Let f (x) = cosx + SiNx, whether a belongs to (0,90), make f (x + a) = f (x + 3a) constant, and prove Please tell me the process and ideas thank you

Let f (x) = cosx + SiNx, whether a belongs to (0,90), make f (x + a) = f (x + 3a) constant, and prove Please tell me the process and ideas thank you

F (x) = radical 2 sin (x + Π / 4)
Let 3a-a = 2 Π a = Π
If (0,90) denotes an angle, then it does not exist
If it's a real number, there's a problem

Let f (x) = cosx, prove that (cosx) '= - SiNx, and find f' (π / 6) and f '(π / 3)
Great Xia, please explain in detail

Cos (x + DX) - cosx = cosx cos DX SiNx sindx - cosx when DX - > 0, [cos (x + DX) - cosx] / DX = [cosx (COS DX-1) - SiNx sindx] / DX - > - SiNx sindx / DX - > - SiNx this is a literal proof. F '(π / 6) = - 1 / 2F' (π / 3) = - radical 3 / 2

Find the integral sign (1 / 2) x (e ^ x) (SiNx cosx) DX
Find the integral sign (1 / 2) x (e ^ x) (SiNx cosx) DX,

First put (e ^ x) (SiNx cosx) into the differential sign D, and change it into the integral sign 1 / 2) XD (e ^ x) (- cosx SiNx), and then distribute the integral

Integral: ∫ SiNx * SiNx / (1 + cosx * cosx) DX
Indefinite integral

Let t = TaNx, then x = arctant, DX = DT / (1 + T & # 178;), SEC & # 178; X = 1 + T & # 178;
So ∫ Sin & # 178; X / (1 + cos & # 178; x) DX = ∫ Tan & # 178; X / (1 + sec & # 178; x) DX
=∫t²/[(1+t²)(2+t²)]dt
=∫[2/(2+t²)-1/(1+t²)]dt
=√2∫d(t/√2)/[1+(t/√2)²]-∫dt/(1+t²)
=√ 2arctan (T / √ 2) - arctan + C (C is an integral constant)
=√2arctan(tanx/√2)-x+C.

Integral of (sinxcosx) / (1 + SiNx ^ 4)

Original formula = ∫ sinxd (SiNx) / [1 + (SiNx) ^ 4]
=(1/2)∫d(sin²x)/[1+(sin²x)²]
=(1 / 2) arctan (Sin & sup2; x) + C (C is an integral constant)

If f (SiNx + cosx) = sinxcosx, then the value of F (0) + F (1) is______ .

Because (SiNx + cosx) ^ 2 = sin ^ 2 (x) + cos ^ 2 (x) + 2sinxcosx = 1 + 2sinxcosx
So f (SiNx + cosx) = sinxcosx = [(SiNx + cosx) ^ 2-1] / 2
So f (x) = (x ^ 2-1) / 2
So f (0) + F (1) = - 1 / 2

Find f (x) given f (SiNx + cosx) = (SiNx + cosx) / (sinxcosx)

Let SiNx + cosx = t, then sinxcosx = (t-1) / 2, so f (T) = 2T / T-1, so f (x) =. Domain

It is known that f (x) = (SiNx + cosx) ^ 2 / 2 + 2sin2x cos ^ 2x

f(x)=(sin^2x+2sinxcosx+cos^2x)/2+2sin2x-(1+cos2x)/2=(1+2sinxcosx)/2+2sin2x-(1+cos2x)/2
=(5sin2x-cos2x)/2
Maybe it can be changed into other forms
Anyway, the method is more flexible

It is known that f (x) = (SiNx + cosx) &# 178 / 2 + 2sin2x cos & # 178; 2x
(1) (2) if f (x) = 2, - π / 4 < x < 3 π / 4, find the value of X

f(x)=(sinx+cosx)²/2+2sin2x-cos²2x?
What's the denominator? Is it two?
f(x)=(sinx+cosx)²/2+2sin(2x)-cos²(2x)
f(x)=(sin²x+2sinxcos+cos²x)/2+sin(2x)-[1-sin²(2x)]
f(x)=(sin²x+cos²x)/2+(1/2)sin(2x)+sin(2x)-1+sin²(2x)
f(x)=1/2+(3/2)sin(2x)-1+sin²(2x)
f(x)=sin²(2x)+(3/2)sin(2x)-1/2
From the definition of sine function, we can know that the definition domain of the function is: X ∈ (- ∞, ∞)
Let: u = sin (2x), obviously: - 1 ≤ u ≤ 1, substituting into the given function, we have:
f(u)=u²+(3/2)u-1/2
f'(u)=2u+3/2
1. Let f '(U) > 0, that is, 2U + 3 / 2 > 0
The solution is: u ＞ - 3 / 4, that is: - 3 / 4 ＜ u ≤ 1
2. Let f '(U) < 0, that is, 2U + 3 / 2 < 0
The solution is: u ＜ - 3 / 4, that is: - 1 ≤ u ＜ - 3 / 4
To sum up, there are:
When u ∈ (- 3 / 4,1], f (U) is a monotone increasing function;
When u ∈ [- 1, - 3 / 4), f (U) is a monotone decreasing function
When u = - 3 / 4, f (U) has a minimum: F (- 3 / 4) = - 3 / 4 & # 178; + (3 / 2) × (- 3 / 4) - 1 / 2 = - 17 / 16
In addition, f (- 1) = (- 1) &# 178; + (3 / 2) × (- 1) - 1 / 2 = - 1, f (1) = 1 & # 178; + (3 / 2) × 1-1 / 2 = 2
So: the range is: F (x) ∈ [- 1,2]
It is known that f (x) = 2, that is, f (U) = 2
u²+(3/2)u-1/2=2
2u²+3u-5=0
(2u+5)(u-1)=0
The solution is: U1 = - 5 / 2, U2 = 1
Because: - 1 ≤ u ≤ 1, u = - 5 / 2 is omitted
If u = 1 is substituted into the previous solution, the following is true:
sin(2x)=1
2X = 2K π + π / 2, where k = 0, ± 1, ± 2 , the same below
The solution is: x = k π + π / 4
Because: - π / 4 < x < 3 π / 4,
The final solution is: x = π / 4

It is known that f (x) = (SiNx + cosx) 22 + 2sin2x − cos22x. (1) find the domain of definition and value of F (x); (2) if f (x) = 2, − π 4 ＜ x ＜ 3 π 4, find the value of X

(1) because 1 + sin2x (1) because 1 + sin2x is not equal to 0, so it's the result of the following (1) because 1 + sin2x (1) (1) because 1 + sin2x2x (1) because 1 + sin2x (1) because 1 + sin2x (1) because 1 + sin2x2x is not equal to 0, so it's the result of the following (1) because 1 + sin2x2x is not equal to 0, so it's the result of the following (1) and 0 < 1 + sin2x2x ≤ 2, so, so f (x (x) ≥ 12) 12. So the definition domain is {x {x | x} x {x {x {x {x \\\\124; x} x} x} x} X - 12 because − π 4 < x < 3 π 4 Take − π 2 ＜ 2x ＜ 3 π 2 so 2x = − π 6 or 2x = 7 π 6 so x = − π 12 or x = 7 π 12 (6 points)