Simplified SiNx square + root triples sinxcosx + 2cosx square

Simplified SiNx square + root triples sinxcosx + 2cosx square

=√3/2sin2x+1+1/2+cos2x/2
The results show that: 1
=sin(2x+π/6)+3/2

Reduction of y = radical 1 + SiNx + radical 1-sinx? What is the range of its value

Y = radical 1 + SiNx + Radix 1-sinx
=Radical (1 + 2sin1 / 2xcos1 / 2x) + radical (1-2sin1 / 2xcos1 / 2x)
=Radical (sin1 / 2x + cos1 / 2x) ^ 2 + radical (sin1 / 2x-cos1 / 2x)
=|sin1/2x+cos1/2x|+|sin1/2x-cos1/2x|
(it depends on the size of X to remove the absolute value)

Simplify the root 1-sinx 1 + SiNx radical 1 + SiNx 1-sinx as the second quadrant

Root 1-sinx 1 + SiNx 1 + SiNx 1-sinx = root [(1-sin ^ x) / (1-sinx) ^ 2] - radical [(1-sin ^ x) / (1 + SiNx) ^ 2] = root [cos ^ X / (1-sinx) ^ 2] - root [cos ^ X / (1 + SiNx) ^ 2] = - cosx / (1-sinx) + cosx / (1 + SiNx) = cosx * [- (1 + SiNx) + (1-sinx)] / (1

Given the function f (x) = the square of sin (2x + quarter π) + (root 3) the square 2x of COS, find the minimum positive period and monotone of F (x) Decreasing interval (2) if x ∈ [- 6 parts π, 6 parts π], find the maximum value of F (x) and the corresponding value of X

f（x）= sin^2(2x+π/4) + √3 cos^2(2x)
= {sin2xcosπ/4+cos2xsinπ/4}^2 + √3 cos^2(2x)
= 1/2(sin2x+cos2x)^2 + √3 cos^2(2x)
= 1/2(1+2sin2xcos2x) + √3 * (cos4x+1)/2
= 1/2 + 1/2sin4x +√3/2 cos4x + √3/2
= sin4xcosπ/3 + cos4xsinπ/3 + (1+√3)/2
= sin(4x+π/3) + (1+√3)/2
Minimum positive period = 2 π / 4 = π / 2
When 4x + π / 3 ∈ (2k π + π / 2,2k π + 3 π / 2), where k ∈ Z, monotonically decreases
Monotone decreasing interval: X ∈ (K π / 2 + π / 24, K π / 2 + 7 π / 24), where k ∈ Z
If x ∈ [- π / 6, π / 6]
4x+π/3∈（-π/3,π)
When 4x + π / 3 = - π / 3, x = - π / 6, the function has the minimum value f (x) min = sin (- π / 3) + (1 + √ 3) / 2 = - √ 3 / 2 + (1 + √ 3) / 2 = 1 / 2
When 4x + π / 3 = π / 2, x = π / 24, the function has the maximum value f (x) max = sin (π / 2) + (1 + √ 3) / 2 = 1 + (1 + √ 3) / 2 = (3 + √ 3) / 2

Find the minimum positive period of the function y = sin (x / 3) cos (x / 3) + root 3 * cos square (x / 3) - root 3 / 2, and write out its image symmetry axis and center

The first multiple angle formula using sine
The second company that uses cosine
Finished the recipe
It's coming out
Y = 0.5sin (2x / 3) + three cos (2x / 3)
=sia(2x/3+π/3)
No, you just hi me

Let f (x) a sin x cosx root 3 a cos square x + radical 3 / 2A + B (a > 0) find monotone decreasing interval?

Because sin2x = 2sinxcosx, cos2x = 2cos ^ 2x-1
f(x)=a/2*sin2x-√3a/2cos2x+b
=asin(2x-π/3)+b
Because a > 0,
-π/2+2kπ

Let f (x) = cos 2 (x + π / 12), G (x) = 1 + 1 / 2sin2x 1. Let x = x0 be a symmetric axis of the image of function y = f (x), and calculate the value of G (x0); 2. Find the monotone increasing interval of function H (x) = f (x) + G (x)

