Simplify 3 / 2cosx - √ 3 / 2sinx Ah, this kind of problem is not very good, please teach and guide~

Simplify 3 / 2cosx - √ 3 / 2sinx Ah, this kind of problem is not very good, please teach and guide~

3/2cosx-√3/2sinx
=√3（√3/2cosx-1/2sinx）
=√3（cos(2k∏+∏/6)cosx-sin(2k∏+∏/6)sinx）
=√3cos(x+2k∏+∏/6)
=√3cos(x+∏/6)

How to convert 2 / 2cosx - √ 2 / 2sinx - √ 2 / 2cosx - √ 2 / 2sinx / √ 2 / 2cosx - √ 2 / 2sinx + √ 2 / 2cosx + √ 2 / 2sinx how to turn into - √ 2sinx / √ 2cosx

How can [(√ 2 / 2) cosx - (√ 2 / 2) SiNx - (√ 2 / 2) SiNx - (√ 2 / 2) SiNx] / [(√ 2 / 2) cosx - (√ 2 / 2) SiNx + (√ 2 / 2) cosx + (√ 2 / 2) SiNx] how to turn into - (√ 2) SiNx / [(√ 2) cosx]
The original formula = - [2 (√ 2 / 2) SiNx] / [2 (√ 2 / 2) cosx] = - [(√ 2) SiNx] / [(√ 2) cosx] = - SiNx / cosx = - TaNx
On the molecular level, (√ 2 / 2) cosx - (√ 2 / 2) cosx = 0; - (√ 2 / 2) SiNx - (√ 2 / 2) SiNx = - 2 (√ 2 / 2) SiNx = - (√ 2) SiNx;
On the denominator, (√ 2 / 2) cosx + (√ 2 / 2) cosx = 2 (√ 2 / 2) cosx = (√ 2) cosx; - (√ 2 / 2) SiNx + (√ 2 / 2) SiNx = 0;

Given the vector a = (2sinx / 2,1 - √ 2cosx / 2) B = (cosx / 2,1 + √ 2cosx / 2) function f (x) = ㏒ (a × b); (1) find the definition domain and value range of function f (x) (2 find the monotone interval of function f (x))

In the title (a × b) can not be used casually, so that the vector product of vector, also known as cross product, is a vector perpendicular to a and B
According to the meaning of this question, it should be the product of quantity, that is, the product of points
a·b = 2sin(x/2)cos(x/2)+[1-√2cos(x/2)] [1+√2cos(x/2)]
= sinx+{1-2[cos(x/2)]^2} = sinx-cosx = √2sin(x-π/4)
f(x)= log[√2sin(x-π/4)]
(1) Definition domain sin (x - π / 4) > 0,2k π

Given the vector a = (2sinx, cosx), B = (√ 3cosx, 2cosx), the function f (x) = a * B + 1 (1) find the maximum value of function f (x) and the set of independent variables X which obtain the maximum value; (2) find the monotone decreasing interval of function f (x)

Given the vector a = (2sinx, cosx), B = ((√ 3) cosx, 2cosx), the function f (x) = a · B + 1, X ∈ R
f(x)=2(√3)sinxcosx+2cos²x+1=(√3)sin2x+cos2x+2
=2[sin2xcos(π/6)+cos2xsin(π/6)]+2=2sin(2x+π/6)+2
When Max f (x) = 4, when 2x + π / 6 = 2K π + π / 2, that is, x = k π + π / 6, the maximum value of F (x) is 4
The simple subtraction interval of F (x): from 2K π + π / 2 ≤ 2x + π / 6 ≤ 2K π + 3 π / 2, the simple subtraction interval of F (x) is: 2K π + π / 2 ≤ 2x + π / 2
kπ+π/6≦x≦kπ+2π/3.

Given the vector a = (cosx, 2sinx), B = (2cosx, √ 3cosx), f (x) = a × B + m) (1), find the minimum positive period of F (x) (2) If the maximum of F (x) on [- π / 6, π / 6] is 3, find the value of M

1)f(x)=2cos^2x+2√3sinxcosx+m
=cos2x+√3sin2x+m+1
=2sin(2x+π/6)+m+1
Minimum positive period T = 2 π / 2 = π
2) When f (x) is on [- π / 6, π / 6], 2x + π / 6 ∈ [- π / 6, π / 2]
Then f (x) ∈ [M, M + 3]
From M + m + 3 = 3, M = 0

Given the vector a = (cosx, 2sinx), vector b = (2cosx, √ 3cosx), f (x) = vector a, · vector B (1) find the minimum positive period and monotone increasing interval of function f (x); (2) translate y = f (x) according to vector m to get the image of y = 2sin2x, and find the vector M

(1) The minimum positive period T = 2 π / 2 = π by, 2K π - π / 2 ≤ 2x + π / 6 ≤ 2K π + π / 2, K π - π / 3 ≤ x ≤ K π + π / 6, increasing interval [K π - π / 3, K π + π / 6], K ∈ Z

Given a vector a = (2cosx, 2sinx), vector b = (cosx, √ 3cosx), function f (x) = a × B Find the minimum positive period sum range of function f (x)

f(x)=2(cosx)^2+2√3sinxcosx=√3sin2x+cos2x+1=2sin(2x+π/6)+1
The minimum positive period is t = 2 π / 2 = π
The minimum value is - 1, the maximum value is 3, and the range is [- 1,3]

Given the vector a = (2cosx, 2sinx), B (√ 3cosx, cosx), if f (x) = a · B - √ 3 If f (α / 2 - π / 6) - f (α / 2 + π / 12) = √ 6 and α belongs to (π / 2, π), calculate α

0

The known function f (x) = sin (π - x) sin (π) 2-x）+cos2x (1) Find the minimum positive period of function f (x); (2) When x ∈ [- π 8，3π 8] Find the monotone interval of function f (x)

When x ∈ [- π 8, 3 π 8], the function f (x) is simple

Given the function f (x) = (COS ^ 2) x + 2sinxcosx - (sin ^ 2) x, find the minimum positive period of y = f (x)

f(x)=(cos^2)x+2sinxcosx-(sin^2)x=(cos^2)x-(sin^2)x+2sinxcosx
=cos2x+sin2x
=√2(sin2x+π/4)
So the minimum positive period T = 2 π / 2 = π