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Given that the line x = π / 6) is a symmetric axis of the image of the function y = asinx bcosx, then what is the symmetry axis of the image of the function y = bsinx + acosx?
X = π / 6 is the axis of symmetry, the principle of left plus right subtraction, y = asinx bcosx shifts π / 2 unit to the left, y = asin (x + π / 2) - bcos (x + π / 2) = acosx + bsinx, π / 6 - π / 2 = - π / 3. Since sin is x, the axis of symmetry is x = - π / 3 + K π
If x = π / 6 is a symmetric axis of the function y = asinx bcosx, then y = bsinx acosx
The maximum or minimum value of y = asin π / 6-bcos π / 6 is determined by asinx + bcosx = √ A2 + b2sin (x + F), and F is determined by TANF = B / A
So y = asinx bcosx = √ A2 + b2sin (x + F), TANF = - B / A, when x = π / 6, y takes the maximum (small) value. Then π / 6 + F = π / 2 + 2K π, the solution f = π / 3 + 2K π
Then y = bsinx acosx = √ A2 + b2sin (x + F1), tanf1 = - A / B, thus F + F1 = 90 ° can be obtained
Therefore, y = bsinx acosx = √ A2 + b2sin (x + π / 6-2k π) = √ A2 + b2sin (x + π / 6-2k π) = √ A2 + b2sin (x + π / 6)
If x + π / 6 = π / 2 + 2K π is required, x = π / 3 + 2K π can be obtained by solving x + π / 6 = π / 2 + 2K π
If {SiNx} is ＋ OSX ∈, then {x} is {2}, then {x} is {0} x ∈ x}? When x = 3 π / 2, there is a minimum value - 2. Find the maximum value of y = asinx-b, and find the corresponding value of X
f=3asinx+bcosx
It can be regarded as (B, 3a) dot multiplication (cosx, SiNx)
When two vectors are collinear, there is a maximum
Because Fmin is obtained at x = 3 π / 2, Fmax is obtained at x = 3 π / 2 - π = π / 2
Fmin = - (9a ^ 2 + B ^ 2) ^ 0.5 = - 2 and B = 0 (because x = 3 π / 2 is on the Y axis), so a = 2 / 3
The original formula is y = 2 / 3 SiNx, obviously the maximum value is 2 / 3 - 2 / 3, corresponding to x = π / 2, 3 π / 2
Reason for finding the maximum and minimum value of function y = asinx + bcosx (a, B are positive numbers)
y=asinx+bcosx
Y = radical (A2 + B2) sin (a + C)
The maximum value is the root sign (A2 + B2)
The minimum value is - radical (A2 + B2)
Given that a, B satisfy A2 + b2-4a + 3 = 0, the maximum value of function f (x) = asinx + bcosx + 1 is denoted as φ (a, b), then the minimum value of φ (a, b) is () A. 1 B. 2 C. 3+1 D. 3
∵ real numbers a, B satisfy A2 + b2-4a + 3 = 0, 2 + B2 = 1, denotes a circle with (2,0) as the center and 1 as the radius. The maximum value of the function f (x) = asinx + bcosx + 1 is φ (a, b) = A2 + B2 + 1, and its geometric meaning is the distance from the origin to the point (a, b) plus 1
If the image of the function f (x) = asinx bcosx is symmetric with respect to the straight line x = π / 4, the parity and symmetry center of the function f (3 / 4 π - x) are determined
The image of F (x) = asinx bcosx is symmetric with respect to the line x = π / 4,
Then: F (0) = f (π / 2)
Namely: - B = a
Then: F (x) = a (SiNx + cosx) = √ 2A * sin (x + π / 4)
f(3π/4-x)=√2asin(3π/4-x+π/4)
=√2asin(π-x)
=√2asinx
Is an odd function, and the center of symmetry is (K π, 0)
Let f (x) = 2 sin 4 / X cos 4 / X - 2 root sign 3 sin square 4 / 4 x + root sign 3. Find the minimum positive period of function f (x)? ruti
f(x)=2sinx/2cosx/2-2√3sinx/4+√3
=sinx/2-2√3[1-cosx/2)/2]+√3
=sinx/2-√3+√3cosx/2+√3
=sinx/2+√3cos2x
=2sin(x/2+π/3)
So the minimum positive period T = 2 π / (1 / 2) = 4 π
Given the function f (x) = root sign 3sin (2x PIA / 6) + 2Sin ^ 2 (x-pia / 12), X ∈, find the set of X that makes the function get the maximum value thank you
F (x) = √ 3sin (2x - π / 6) + 2Sin 2 (x - π / 12) = √ 3sin (2x - π / 6) + 1-cos (2x - π / 6) = 2Sin [(2x - π / 6) - π / 6] + 1 = 2Sin (2x - π / 3) + 1 makes the function get the maximum value, that is, 2x - π / 3 = 2K π + π / 2x = {x| K π + 5 π / 12, K ∈ Z}
Given the function f (x) = 2sinx / 4cosx / 4-2 radical sign 3sin square X / 4 + root sign 3, X belongs to r.1. Find the minimum positive period and maximum value of F (x) If G (x) = f (x + 3 / π), judge the parity of function g (x) and explain the reason
f(x)=2sinx/4cosx/4+√3[1-2(sinx/4)^2]
=sin(x/2)+√3cos(x/2)
=2sin(x/2+π/3)
The minimum positive period T = 2 π / (1 / 2) = 4 π
F (x) min = - 2 F (x) max = 2
g(x)=2sin[(x+π/3)/2+π/3)=2sin(x/2+π/2)=2cos(x/2)
Because g (- x) = 2cos (- X / 2) = 2cos (x / 2) = g (x)
So g (x) is an even function
Let P: X belong to [π / 4, π / 2], Q belong to | f (x) - M | less than 3, P is Q A sufficient condition is given to find the value range of real number M
f(x)=2sin^2(π/4+x)-(√3)cos2x-1=1-cos(π/2+2x)-(√3)cos2x-1=sin2x-(√3)cos2x=2sin(2x-π/3),x∈[π/4,π/2],2x-π/3∈[π/6,2π/3],sin(2x-π/3∈[1/2,1],f(x)∈[1,2],|f(x)-m|
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