(1)
f(x)＝cos²(x＋π/12)＝1/2[1＋cos(2x＋π/6)]
∵ x = x0 is a symmetric axis of the image of the function y = f (x)
∴2x0＋π/6＝kπ
That is, 2x0 = k π - π / 6 (K ∈ z)
∴g(x0)＝1＋1/2sin2x0＝1＋1/2sin(kπ－π/6)
When k is even, G (x0) = 1 + 1 / 2Sin (- π / 6) = 1-1 / 4 = 3 / 4
When k is odd, G (x0) = 1 + 1 / 2Sin (π / 6) = 1 + 1 / 4 = 5 / 4
(2)
h(x)＝f(x)＋g(x)
＝1/2[1＋cos(2x＋π/6)]＋1＋1/2sin2x
＝1/2[cos(2x＋π/6)＋sin2x]＋3/2
＝1/2(√3/2•cos2x＋1/2sin2x)＋3/2
＝1/2sin(2x＋π/3)＋3/2
When 2K π - π / 2 ≤ 2x + π / 3 ≤ 2K π + π / 2
That is, K π - 5 π / 12 ≤ x ≤ K π + π / 12 (K ∈ z)
The function H (x) = 1 / 2Sin (2x + π / 3) + 3 / 2 is an increasing function
So the monotone increasing interval of function H (x) is [K π - 5 π / 12, K π + π / 12] (K ∈ z)

Let f (x) = [cos (x + π / 12)] ^ 2, G (x) = 1 + 1 / 2sin2x (1) Let x = x0 be a symmetric axis of the image of the function y = f (x), and find the value of G (2x0); (2) find the value range of the function H (x) = f (x) + G (x), X ∈ [0, π / 4]

F (x) = [cos (x + π / 12)] ^ 2 = [1 + cos {2 (2x + π / 6)}] {2 (x = 5 π / 12 axis of symmetry) is replaced by G (x) = 1 + 1 / 2sin2x = 5 / 4 (2) H (x) = [cos (x + π / 12)] ^ 2 + 1 + 1 / 2sin2x = 3 / 2 + 1 / 4sin2x + √ 3 / 4cos2x = 2 / 3 + 1 / 2Sin (π / 3 + 2x) as for the range, you can play ball

Let f (x) = cos 2 x, G (x) = 1 + 1 / 2sin2x (1) If the point a (a, y) (a ∈ [0, π / 4]) is the common point of the image of the function f (x) and G (x) (2) Let x = x0 be a symmetric axis of the image of the function y = f (x), and find the value of G (2x0) (3) Find the value range of function H (x) = f (x) + G (x), X ∈ [0, π / 4]

(1)
f(x)=cos²x
g(x)=1+(1/2)sin2x
f(x)=g(x)
(cosx)^2 =1+(1/2)sin2x
(cos2x +1)/2 = 1+(1/2)sin2x
sin2x- cos2x + 1 =0
√2sin(2x-π/4) =-1
2x-π/4 = -π/4
x = 0
ie a=0
(2)
y=f(x)
= (cosx)^2
(cos(x+x0))^2 = (cos(x-x0))^2
put x=x0
(cos(2x0))^2 = 1
2x0 = kπ
g(2x0) = 1+ (1/2)sin(kπ)
=1
(3)
h(x) =f(x) +g(x)
= (cosx)^2+ 1+(1/2)sin2x
= (1/2 )cos2x+(1/2)sin2x + 3/2
= (√2/2)(sin2x+π/4) + 3/2
The value range of H (x): X ∈ [0, π / 4]
=[3/2,3/2+√2/2]

The known function f (x) = 2sinxcos (x + π) 3) + 3cos2x+1 2sin2x． (1) Find the minimum positive period of function f (x) (2) Find the maximum and minimum value of function f (x); (3) Write the monotone increasing interval of function f (x)

f(x)=2sinxcos(x+π
3) +
3cos2x+1
2sin2x
=2sinx（cosxcosπ
3-sinxsinπ
3）+
3cos2x+1
2sin2x
=sinxcosx-
3sin2x+
3cos2x+1
2sin2x
=sin2x+
3cos2x
=2sin（2x+π
3），
(1) Because t = 2 π
2 = π, so the minimum positive period of F (x) is π;
(2) From - 1 ≤ sin (2x + π)
3) The results show that - 2 ≤ f (x) ≤ 2,
Then the maximum value of function f (x) is 2 and the minimum value is - 2;
(3) Let 2K π - π
2≤2x+π
3≤2kπ+π
2，
The solution is k π - 5 π
12≤x≤kπ+π
12，
Then the monotone increasing interval of F (x) is: [K π - 5 π
12，kπ+π
12]．