Let f (x) = (SiNx) ^ 2 + 2sinxcosx + 3 (cosx) ^ 2, X belongs to r if f (x / 2) = 11 / 5 and 0

Let f (x) = (SiNx) ^ 2 + 2sinxcosx + 3 (cosx) ^ 2, X belongs to r if f (x / 2) = 11 / 5 and 0

According to the trigonometric function formula, f (x) = sin2x + cos2x + 2 is obtained
So f (x / 2) = SiNx + cosx + 2 = 11 / 5
So SiNx + cosx = 1 / 5
The above formula is deformed below
SINX+COSX=1/5 (1)
SINX^2+COSX^2=1 (2)
SINX=4/5
COSX= -3/5
TANX= SINX/COSX=-4/3

What is the minimum value of the function f (x) = SiNx ^ 4 + 2sinxcosx + cosx ^ 4? Help, everyone~~~~~~~~~·

F (x) = SiNx ^ 4 + 2sinxcosx + cosx ^ 4 = (sin? - x + cos? - x) - 2Sin? Xcos? X + 2sinxcosx = 1-2sin? Xcos? X + 2sinxcosx = - 1 / 2 (sin? 2x-2sin2x) + 1 = - 1 / 2 (sin2x-1sin2x) + 3 / 2 when sin2x = - 1, f (x) gets the minimum

Known function y = (SiNx) square + 2sinxcosx - (cosx) square, X belongs to R (1) find the minimum positive period of the function (2) find the maximum value of the function

Y = (SiNx) squared + 2sinxcosx - (cosx) squared
=sin2x-cos2x
=√2sin(2x-π/4)
therefore
(1) The period is π
(2) Maximum = √ 2

High and middle trigonometric function: if 0 < x ≤ π / 6, find the value range of function f (x) = 2sinxcosx + 2 √ 3cos? X - √ 3

f(x)=2sinxcosx+2√3cos^2-√3
=sin2x+√3(2cos^2-1)
=sin2x+√3cos2x
=2sin(2x+60°)
60°

Given the function f (x) = 2 √ 3cos ^ x-2sinxcosx - √ 3 (1), find the monotone decreasing interval of the function (2) If the function image is shifted to the left by π / 3, and then the abscissa of all points is reduced to 1 / 2 of the original value, the analytic expression of the function image g (x) obtained shall be written (3) The range of values in the interval [- π / 8, π / 8]

In this paper, the author of this paper, f (x) = 2 √ 3cos ^ X-2 SiNx cosx x-3 = √ 3 (2cos ^ x-1) - sin2x = 3 (2cos ^ x-1) - sin2x = (3 / 3 / 2cos2x-1 / 2sin22x) = 2Sin (π / 3-2x) 1 ∵ f (x) = 2Sin (π / 3-2x) ∵ f (x) = 2Sin (π / 3-2x) when π / 2 + 2K π < π / 3-2x < 3 π / 2 + 2K π, that is - π / 2-k π π / 2-k π / 2-k π / 2-k π / 2-k π / 2-k x ＜ - π / 12 - K π function

Given vector a = (2, sin), vector b = (SiNx square, 2cosx). Function f (x) = vector a multiplies vector B to find monotone increasing interval of F (x)

f(x)=ab=2(sinx)^2+2sinxcosx=(1-cos2x)+sin2x=√2sin(2x-π/4)+1
2kπ-π/2

Given the function f (x) = 2cosx squared + sin2x-1, find the period and monotone increasing interval (1) Finding period and monotone increasing interval (2) If x ∈ [0, π / 2], find the maximum and minimum of F (x)

（1）f(x)=2cos^2x+sin2x-1=cos2x+1+sin2x-1=sin2x+cos2x=√2sin(2x+π/4)
∴T=2π/2=π
From - π / 2 + 2K π

F (x) = sin (2x + Pai / 3) + root sign 3cos (2x + Pai / 3) minus interval period if Pai / 3 is changed to a, 0

F (x) = 2Sin (2x + row / 3 + row / 3) = 2Sin (2x + 2 row / 3)
Cycle is 2 rows / 2 = rows
The minus interval is 2K rows + rows / 2

The monotone decreasing interval of the function y = cos (- 2x + π / 3) is

y=cos(-2x+π/3)=cos(2x-π/3)
Decline interval:
2kπ

Find the monotone decreasing interval of function y = Log1 / 2 [cos (- X / 3 + Pai / 4)]

cos(-x/3+pai/4)=cos(x/3-/4)>0
2kπ-π/2