Let f (x) = √ 3sin ^ 2x + sinxcosx - √ 3 / 2 (x ∈ R). (1) if x ∈ (0, π / 2), find the maximum value of F (x)

Let f (x) = √ 3sin ^ 2x + sinxcosx - √ 3 / 2 (x ∈ R). (1) if x ∈ (0, π / 2), find the maximum value of F (x)

F (x) = radical 3sin ^ 2x + sinxcosx - radical 3 / 2
=sinxcosx-√3/2*(1-2sin²x)
=(1/2)sin2x-(√3/2)cos2x
=sin(2x)*cos(π/3)-cos(2x)sin(π/3)
=sin(2x-π/3)
For example, 0

Let f (x) = sinxcosx + cos? X ① Find the minimum positive period of F (x); ② when x ∈ [0, π / 2], find the maximum and minimum value of function f (x)

F (x) = sinxcos x + cos ^ 2 = 1 / 2 sin2x + 1 / 2 (cos2x + 1) = 1 / 2 (sin2x + cos2x) + 1 / 2 = root 2 / 2Sin (2x + π / 4) + 1 / 2 = root 2 / 2Sin (2x + π / 4) + 1 / 2, so the period is π; when x ∈ [0, π / 2]; 2x + π / 4 ∈ [π / 4,3 π / 4] then when 2x + π / 4 = π / 2, f (x) has the maximum value, f (x) has the maximum value, f (x) has the maximum value when 2x + π / 4 = π / 2 = π / 2 = π / 2 = π / 2, f (x) has for

The maximum value of the function y = sinxcosx cos? X is

According to the angle doubling formula: sin2x = 2sinxcosx, cos2x = 2cos? X-1, we get that: sinxcosx = (1 / 2) sin2x, cos? X = (1 + cos2x) / 2. Therefore, y = (1 / 2) sin2x - (1 / 2) cos2x-1 / 2 = (√ 2 / 2) sin (2x - π / 4) - 1 / 2 when sin (2x - π / 4) = 1, y (max) = (√ 2-1) / 2

The function y = 9 (1 / 3) cos? X + (radical 3 / 2) sinxcosx + 1 is known (1) The set of uniform variables X when the function y takes the maximum value (2) The image of this function is obtained from the image of y = SiNx (x ∈ R)

-3

Finding the maximum value of the function y = cos? X-sinxcosx

X-sinxcosx = 1 / 2 (1 + cos2x) - 1 / 2sin22x = 1 / 2cos2x-1 / 2sin22x + 1 / 2 / 2 (1 / 2cos2x-1 / 2sin22x + 1 / 2 = √ 2 / 2 (√ 2 / 2 * cos2x - √ 2 / 2 * sin2x) + 1 / 2 = √ 2 / 2cos (2x + π / 4) + 1 / 2 (1 / 2) when 2x + π / 4 = 2K π + π / 2, K ∈ Z, y gets the maximum value (√ 2 + 1) / 2 when 2x + π / 4 = 4 = 2x + π / 4 = 4 = 2 = 2x + π / 4 = 4 = 2 + 2 + 2 + 2 + 2 + 2 + 2 in the case of 2K π - π / 2, K ∈ Z, y

Given the function y = 1 / 2 cos 2 x + 2 / 3 sinxcosx + 1 [x ∈ R], find the maximum value of the function and the set of corresponding independent variable x

y=1/2cos²x+√3/2sinxcosx+1
=1/4(1+cos2x)+√3/4*sin2x+1
=1/2(√3/2sin2x+1/2cos2x)+5/4
=1/2sin(2x+π/6)+5/4
When 2x + π / 6 = 2K π - π / 2
When x = k π - π / 3, K ∈ Z,
The minimum value of F (x) is 1 / 2-5 / 4 = - 3 / 4
The set of X is {x} x = k π - π / 3, K ∈ Z}

Let f (x) = cos 2 x + sinxcosx (x ∈ R) (1) find the value of F (3 π / 8) (2) find the monotone interval of F (x) No need to ask

f(x)=cos²x+sinxcosx
=(cos2x+1)/2+sinxcosx
=1/2cos2x+1/2+1/2sin2x
=√2/2sin(2x+π/4)+1/2
f(3π/8)=√2/2*sinπ+1/2=1/2
Increase interval first
Let 2K π - π / 2

Let f (x) = cos ^ 2x - √ 3 sinxcosx + 1 find monotone increasing interval

f(x)＝cos^2x－√3 sinxcosx＋1
=（cos2x+1）/2-(√3/2)sin2x+1
=(1/2)cos2x-(√3/2)sin2x+3/2;
=cos(2x+π/3)+3/2;
 when increasing monotonically, - π + 2K π ≤ 2x + π / 3 ≤ π + 2K π (K ∈ z)
-4π/3+2kπ≤2x≤2π/3+2kπ;
-2π/3+kπ≤x≤π/3+kπ;
So the monotone increasing interval is [- 2 π / 3 + K π, π / 3 + K π] (K ∈ z)
If there is anything you don't understand, you can ask,

Given the function f (x) = 2A (COS ^ 2x + sinxcosx) + B 1, when a = 1, find the period of F (x) and monotone increasing interval 2, when Let f (x) = 2A (COS ^ 2x + sinxcosx) + B 1. When a = 1, find the period and monotone increasing interval of F (x) 2. When a is not equal to 0 and X belongs to [0, π / 2], the maximum value of F (x) is 4 and the minimum value is 3 Second question

When a = 1, f (x) = 2 (COS ^ 2x + sinxcosx) + B = 1 + cos2x + sin2x + B = sin (2x + π / 4) + B
The period π monotonically increasing interval of F (x) [K π - 3 π / 8, K π + π / 8]

Given that the function f (x) = sinxcosx + cos Λ 2x (x belongs to R), (1) find the monotone interval of function f (x), (2) when x ∈ 10,2 ∧ L, find the maximum value of function f (x)

(1) F (x) = 1 / 2sin2x + (1 + cos2x) / 2 = (radical 2) / 2Sin (2x + π / 4) + 1 / 2
-π/2+2kπ