It is known that the definition domain of the function f (x) = sin (x + θ) + cos (x + θ) is R. (1) when θ = 0, find the monotone increasing interval of F (x). (2) if θ∈ It is known that the domain of definition of the function f (x) = sin (x + θ) + cos (x + θ) is R. (2) if θ∈ (0, π) and SiNx is not equal to 0, f (x) is even when θ is a value

It is known that the definition domain of the function f (x) = sin (x + θ) + cos (x + θ) is R. (1) when θ = 0, find the monotone increasing interval of F (x). (2) if θ∈ It is known that the domain of definition of the function f (x) = sin (x + θ) + cos (x + θ) is R. (2) if θ∈ (0, π) and SiNx is not equal to 0, f (x) is even when θ is a value

1. When θ = 0, f = SiNx + cosx. F '= cosx - SiNx, then 2K * π - 3 / 4 * π
When θ = 0, f = SiNx + cosx. F '= - cosx + SiNx, then 2K * 180
Monotone interval of y = 3cos (3x-4 π)
Because the monotone interval of y = cos (x / 2 + π / 3) I did in the previous problem, it is assumed that 2K π - π ≤ (x / 2 + π / 3) ≤ 2K π, and the monotone increment is obtained. In this problem, it is assumed that 2K π + π ≤ (3x-4 π) ≤ 2K π + 2 π, and the monotone increment is obtained
I want to ask why they are different
2kπ-π≤(x/2+π/3）≤2kπ
2kπ+π≤(3x-4π）≤2kπ+2π
The two approaches are consistent,
Because the monotone increasing interval of cosine can be written as [2K π + π, 2K π + 2 π]
It can also be written as [2K π - π, 2K π]
Finding monotone interval of function y = 3cos (2x - π / 3)
The increasing range of cosx is [- π + 2K π, 2K π], and the decreasing range is [2K π, π + 2K π]
So let - π + 2K π
Because cosx increases on (2kpi pi / 2,2kpi) and decreases on [2kpi, 2kpi + pi / 2]
Therefore, according to relevant principles, the function increases when 2x pi / 3 belongs to (2kpi pi / 2,2kpi), and decreases when 2x pi / 3 belongs to [2kpi, 2kpi + pi / 2]
The range of solution x is monotone interval
Finally, we get that the original function increases on (kpi-pi / 12, KPI + pi / 6) and expands on [KPI + pi / 6, KPI + 5pi / 12]
Because cosx increases on (2kpi pi / 2,2kpi) and decreases on [2kpi, 2kpi + pi / 2]
Therefore, according to relevant principles, the function increases when 2x pi / 3 belongs to (2kpi pi / 2,2kpi), and decreases when 2x pi / 3 belongs to [2kpi, 2kpi + pi / 2]
The range of solution x is monotone interval
Finally, we get that the original function increases on (kpi-pi / 12, KPI + pi / 6) and decreases on [KPI + pi / 6, KPI + 5pi / 12] (k belongs to Z)
Function y = 3cos (2x tt / 3), X belongs to real number, in what interval is a decreasing function
How to do it
Y＝3cos（2x-∏/3）
＝3cos[2(x-∏/6)]
It is derived from the transformation of y = cosx
1. Move horizontally to the left Π / 6:
2. Horizontal compression 1 / 2:
3. Vertical stretch 3 times
The former coefficient is vertical stretching, which does not affect the change of domain; the common horizontal movement and compression directly affect the change of domain
