Given the vector a = (2cosx / 2, Tan (x / 2 + π / 4)), B = (√ 2Sin (x / 2 + π / 4), Tan (x / 2 - π / 4)), Let f (x) = vector a * vector B, find the range of value of F (x), the minimum positive period

Given the vector a = (2cosx / 2, Tan (x / 2 + π / 4)), B = (√ 2Sin (x / 2 + π / 4), Tan (x / 2 - π / 4)), Let f (x) = vector a * vector B, find the range of value of F (x), the minimum positive period

F (x) = vector a * vector b
=2cosx/2*√2sin(x/2+π/4)+tan(x/2+π/4)*tan(x/2-π/4)
=2cosx/2*(sinx/2+cosx/2)+1
=2sinx/2cosx/2+2(cosx/2)^2
=sinx+cosx+2
=√2sin(x+π/4)+2
The value range of F (x) [2 - √ 2,2 + √ 2]
Minimum positive period T = 2 π

Given vector M = (Radix 3sin (x / 4), 1), vector n = (COS (x / 4), cos ^ 2 (x / 4)) f (x) = M.N (2) Let f (x) = M.N. in △ ABC, the opposite side of ABC is ABC, satisfying (2a-c) CoSb = bcosc. Find the value range of F (a) I hope to give the detailed process. The answer is (1,3 / 2) It's 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 concluded that
(2sinA-sinC)cosB=sinBcosC
2sinAcosB=sinBcosC+sinCcosB=sin(B+C)
∵ a, B, C are the three inner angles of the 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 the independent variable a of the 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 sine function y = SiNx, it can be concluded that:
sin(A/2+π/6) ∈ （1/2 ,1）
f(A) ∈ （1,3/2）

Known vector M = (√ 3sin (x / 4), 1), n = (COS (x / 4) online and so on! Let m = (√ 3sin (x / 4), 1), n = (COS (x / 4), cos Λ 2 (x / 4)). Let f (x) = m · n (1) If f (x) = 3 / 2, find the value of COS (2 π / 3-x) My m * n is not equal to one! Will copy!

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)
It is known that m * n = 3 / 2
sin(x/2 + π/6)+(1/2)=3/2
sin(x/2 + π/6)=1
Thus: cos (2 π / 3 - x) = - cos [π - (2 π / 3 - x)] = - cos (x + π / 3)
=-cos[2*(x/2 + π/6)]=-[1 - 2sin^(x/2 + π/6)]=2*1^2 -1=1
As like as two peas, I would not deny that the copy is my laider, but even if your topic is m*n, it is not the same as the one of others. The most part of the steps is exactly the same. Typing is a troublesome matter. And I also provide reference materials, and if your math level is not bad, You should be able to follow the ideas in front of you to finish directly, the last five lines you can not read, I also revised, you do not accept it

What is the equation of an axis of symmetry of the image of the function y = sin (3x + π / 3) cos (x - π / 6) + cos (3x + π / 3) cos (x + π / 3)?

cos（x+π/3）=sin(π/2-x-π/3)=sin（π/6-x）=-sin（x-π/6）y=sin（3x+π/3）cos（x-π/6）+cos（3x+π/3）cos（x+π/3）=sin（3x+π/3）cos（x-π/6）-cos（3x+π/3）sin（x-π/6）=sin（3x+π/3-x+π/6）=sin（...

A symmetric axis equation for the image of the function y = SiN x / 2 + (√ 3) cos X / 2 is

y=sin x／2 ＋（√3）cos x／2
=2sin(1/2*sinx/2+cos x／2*√3/2)
=2sin(sinx/2*cosπ/3+cos x／2*sinπ/3)
=2sin(x/2+π/3)
Let X / 2 + π / 3 = k π + π / 2, then x = 2K π + π / 3
Let k = 0, x = π / 3 be an axis of symmetry

Function y = sin (2x + π / 3) the symmetry axis equation of the image is, to be explained in detail

The symmetry axis of y = SiNx is x = k π + π / 2, K ∈ Z
When x = k π + π / 2, y reaches the maximum value
Find the symmetry axis of the image of function y = sin (2x + π / 3)
It takes 2x + π / 3 as a whole, which makes y get the maximum value
ν from 2x + π / 3 = k π + π / 2
2X = k π + π / 6
The symmetry axis is x = k π / 2 + π / 12, K ∈ Z

A symmetric axis equation of the image of the function f (x) = sin (2x-3 / π) is

Is it f (x) = sin (2x - π / 3)?
When 2x - π / 3 = k π + π / 2, that is, x = k π / 2 + 5 π / 12, K ∈ Z is the symmetry axis of F (x)
So suppose k = 0, then x = 5 π / 12 is an axis of symmetry of F (x)

The symmetry axis equation and symmetry center of the image with function y = sin (2x - π / 6) are obtained Process! Thank you

2x-π/6=π/2+kπ
So the symmetry axis equation x = π / 3 + K π / 2
2x-π/6=π+kπ
So the center of symmetry is [7 / 12 π) + K π / 2,0]

The function y = sin (2x + 5 π / 2) image symmetry axis equation is No need to copy it. Thank you, A simple method

The symmetry axis of y = SiNx is x = π / 2 + K π (K ∈ z);
So 2x + 5 π / 2 = π / 2 + K π (K ∈ z)
So the axis of symmetry is x = k π / 2 (K ∈ z)
It's my pleasure to answer your questions and skyhunter 002 to answer your questions
If there is anything you don't understand, you can ask,

Let f (x) = √ 3cos2 ω x + sin ω xcos ω x + a (where ω > 0, a ∈ R), and the abscissa of the first highest point of the image of F (x) on the right side of the y-axis is π / 6. (1) find the value of ω; (2) if the minimum value of F (x) on the interval [- π / 3,5 π / 6] is √ 3 ω=1/2 a=（√3+1）/2

First of all, your title may be wrong, the middle should be a coefficient of 2. The coefficient of the last term should also have a 2; (1) f (x) = √ 3cos2 ω x + 2Sin ω xcos ω x + 2A = 2Sin (2 ω x + π / 3) + 2AF (x) the abscissa of the first highest point on the right side of the y-axis is π / 6, thus 2 ω x π / 6 + π / 3 = π / 2, ω = 1