Let f (x) = √ 3 / 2 - √ 3sin ^ 2wx sinwxcoswx, and the distance from a symmetry center of the image to the nearest symmetry axis is π / 4 1. Find w 2. Find the maximum and minimum of F (x) on [π, π / 2]

Let f (x) = √ 3 / 2 - √ 3sin ^ 2wx sinwxcoswx, and the distance from a symmetry center of the image to the nearest symmetry axis is π / 4 1. Find w 2. Find the maximum and minimum of F (x) on [π, π / 2]

f(x)=√3/2-√3sin²wx-sinwxcoswx
=(√3/2)*(1- 2sin² wx）- (1/2)*2sinwxcoswx
=(√3/2)*cos(2wx）- (1/2)*sin(2wx)
=cos(2wx)*cos(π/6) - sin(2wx)*sin(π/6)
=cos(2wx + π/6)
If the distance from a symmetry center of the function image to the nearest symmetry axis is known to be π / 4, then there are:
Minimum positive period T = 2 π / (2W) = 4 * π / 4
The solution is: w = 1
Then the analytic expression of the function can be written as: F (x) = cos (2x + π / 6)
If π / 2 ≤ x ≤ π, then: π ≤ 2x ≤ 2 π
Then there are: 7 π / 6 ≤ 2x + π / 6 ≤ 13 π / 6
So when 2x + π / 6 = 2 π, that is, x = 11 π / 12, the maximum value of the function is 1;
When 2x + π / 6 = 7 π / 6, that is, x = π / 2, the minimum value of the function is - (√ 3) / 2
Given Sina + cosa = 3 / 5, a is the second quadrant angle, find Tana
Because Sina + cosa = 3 / 5,
So (Sina + COSA) &# 178; = (Sina) &# 178; + (COSA) &# 178; + 2sina * cosa
=1+2sina*cosa=9/25
So 2sina * cosa = (9-25) / 25 = - 16 / 25
So (Sina COSA) & # 178; = (Sina) & # 178; + (COSA) & # 178; - 2sina * cosa
=1+16/25=41/25
Because a is the second quadrant,
So Sina > 0, cosa0
So Sina cosa = √ 41 / 5
So [sincosa] = (3 + / sincosa) + (a] = (3 + / sincosa) + (3 + / sincosa)]
So cosa = [(Sina + COSA) - (Sina COSA)] / 2 = (3 - √ 41) / 10
So Tana = Sina / cosa = - (25 + 3 √ 41) / 16
Let us know that y is a positive scale function with respect to Z, and the scale coefficient is 2; Z is an inverse scale function with respect to x, and the scale coefficient is - 3
(1) Write out the analytic expressions of the positive proportion function and the inverse proportion function of this function
(2) Find the value of X and y when z = 5
(3) Find the analytic expression of Y with respect to X. is this function an inverse proportion function?
There is a great reward
One
y=2z
z=-3/x
Two
When z = 5
y=2*5=10
5=-3/x
x=-3/5
Three
z=y/2
z=-3/x
So y / 2 = - 3 / X
y=-6/x
It's an inverse scale function
Given that a focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 / 2 = 1 is (2,0), then the equation of the ellipse is
The focus of the ellipse is on the x-axis
c=2,b²=2
Then: A & # 178; = B & # 178; + C & # 178; = 6
So the equation of ellipse is:
X square / 6 + y square / 2 = 1
sina>0,tana
Sina > 0 first and second quadrants
tana
It is known that y is the positive proportional function of Z, and the proportional coefficient is - 3; Z is the inverse proportional function of X, and the proportional coefficient is 4. Try to judge whether y is the inverse proportional function of X? If so, ask for the proportional coefficient
Y is the inverse function of X
According to the meaning of the title, y = - 3Z, z = 4 / X
So, y = - 3 (4 / x) = - 12 / x,
The scale factor is - 12
From the meaning of the question: y = 2Z z = - 3 / X when z = 5, x = - 3 / 5, y = 10, y = 2Z = 2 * (- 3 / x) = - 6 / x, that is, y = - 6 / X is an inverse proportional function. 1.y=2z z=-
It's - 12
Because y is a positive scaling function with respect to Z, the scaling factor is - 3,
So y = - 3Z
And because Z is the inverse scale function of X, the scale factor is 4
therefore
z=4/x
Bring z = 4 / X to y = - 3Z
Y = - 12
The scale factor is - 12
Given that the center of the ellipse is at the origin, the focus is on the x-axis, the eccentricity is 1 / 2, and the focal length is 8, what is the equation of the ellipse?
The focus is on the x-axis and the eccentricity is 1 / 2,
Then C / a = 1 / 2
The focal length is 8, which means 2C = 8
So C = 4, a = 8
Then B ^ 2 = a ^ 2-C ^ 2 = 48
So the elliptic equation is x ^ 2 / 64 + y ^ 2 / 48 = 1
Given that a is the fourth quadrant angle, Tana = - 5 / 12, then cosa =?
A is the fourth quadrant angle and the cosine is positive,
secA=√[1+(tanA)^2]=13/12,
cosA=12/13.
1、 Find out the function y = - X & # 178; + 2x + 3
In the range (1) x ∈ 0 ≤ x ≤ 3 (2) x ∈ {- 1, - 2,0,1,2} 2. Draw the following function image f (x) x & # 178; + 4x, X ≥ 0; - X & # 178; + 4x, x < 0 (this is a piecewise function)
Draw the following function image y = | X & # 178; - 2x | y = x & # 178; - | 2x | (quick sketch is OK)
Y & nbsp; = & nbsp; - (X & nbsp; - & nbsp; 1) & # 178; + & nbsp; 4, & nbsp; axis of symmetry X & nbsp; = & nbsp; 1, & nbsp; opening downward, & nbsp; vertex (1, & nbsp; 4) (1) x & nbsp; = & nbsp; 1, & nbsp; Y & nbsp; = & nbsp; 4x & nbsp; = & nbsp; 3, & nbsp; Y & nbsp; = & nbsp; 0 range: & nbsp
The standard equation of ellipse with focus on x-axis, focal length equal to 4 and passing through point P (3, − 26) is______ .
Let the equation of ellipse be x2a2 + y2b2 = 1 (a ＞ B ＞ 0) ∵ focal length equal to 4, and the ellipse passes through the point P (3, − 26) ∵ C = A2 − B2 = 232a2 + (− 26) 2B2 = 1. The solution is A2 = 36, B2 = 32 (rounding off). Therefore, the standard equation of ellipse is x236 + y232 = 1. So the answer is: x236 + y232 = 1