Where cosx is decreasing on [2K Π, 2K Π + Π], that is, 2K Π ≤ x ≤ 2K Π + Π;
We can directly substitute the required function after cos as X of the above inequality, and then solve it
2k∏≤2x-∏/3≤2k∏+∏
2k∏+∏/3≤2x≤2k∏+4∏/3
k∏+∏/6≤x≤k∏+2∏/3
That is, the decreasing interval is: [K Π + Π / 6, K Π + 2 Π / 3] (K ∈ z)
The decreasing interval of function y = 3cos (- 2x + π / 4) x ∈ [- π, π]
When the function monotonically decreases, there are: 2K π, - 2x + π / 4 2K π + π (k is any integer),
In this paper, we sort out: - K π - (3 π / 8) x, - K π + (π / 8) and take its common interval with [- π, π],
That is to say, the function subtraction interval: [- π, (- 7 π / 8)] ∪ [(- 3 π / 8), (π / 8)] ∪ [(5 π / 8), π]
The decreasing interval of the function y = 3cos ((π / 3) - 2x) is
A.[kπ-（π/2）,kπ+（5π/12）] (k∈z)
B.[kπ+（5π/12）,kπ+（11π/12）]（k∈z)
C.[kπ-（π/3）,kπ+（π/6）]（k∈z)
D.[kπ+（π/6）,kπ+（2π/3）]（k∈z)
Choose C
First, the algebraic expression in brackets belongs to (0, π) closed interval,
Then the equation is solved to get [- (π / 3), + (π / 6)]
Because the period K π
So choose C
（2-cos^2x)(2+tan^2x)=(1+2tan^2x)(2-sin^2x)
Original equivalent to: (2-cos ^ 2x) / (2-sin ^ 2x) = (1 + 2tan ^ 2x) / (2 + Tan ^ 2x)
Left = (1 + sin ^ 2x) / (1 + cos ^ 2x)
Right = (COS ^ 2x + 2Sin ^ 2x) / (2cos ^ 2x + sin ^ 2x) = (1 + sin ^ 2x) / (1 + cos ^ 2x) = left
Proof of the original formula
（2-cos^2x)(2+tan^2x)=(1+2tan^2x)(2-sin^2x)
4-2cos²x+2tan²x-sin²x=2+4tan²x-sin²x- 2sin²xsin²x/cos²x
Simplify
1-cos & # 178; X = Tan & # 178; x-sin & # 178; xsin & # 178; X... expansion
（2-cos^2x)(2+tan^2x)=(1+2tan^2x)(2-sin^2x)
4-2cos²x+2tan²x-sin²x=2+4tan²x-sin²x- 2sin²xsin²x/cos²x
Simplify
1-cos²x=tan²x- sin²xsin²x/cos²x
sin²x=tan²x(1-sin²x)
sin²x=tan²x cos²x
sin²x=sin²x
The equation is always closed
Sin (2x) / 1 + cos (2x) = 3 / 4, solving the math problem of Tan (x / 2)
2sinxcosx/2cos^2x=3/4tanx=3/4tan2A=2tanA/（1-tanA^2）3/4=2tan(x/2)/(1-tan^2(x/2))3-3tan^2(x/2)=8tan(x/2)3tan^2(x/2)+8tan(x/2)-3=0(3tan(x/2)-1)(tan(x/2)+3)=0tan(x/2)=1/2tan(x/2)=-3
sin(2x)/1+cos(2x)=3/4
=tanx
tanx=2tan(x/2)/(1-tan(x/2)^2)=3/4
2tan(x/2)=3/4-3/4tan(x/2)^2
8tan(x/2)=3-3tan(x/2)^2
3tan(x/2)^2+8tan(x/2)-3=0
Tan (x / 2) = - 3 or 1 / 3
Oh, I forgot all about it
It is proved that: (sin 2x / (1-cos 2x)) · (SiN x / (1 + SiN x)) = Tan (π / 4-x / 2)
[sin 2x /(1-cos 2x)]·[sin x /(1+sin x)]=2sinx*cosx*sinx/[2(sinx)^2*(1+sinx)]=cosx/(1+sinx)=[(cosx/2)^2-(sinx/2)^2]/[(cosx/2)^2+(sinx/2)^2+2cosx/2*sinx/2]=[(cosx/2+sinx/2)*(cosx/2-sinx/2)]/[(cosx/2+si...
Given that the elliptic equation 3x + 4Y = 12, the straight line L passes through the right focus F of the ellipse and the slope is 1, the chord length of the intersection of the equations connected by straight lines can be obtained
c=1,F(1,0)
L:y=x-1
3x^2+4y^2=12
3x^2+4*(x-1)^2=12
7x^2-8x-8=0
x1+x2=8/7,x1*x2=-8/7
(y1-y2)^2=(x1-x2)^2=(x1+x2)^2-4x1*x2=288/49
(x1-x2)^2+(y1-y2)^2=576/49
Chord length = 24 / 7