（sin2x+cosx）/2cosx-2cos^2x-sinx

（sin2x+cosx）/2cosx-2cos^2x-sinx

sin2x=2sinxcosx
Original formula = (2sinxcosx+cosx) /2cosx-2cos^2x-sinx
=(2sinx+1)/2-2cos^2x-sinx
=sinx+1/2-2cos^2x-sinx
=1/2-2cos^2x

0

In the case of x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, x = 1, X

Given the vector a = (2, SiNx), B = (sin ^ 2x, 2cosx), the function f (x) a multiplies B. find the monotone interval of F (x)

(x) = a (x) = A.B = 2.sinx + 2 SiNx cosx = 1-cos2x + SiNx = 1-cos2x + sin2x = √ 2 [sin (2x) * cos (π / 4) - cos (2x) * sin (π / 4)] + 1 = √ 2Sin (2x - π / 4) + 1 (1) increase interval 2K π - π / 2 ≤ 2x - π / 4 ≤ 2K π / 4 ≤ 2K π + π / 22K π - π / 4 ≤ 2x ≤ 2K π + 3 π / 4K π - π / 4 ≤ 2x ≤ 2K π + 3 π / 4K π - π - π / 8 ≤ 8 ≤ 8 ≤ 8 ≤ 8 ≤ 8 ≤ 8 ≤ 8 ≤ 8 ≤ 8 x ≤ K π + 3 π / 8 increasing interval [K

It is known that f (x) = sin ^ 2x + 2sinxcosx + 3cos ^ 2x x belongs to the (0, π) solution (1) The maximum value of function f (x) is to find the increasing interval of the value of X (2) function when the function f (x) takes the maximum value

F (x) = sin ^ 2x + 2x + 2sinxcos x + 3cos ^ 2x = (1-cos2x) / 2 + sin2x + 3 / 2 (1 + cos2x) = cos2x + sin2x + 2 = root number 2Sin (2x + π / 4) + 2 (2 (1) x belongs to (0, π), 2x + π / 4 belongs to (π / 4,9 π / 4), f (x) maximum value is the root number, 2 + 2,2x + π / 4 = π / 2 = π / 2, when x = π / 8, this function has the maximum maximum when x = π / 8, this function has the maximum maximum when x = π / 8, when x = π / 8, this function has the maximum value of this function has the maximum value of the root number value (2) π / 4

y=sin^2x+2sinxcosx-3cos^2x ·······I can do it again when I'm dizzy --

What do you want? Why is there no problem?

Value range of y = sin ^ 2x + 2sinxcosx + 3cos ^ 2x?

y=sin^2x+2sinxcosx+3cos^2x
=1-cos^2x+sin2x+3cos^2x
=sin2x+2cos^2x+1
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
So the range is [2 - √ 2,2 + √ 2]

(1 + 2sinxcosx + cos ^ 2x - Sin ^ 2x) + 1 How did you get to this step [(cosx + SiNx) (cosx + SiNx + cosx SiNx] + 1? Which formula was used?

The solution of the solution is (1 + 2 SiNx cosx + cos ^ 2x-sin ^ 2x) + 1 = (COS ^ 2x + sin ^ 2x + 2x + 2sinxcos x + cos ^ 2x-sin ^ 2x) + 1 = (COS ^ 2x + 2sinxcos x + sin ^ 2x + cos ^ 2x + cos ^ 2x-sin ^ 2x) + 1 = [(cosx + SiNx) 2 + cos ^ 2x-sin ^ 2x) + 1 = [(cosx + SiNx) x + 2 + (cosx-sinx) (cosx + SiNx) (cosx + SiNx) (cosx + SiNx) (cosx + SiNx)]......... (cosx + SiNx + SiNx + SiNx + SiNx) (cosx + SiNx + SiNx + SiNx + cosx + SiNx + SiNx it's a good idea

(SiNx + 3cosx) / (3cosx SiNx) = 5, then the value of sin ^ 2x sinxcosx is

The problem of homogeneous expression evaluation. ∵ (SiNx + 3cosx) / (3cosx SiNx) = 5  (TaNx + 3) / (3-tanx) = 5, the solution is: TaNx = 2  sin  x-sinxcosx = (sin  x-sinxcosx) / 1 = (sin  x-sinxcosx) / (sin ∩ x-sinx-sinx) = (Tan ｛ x-sinx) / (sin

If SiNx + sin ^ 2x = 1, then cos ^ 2x SiNx =? emergency ...

Because sin ^ 2x + cos ^ 2x = 1,
sinx+sin^2x=1
So cos ^ 2x = SiNx
So cos ^ 2x SiNx = SiNx SiNx = 0

How can SiNx be equal to sin (x / 2)

Because sinx=2sin (x/2) cos (x/2)
So if we want SiNx = sin (x / 2)
As long as 2Sin (x / 2) cos (x / 2) = sin (x / 2)
sin（x/2）[2cos（x/2）-1]=0
So sin (x / 2) = 0 or cos (x / 2) = 1 / 2
So x = 2K school or x = 2K school + or - 1 / 3 school