Known vector M = (root 3sinx / 4,1), vector n = (cosx / 4, COS) ² X / 4) function f (x) = m * n, In the acute angle △ 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 F (2b)

Known vector M = (root 3sinx / 4,1), vector n = (cosx / 4, COS) ² X / 4) function f (x) = m * n, In the acute angle △ 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 F (2b)

1. Firstly, f (x) = m * n is transformed into a functional formula, f (x) = m * n = √ 3sinxcosx / 16 + cos ² X / 4, according to the double angle formula, it is equal to sin2x of three times the root of 32 / 32 plus cos2x of eight plus 1, and extract the common factor to obtain sin2x of three times the root of 16 / 1 in brackets plus

Known angle α ∈ (0, π), vector M = (2, COS) α), n=(cos α², 1) , and Mn = 1, f (x) = root 3sinx + cosx (1) Angle finding α Size of (2) Find function f (x)+ α) Monotone decreasing interval of

Mn = 1,2cos ^ 2A + cosa = 12cos ^ 2A + cosa-1 = 0 (2cosa-1) (COSA + 1) = 0cosa = 1 / 2, cosa = - 1 (rounded off, because α ∈(0,π),)a=π/3f(x)=√3sinx+cosx=2(√3/2sinx+1/2cosx)=2sin(x+π/6)f(x+ α)= 2Sin (x + π / 3 + π / 6) = 2Sin (x + π / 2) = 2cosx, so

Given vector M = (root 3sinx / 4,1), vector n = (cosx / 4, cos ^ 2 x / 4) if vector m is perpendicular to vector n, find cos (2 π / 3-x)

m•n=√3sin(x/4)cos(x/4)+cos ² (x/4)
=(√3/2)sin(x/2)+(1/2)cos(x/2)+1/2
=cos(x/2-π/3)+1/2
Because vector m is perpendicular to vector n, cos (x / 2 - π / 3) + 1 / 2 = 0,
cos(x/2-π/3)=-1/2
So cos (2 π / 3-x) = cos (X-2 π / 3)
= cos2(x/2-π/3)=2 cos ² (x/2-π/3)-1
=-1/2.

Given vector M = (root sign 3sinx / 4,1), vector n = (cosx / 4, cos ^ X / 4), if vector m * n = 1, find the value of COS (2 π / 3-x)

m.n=1
(√3sin(x/4),1). (cos(x/4),(cos(x/4))^2)=1
√3sin(x/4). (cos(x/4)+ (cos(x/4))^2 =1
(√3/2)sin(x/2) +( cos(x/2) +1) /2=1
((√3/2)sin(x/2)+(1/2)cosx/2) = 1/2
cos(π/3 -x/2) = 1/2
[cos(π/3 -x/2)]^2 = 1/4
(cos(2π/3 -x) +1)/2 = 1/4
cos(2π/3 -x) = -1/2

Known vector OA = (COS) α, sin α), among α ∈ [- π, 0], vector M = (2,1), vector n = (0, -√ 5), and vector m ⊥ (vector OA vector n) (1) Find vector OA (2) If cos( β- π)=√2/10,0< β< π. Find cos (2) α-β)

(1) (vector OA vector n) = (COSA, Sina + √ 5) because m ⊥ (oa-n), then m · (oa-n) = 2cosa + Sina + √ 5 = 0 because Sina ^ 2 + cosa ^ 2 = 1, α ∈ [- π, 0], so Sina < 0cosa=- ½ (√ 5 + Sina), substitute it into the equation in the second line, and simplify it to: 5sina ^ 2 + 2 √ 5sina + 1 = 0s

Known vector M = (COS) θ, Sin θ) And n (root 2 - Sin θ, Cos θ), And IM + NL = find cos( θ/ 2+π/8)

Known vector M = (COS) θ, sin θ) Sum vector n = (√ 2-sin) θ, cos θ),θ ∈ (Π, 2 Π), and M + n = 8 √ 2 / 5, find cos( θ/ 2 + π / 8). [solution] because M = (COS) θ, sin θ) And N = (√ 2-sin) θ, cos θ) So m + n = (√ 2-sin) θ+ cos θ, sin θ+ cos θ) So | m + n | = under the root sign