The maximum value of y = sin ^ 2x + sinxcosx + 3cos ^ 2x

The maximum value of y = sin ^ 2x + sinxcosx + 3cos ^ 2x

y=(sinx)^2+sinxcosx+3(cosx)^2
=2(cosx)^2+(1/2)sin2x+1
=(1/2)sin2x+cos2x+2
=(√5/2)sin(2x+p)+2
The maximum value is 2 + √ 5 / 2, and the minimum value is 2 - √ 5 / 2

Simplify 3cos ^ 2x + 2cosxsinx + sin ^ 2x

3cos^2x+2cosxsinx+sin^2x
=cos^2x+sin^2x+2cosxsinx +2cos^2x
=1+sin2x+2cos^2x
=1+sin2x+1+cos2x
=2+sin2x+cos2x
=2+√2(√2/2sin2x+√2/2cos2x)
=sin(2x+π/4)+2

How to simplify sin ^ 2x + 2sinx + 3cos ^ 2x

It should be 2sinxcosx in the middle
The original formula = (sin? X + cos? X) + sin2x + 2cos? X-1 + 1
=1+sin2x+cos2x+1
=√2(√2/2*sin2x+√2/2cos2x)+2
=√2(sin2xcosπ/4+cos2xsinπ/4)+2
=√2sin(2x+π/4)+2

F (x) = sin ^ 2x + sinxcosx-1 / 2, simplify

f(x)=(1-cos2x)/2+1/2*sin2x-1/2
=1/2(sin2x-cos2x)
=√2/2*(sin2x*√2/2-cos2x*√2/2)
=√2/2(sin2xcosπ/4-cos2xsinπ/4)
=√2/2sin(2x-π/4)

Given the function f (x) = 5sinxcosx-5 √ 3cos? X + 5 / 2 √ 3 (x ∈ R) for solving T-monotone interval symmetric axisymmetry

f(x)=5sinxcosx-5√3 cos^2 x+5√3/2=5sin2x/2-5√3[(1+cos2x)/2]+5√3/2
=5sin2x/2-5√3cos2x/2
=5*sin(2x-п/3)
So the minimum positive period of the function is 2 п / 2 = п
2kп-п/2≤2x-п/3≤2kп+п/2
2kп-п/6≤2x≤2kп+5п/6
kп-п/12≤2x≤kп+5п/12
So the monotone increasing interval of the function is [K п - п / 12'k п + 5 п / 12]
2kп-п≤2x-п/3≤2kп-п/2
2kп-2п/3≤2x≤2kп-п/6
kп-п/3≤x≤kп-п/12;
2kп+п/2≤2x-п/3≤2kп+п
2kп+п5/6≤2x≤2kп+4п/3
kп+п5/12≤2x≤kп+2п/3
So the decreasing interval of the function is [K п - п / 3'k п - п / 12], [K п + п 5 / 12'k п + 2 п / 3]
2x-п/3=2kп
x=kп+5п/12
So the axis of symmetry of the function is x = k п + 5 п / 12

Known function y = 3cos (2x - π / 6) to find the center and axis of symmetry

Symmetry center (K π / 2 + π / 3,0)
The axis of symmetry x = k π / 2 + π / 12, where k is an integer
The center of symmetry is generally the equilibrium position
The value of X when the axis of symmetry is generally the maximum or minimum

According to the following conditions: F (x) = sin (x + Π / 4) + 2Sin (x - Π / 4) - 4cos2x + 3cos (x + 3 Π / 4) (1) x = Π / 4 (2) x = 3 Π / 4 (1) x=∏/4 (2) x=3∏/4

f(x)=sin(x+∏/4)+2sin(x-∏/4)-4cos2x+3cos(x+3∏/4)
(1) When x = Π / 4,
x+π/4=π/2.sin(x+∏/4)=1
x-π/4=0.sin(x-∏/4)=0
2x=π/2.cos2x=0
x+3π/4=π.cos(x+3∏/4)=-1
f(x)=1+0-0-3=-2
(2) When x = 3 Π / 4
x+π/4=π.sin(x+∏/4)=0
x-π/4=π/2.sin(x-∏/4)=1
2x=3π/2.cos2x=0
x+3π/4=3π/2.cos(x+3∏/4)=0
f(x)=0+2-0+0=2

The value of the function f (x) = sin (x + π / 4) + 2Sin (x - π / 4) - 4cos2x + 3cos (x + 3 π / 4) is obtained according to the following conditions: (1) x = π / 4 (2) x = 3 π / 4

(1)f(x)=sin(π/4+π/4)+2sin(π/4-π/4)-4cosπ/2+3cos(π/4+3π/4)=sin(π/2)+2sin0-4*0+3cosπ
=1+0-4*0+3*(-1)=-2
(2) Substituting x = 3 π / 4, f (x) = 0 + 2-4 * 0 + 3 * 0 = 2

(2 / 2) amplitude, period, initial phase, how to translate to get y = 2Sin (2x + Pai / 3)

First, shift y = SiNx to the left / 3 units
Then the abscissa becomes half of the original
Then the ordinate becomes twice the original
This is the basic problem of trigonometric function image

The known function y = 3 / 2Sin (x / 2 + π / 6) (1) indicates the amplitude, period, initial phase, frequency and monotone interval of this function

A=3/2
T=2π/(1/2)=4π
The initial phase is 3 / 2 sin π / 6 = 3 / 4
Frequency f = 1 / T = 1 / (4 π)
X / 2 + π / 6 increases monotonically on (- π / 2 + 2K π, π / 2 + 2K π), that is, when x is on (- 4 / 3 π + 4K π, 2 / 3 π + 4K π)