It is known that the functions f (x) = asin (KX + π / 3) and φ (x) = btan (KX - π / 3), k > 0, and a; B belong to R. the sum of the minimum positive periods of the two functions is 3 π / 2, and f (π / 2) = φ (π / 2), f (π / 4) = - √ 3 φ (π / 4) + 1, find the analytic expressions of two functions

It is known that the functions f (x) = asin (KX + π / 3) and φ (x) = btan (KX - π / 3), k > 0, and a; B belong to R. the sum of the minimum positive periods of the two functions is 3 π / 2, and f (π / 2) = φ (π / 2), f (π / 4) = - √ 3 φ (π / 4) + 1, find the analytic expressions of two functions

According to the meaning of the title
∵k＞0∴2π/k+π/k＝3π/2
So k = 2
And ∵ f (π / 2) = φ (π / 2), f (π / 4) = - √ 3 φ (π / 4) + 1
∴a-2b＝0①,a+2b＝2②
A = 1, B = 1 / 2 for simultaneous solutions
So f (x) = sin (2x + π / 3), φ (x) = 1 / 2tan (2x - π / 3)
Note: for reference only!

It is known that the minimum positive period of the function f (x) = asin ω x + bcos ω x) defined on R (where ω > 0, a > 0, b > 0) is π
If f (π / 4) = root 3 and the maximum value of F (x) is 2, write the expression of F (x)?
I mainly don't know how to use "and the maximum value of F (x) is 2", so I can't get B

For the function f (x) = asin ω x + bcos ω x + 1, the minimum positive period is π, the maximum is 3, and f (π 6) = 3 + 1 (AB ≠ 0), the analytic expression of F & nbsp; (x) is obtained

F (x) = asin ω x + bcos ω x + 1 = A2 + b2sin (ω x + ϕ) + 1, the minimum positive period is π, the maximum is 3, and f (π 6) = 3 + 1 (AB ≠ 0), so 2 π ω = π, ω = 2, A2 + B2 + 1 = 3, asin π 3 + bcos π 3 + 1 = 3 + 1, a = 1, B = 3, so f (x) = sin2x + 3cos2x + 1

Given the functions f (x) = asin (KX + π / 3), G (x) = btan (KX - π / 3), k > 0, the sum of their periods is 3 π / 2, and f (π / 2) = g (π / 2), f (π / 4)=-

T1 + T2 = 2 π / K + 2 π / k = 4 π / k = 3 π / 2, k = 8 / 3
f(x)=asin((8/3)x+π/3),g(x)=btan((8/3)x-π/3)
f(π/2)=-(1/2)a=g(x)=0,a=0,
f(π/4)=0

It is known that the sum of the minimum positive periods of F (x) = asin (ω χ + π / 3), G (x) = btan (ω χ - π / 3) (ω > 0) is 3 π / 2, and f (π / 2) = g (π / 2)
F (π / 4) + √ 3G (π / 4) = 1, find the analytic expression of F (x) g (x)
Finding the analytic expressions of F (x) and G (x)

From F (π / 2) = g (π / 2) to - asin (π / 3) = - btan (π / 3) to a = 2BF (x) = 2bsin (2 χ + π / 3), G (x) = btan (2 χ - π / 3) (ω > 0) from F (π / 4) + √ 3G (π / 4) = 1 to 2bcos π / 3 + √ 3

Given the function f (x) = | SiNx |, if the image of the function f (x) = | SiNx |, and the line y = KX (k > 0) have and only have three common points, the abscissa of the three common points is the same
It is proved that cos α / (sin α + sin 3 α) = (1 + α ^ 2) / 4 α

It can be seen from the graph that the line is tangent to the function image at the third intersection (the function on this interval is f (x) = - SiNx), so k = - Sina / a = cosa, because sin ^ 2 (a) + cos ^ 2 (a) = 1, so cosa = 1 / √ (a ^ 2 + 1), Sina = A / √ (a ^ 2 + 1), so cosa / (Sina + sin3a) = cosa / (2Sin (2a) COSA) = 1 / (4sina

Given the function f (x) = absolute value SiNx, if the function and y = KX (k > 0) have only three common points, the maximum abscissa of the common point is a
Verification: cosa / (Sina + sin3a) = (1 + A & # 178;) / 4A
Thank you for your advice

Obviously, the origin (0,0) is a common point
And there is a common point on the interval (π / 2, π)
And there is a common point on the interval (π, 3 π / 2)
The common point on the interval (π, 3 π / 2) is formed by the tangent of the line y = KX and f (x)
It is easy to know that the abscissa of this common point is the largest
That is Ka = | Sina|
Because π

If the image of function f (x) = | SiNx | and the line y = KX have only three common points, and their abscissa are α, β, γ (α)
Fifteenth subjects of the first mock exam of mathematics in Hefei in 2012

You're wrong. Because the straight line is tangent to the curve, γ won't be equal to & nbsp; 3 π / 2 & nbsp; and it's a little smaller than & nbsp; 3 π / 2 & nbsp; (γ is about 4.50, 3 π / 2 is about 4.71)

If the function f (x) = | SiNx | has only three common points with the line y = KX,
And the abscissa are α, β, and γ (α < β < γ). The following conclusions are given: ① k = - cos γ; ② γ ∈ (π, 3 / 2 π); ③ γ = Tan γ; ④ sin γ = 2 γ / (1 + γ 2)——

As shown in the figure, we can see that f (x) = | SiNx | is a periodic function with period π. When f (x) and the line y = KX have only three common points, there are γ∈ (π, 3 π / 2) & nbsp; & nbsp; & nbsp; & nbsp; = & gt; & nbsp; 2. Correctly, when γ∈ (π, 3 π / 2), there are f (x) = - SiNx, F & 39; (x) = - cosx at the third intersection

By observing the images of the functions y = SiNx and y = KX at 0 ≤ x ≤ π / 2, we obtain an inequality that SiNx ≥ KX is constant for 0 ≤ x ≤ π / 2. We use this inequality to solve the problem: for any acute angle △ ABC, Sina + SINB + sinc > m holds, then the maximum value of M is?

The inequality can be: SiN x > = 2 / π * x, 0