/2
Then f (a) = sin (A / 2 + π / 6) + 1 / 2
It is known that: (2a-c) CoSb = bcosc,
According to the sine theorem: A / Sina = B / SINB = C / sinc, it can be obtained that:
(2sinA-sinC)cosB=sinBcosC
2sinAcosB=sinBcosC+sinCcosB=sin(B+C)
∵ a, B and C are the three internal angles of a triangle, and there must be a = π - b-c
‡ Sina = sin (B + C), and Sina > 0
2sinAcosB=sinA
cosB=1/2
B=π/3
∴A+C=2π/3
The value range of a is a ∈ (0,2 π / 3)
That is, the range of independent variable a of function f (a) = sin (A / 2 + π / 6) + 1 / 2 is (0,2 π / 3)
A/2+π/6 ∈ (π/6 ,π/2)
According to the image of the basic sinusoidal function y = SiNx, it can be obtained that:
sin(A/2+π/6) ∈ (1/2 ,1)
f(A) ∈ (1,3/2)

Known vector M = (root 3sinx / 4,1), n = (cosx / 4, square X / 4 of COS), f (x) = m * n 1. If f (x) = 1, find the value of COS (2 Pie / 3-x) 2. In triangle ABC, the opposite sides of angles a, B and C are a, B and C respectively, and meet acosc + 1 / 2C = B. find the value range of function f (b)

f(x)=cos(π/3-x/2)+1/2
f(x)=cos(π/3-x/2)+1/2=1
cos(π/3-x/2)=1/2
cos(2π/3-x)=-1/2
f(B)=cos(π/3-B/2)+1/2
acosC+1/2c=b
b^2+c^2-a^2=-bc
A=2π/3
0

Given the vector OA = [cos @, sin2] vector M = [2,1] n = [0, -√ 5] and M is perpendicular to [oa-n], find the vector OA, please

OA-n=(cos α, sin α √5) m(OA-n)=0 2cos α sin α √5=0 √5sin( α arctan2) √5=0 sin( α arctan2)=-1 α ∈[-π,0] ∴ α=- π/2-arctan2 cos α=- 1/√5 sin α=- 2 / √ 5 OA = (- 1 / √ 5, - 2 / √ 5) if cos( β- π)=√2/10 cos β=- √2/10 cos(2π- β) =- √ 2 / 10 the second question is whether you have typed the title wrong

Known vector m=(sin θ, 2cos θ), n=( 3,−1 2) , if m⊥ n. Then sin2 θ The value of is______

because
m⊥
n. So
m•
n=0
Namely:
3sin θ −cos θ= 0
So sin( θ- π
6) = 0, i.e θ- π
6=kπ   k∈Z,
two θ= 2kπ+π
three
∴sin2 θ=
three
two
So the answer is:
three
two

Known vector M = (COS) α- √2/3,-1),n=(sin α, 1) , m and N are collinear vectors, and α Find sin2 with ∈ (- π / 2,0) α/ (sin α- cos α Known vector M = (COS) α- √2/3,-1),n=(sin α, 1) , m and N are collinear vectors, and α Find sin2 with ∈ (- π / 2,0) α/ (sin α- cos α) The value of the

M and N are collinear vectors = > (COS) α- √2/3)/(-1)=sin α/ 1-cos α+ √2/3 = sin α √2/3 = (sin α+ cos α) 2/9 = 1+ sin2 α sin2 α = - 7/9tan2 α = - 7/(4√2)-cos α+ √2/3 = sin α √2/3 = (sin α+ cos α) . (sin α- cos α)/ (sin α- cos α)=...

Known vector m=( 3sinx 4,1), n=(cosx 4,cos2x 4) , f (x) = m• n. In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and meet (2a-c) CoSb = bcosc. Find the value range of function f (a)

Because (2a-c) CoSb = bcosc, the sine theorem gives (2sina sinc) CoSb = sinbcosc
So 2sinacosb sinccosb = sinbcosc
So 2sinacosb = sin (B + C)
Because a + B + C = π
So sin (B + C) = Sina, and Sina ≠ 0
So CoSb = 1
2,B=π
three
So 0 < a < 2 π
three
So π
6<A
2+π
6<π
2,1
2<sin(A
2+π
6)<1
And because f (x) =
m•
n=sin(x
2+π
6)+1
two
So f (a) = sin (a)
2+π
6)+1
two
Therefore, the value range of function f (a) is (1,3)
2)