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The minimum positive period of function f (x) = TaNx (COS (x / 2) ^ 4-sin (x / 2) ^ 4)
f(x)=tanx[cos^4(x/2)-sin^4(x/2)]
=tanx{[cos^2(x/2)+sim^2(x/2)][cos^2(x/2)-sin^2(x/2)]}'
=tanx*cosx
=sinx
The minimum positive period of F (x) is 2 π
If the known function f (x) satisfies f (TaNx) = 1 / sin * 2xcos * 2x, then the analytic formula of F (x) is Why can't we change the one above into sin * 2x + cos * 2x and divide cos * 2x up and down?
Because the value of cosx ^ 2 is uncertain, we can't divide cos * 2x directly, let TaNx = T1 / sin * 2xcos * 2x = (SiNx ^ 2 + cosx ^ 2) * (SiNx ^ 2 + cosx ^ 2) / 2sinxcosx (cosx ^ 2-sinx ^ 2) and divide cosx ^ 2 * cosx ^ 2 to get f (TaNx) = (TaNx ^ 2 + 1) (TaNx ^ 2 + 1) / 2tanx (1-tanx ^)
The known function f (x) = (1 + 1 / TaNx) sin ^ 2x + m sin (x + π / 4) sin (x - π / 4)
sin(x+π/4)sin(x-π/4) = [cos(π/2)-cos(2x)] /2=-cos(2x) /2(1+1\tanx)sin^2x=[(sinx+cosx)/sinx]*sin^2x=sinx(sinx+cosx)=sin^2x+sin2x/2=(1-cos2x)/2+sin2x/2f(x)=sin2x/2+(-m-1)/2*cos2x+1/2
F (x) = (LNX) (TaNx) e ^ sin ^ 2x, then what function is f (x)
F (x) is an elementary function
Let f (x) be a second-order differentiable function, and f (TaNx) = 1 + sin ^ 2x / cos ^ 2x, find f '' (x) Yes (1 + sin ^ 2 x) cos ^ 2 x
f(tanx)=1+tan^2x
So f (x) = 1 + x ^ 2
f'(x)=2x
f''(x)=2
The known function f (x) = sin ^ Wx + √ 3 coswx.cos (π / 2-wx) (W > 0) and the function y = f (x), the distance between two adjacent symmetrical axes is π 2 (1) find the value of W and monotonically increasing interval of F (x) (2) in △ ABC, a, B, C are opposite sides of angles a, B and C respectively, if a = radical 3, B = radical 2, f (a) = 1, find angle C
Cos (π / 2-wx) = sin (Wx), so f (x) = sin ^ 2wx + root 3coswx sin (Wx) so = half (root 3 + 2) times sin ^ 2wx, because the distance between two adjacent symmetry axes is π
So w = 1) find the value of W and the monotone increasing interval of F (x)
Find the angle a by F (a) = 3
The function f (x) = sin ^ 2wx + radical 3coswx cos (Wu / 2) is known (1) (2) in the triangle ABC, a, B, C are the opposite sides of angles a, B and C respectively. If a = root 3, B = root 2, f (a) = 3 / 2, find the angle c quickly and send it to me. Please,
In this paper, we get the following results: w (x) = (1-cos2wx) / 2 + √ 3sinwxsin (Wx + π / 2) = (1-cos2wx) / 2 + √ 3sinwx x = 1 / 2-1 / 2 * cos2wx + √ 3 / 2 * sin2wx = 1 / 2 + (3 / 2 * sin2wx-1 / 2 * cos2wx = 1 / 2 + (3 / 2 * sin2wx-1 / 2 * cos2wx) = 1 / 2 + sin (2wx-π / 6) t = 2 π / (2W) = π, we get w = 1, so f (x) = 1 / 2 + sin (2x 2x 2X 2x 2x, so f (x) = 1 / 2 + sin2wx = 1 / 2 + sin (2x 2x- π / 6)
Let f (x) = radical 3 * cos ^ 2 * Wx + sinwxcoswx + a (where w > 0, alpha belongs to R), and the abscissa of the first high point of the image of F (x) on the right side of the Y axis is Wu / 6 (1) Find the value of W (2) If the minimum value of F (x) on the interval [- Wu / 3,5 μ / 6] is the radical 3, find the value of A
W = 0.5 A = root of two plus three plus one
Given the vector a = (Radix 3sinwx, coswx), B = (coswx, - coswx) (W > 0), and the function f (x) = AB + 1 / 2, the distance between the two adjacent symmetry axes is π / 4 (1) : find the monotone increasing interval of function f (x) (2) : if cosx > = 1 / 2, X ∈ (0, π), and f (x) = m has and only one real root, then find the value of real number M
Given the vector a = (Radix 3sinwx, coswx), B = (coswx, - coswx) (W > 0), and the function f (x) = AB + 1 / 2, the distance between the two adjacent symmetry axes is π / 4
(1) : find the monotone increasing interval of function f (x)
(2) If cosx > = 1 / 2, X ∈ (0, π), and f (x) = m has and only one real root, then the value of real number m is obtained. (1) analysis: ∵ vector a = (Radix 3sinwx, coswx), B = (coswx, - coswx) (W > 0)
And ∵ the distance between two adjacent symmetry axes of the image with F (x) = AB + 1 / 2 is π / 4
∴f(x)=a.b+1/2=√3sinωxcosωx-cos²ωx+1/2=√3/2sin2ωx-1/2cos2ωx=sin(2ωx-π/6)
∴T/2=π/4==>T=π/2==>2ω=2π/(π/2)=4
∴f(x)=sin(4x-π/6)
The monotonic increasing interval of function f (x) is as follows:
-π/2+2kπ＜=4x-π/6＜=π/2+2kπ==>kπ/2-π/12<=x<=kπ/2+π/6
(2) Analysis: ∵ cosx > = 1 / 2, X ∈ (0, π), and f (x) = m has and only one real root
This question has a question, please check the title
When - 1 < = m < = 1, X ∈ (0, π), f (x) = m has at least two roots
Vector a = (coswx + Radix 3sinwx, 1), B = (f (x), coswx), where w > 0, and a / / B, the distance between adjacent symmetry axes of the image of F (x) is 3 / 2 π (1) Find the value of W (2) find the symmetric axis equation and monotone interval of F (x)
(1) ∵ vector a ⊥ vector b
ν vector a · vector b = 0
∵-f(x)+(coswx+√3sinwx)coswx=0
f(x)=(coswx+√3sinwx)coswx
=Cos ^ 2wx + √ 3sinwxcoswx
=1/2(1+cos2wx)+√3/2sin2wx
= 1/2+1/2cos2wx+√3/2sin2wx
= sin(2wx+π/6）+1/2
And ∵ f (x), the distance between two adjacent symmetry axes is 3 π / 2
∴T=3π
∴2π/2w=3π
∴w=1/3
（2）f(x）= sin(2/3x+π/6）+1/2
π/2+2kπ≤2/3x+π/6≤3π/2+2kπ,k∈Z
The solution is: π / 2 + 3K π ≤ x ≤ 2 π + 3K π, K ∈ Z
The monotone decreasing interval of function f (x) on [- 2 π, 2 π] is [- 2 π, - π] ∪ [π / 2,2 π], K ∈ Z
RELATED INFORMATIONS
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