Given cos (x-4 points, PAI) = 1 / 4, the value of sin2x is?

Given cos (x-4 points, PAI) = 1 / 4, the value of sin2x is?

cos(π/4-x)=cos(x-π/4)=1/4
Then sin2x
=cos(π/2-2x)
=2cos²(π/4-x)-1
=-7/8
It is known that 0 ＜ x ＜ Pai / 2, which is reduced to: LG [cos xtan x + 1-2sin ^ 2 (PAI / 2)] + LG [followed by 2cos (x-pai / 4)] - LG (1 + sin2x)
It is known that 0 ＜ x ＜ Pai / 2, which is reduced to LG [cos xtan x + 1-2sin ^ 2 (PAI / 2)] + LG [follow 2cos (x-pai / 4)] - LG (1 + sin2x)
lg[cosxtanx+1-2sin²(x/2)]+lg[√2cos(x-π/4)]-lg(1+sin2x)
=lg(sinx+cosx)+lg(cosx+sinx)-lg[(sinx+cosx) ²]
=lg1=0
If the equation x2 | a | 1 + y2a + 3 = 1 represents an ellipse with focus on the x-axis, then the value range of the real number a is______ .
∵ equation x2 | a | 1 + y2a + 3 = 1 represents the ellipse with focus on the x-axis, ∵ a | 1 | a + 3 ﹥ 0, and the solution is - 3 | a | 2
If the equation ((x ^ 2) / M) + ((y ^ 2) / 2) = 1 represents an ellipse with focus on the X axis, then the value range of the real number m is?
M>2
As shown in the figure, in the rectangular coordinate system, O is the coordinate origin. The image with known inverse scale function y = K / X (k > 0) passes through point a (2, m), passes through point a as ab ⊥ X axis at point B,
And the area of △ AOB is 1 / 2
(1) The summation of K;
(2) Point C (x, y) is on the image of inverse scale function y = K / x, and the value range of function value y is obtained when 1 ≤ x ≤ 3
(3) The line L passing through the origin o intersects with the image of inverse scale function y = K / X at P and Q. try to write out the minimum PQ length of the line according to the image
1.△AOB=2*m/2=1/2
m=1/2
k=m/2=1/4
2.y=k/x=1/4x
1/4≦y≦1/12
3.(√5)/2
Sabc=(1/2)*2*m=1/2
M = 1 / 2 and K = 1 for a (2,1 / 2) equation
When x = 1, y = 1. When x = 3, y = 1 / 3, so the range of Y is 1 / 3 to 1
The shortest distance is 2 times the root 2. That is, when l equation is y = X. Follow up: can the third question be more detailed
In the same plane rectangular coordinate system, the area of △ AOB is ()
A. 2B. 6C. 10D. 8
From the problem meaning: y = − 8xy = − x + 2, the solution is x = − 2Y = 4, x = 4Y = − 2; a (- 2,4), B (4, - 2). As shown in the figure: because the intersection coordinate C (0,2) of the first-order function y = - x + 2 and Y axis, OC = 2; so s △ AOB = s △ AOC + s △ cob = 12 × 2 × 2 + 12 × 2 × 4 = 6, so choose B
In a plane rectangular coordinate system, there is a first-order function and an inverse proportion function intersecting at two points. Why are these two points symmetrical about the origin? It's best to have a graphic explanation,
It's a positive scale function
To be exact, a positive scale function and an inverse scale function intersect at two points, which will be symmetrical about the origin
The inverse scale function is an odd function, symmetrical about the origin, while the positive scale function passes through the origin, so no matter how the slope changes, as long as the line passing through the origin intersects with the scale function at two points, these two points will be symmetrical about the origin!
Don't know how to ask me!
Amend as follows
If you want to prove it, it is as follows:
Idea: the intersection of two functions is actually the root of the equation when they are equivalent. Then you just need to get that the root of the equation must be symmetric.
Positive scale function: F (x) = ax
Inverse scale function: F (x) = B / X
Equivalent equation: ax = B / X
So ax ^ 2-B = 0
The equation obviously has two symmetric roots. Or the function f (x) = ax ^ 2-B and X-axis intersection... Expansion
Amend as follows
If you want to prove it, it is as follows:
Idea: the intersection of two functions is actually the root of the equation when they are equivalent. Then you just need to get that the root of the equation must be symmetric.
Positive scale function: F (x) = ax
Inverse scale function: F (x) = B / X
Equivalent equation: ax = B / X
So ax ^ 2-B = 0
The equation obviously has two symmetric roots. In other words, the function f (x) = ax ^ 2-B is symmetric at the intersection of X axis. There are also symmetric intersections of two functions. Put it away
y=ax
y=b/x
Intersection means that the two equations can be simultaneous, eliminating y to find the intersection point
The results are as follows
x^2=b/a
The solution of X is a pair of opposite numbers, and these two functions are odd functions, so the (x, y) point is symmetric about the origin.
Both positive scale function and inverse scale function are odd functions
Their figures are symmetrical about the origin
So their intersections are also symmetrical about the origin
The application of inverse proportion function
The total amount of earthwork to be transported in the new reservoir is 4 × 10 ^ 4m & sup 3;, and a transportation company undertakes the task of transporting earthwork in this project
(1) What is the functional relationship between the average daily workload V (M & sup3; / day) and the time t (days) required to complete the transportation task?
(2) How many days will it take for the transportation company to send 20 trucks to transport 100 M & sup3; of earth and stone every day?
(3) Eight days after the project, due to the need of schedule, the remaining transportation task must be completed four days in advance. How many more trucks will the company need to send to complete the task on time?
Let v = K / T according to the meaning: k = 4 × 10 ^ 4, so v = 4 × 10 ^ 4 / T (t ＞ 0). Second question: when v = 100m & sup3; 100 = 4 × 10 ^ 4 / T, so t = 400 days (if a car transports 100m & sup3; every day, then when v = 2000, t = 20 days). Third question: 20-8 = 12 days, 4 × 10 ^ 4-8 * 20 * 100 = 24000m & sup3
Second grade mathematics problem [image and properties of inverse proportion function]
In the same plane rectangular coordinate system, if there is an intersection (2, - 1) between the image of positive scale function y = K1X and inverse scale function y = K2 / x, then K1=______ ,k2=______ The graphs of these two functions are shown in section______ There is also an intersection in the quadrant whose coordinates are_____ .
k1=-1/2
k2=-2
Beta Quadrant
-2,1
Mathematics in grade two. About inverse proportion function
The line L parallel to the line y = x does not pass through the fourth quadrant and intersects the image of the function y = 3 / X (x > 0) with the point A. through the point a, make ab ⊥ Y axis at the point C, and the perimeter of the quadrilateral ABCD is 8. The analytical formula of L is obtained
Sunian tea:
I think your title is wrong
Through point a, make ab ⊥ Y axis at point C, the perimeter of quadrilateral ABCD is 8?
It should be: ab ⊥ Y axis at point B through point a, AC ⊥ X axis at point C, the perimeter of the quadrilateral aboc is 8
Let the coordinates of point a be (x, y)
2 (x + y) = 8
x+y=4
y=-x+4
That is, the coordinates of point a are (x, - x + 4)
∵ point a is on the image of the function y = 3 / X (x > 0) (substituting the coordinates of the point into the analytical formula)
∴-x+4=3/x
x(-x+4)=3
-x²+4x-3=0
x²-4x+3=0
(x-1)(x-3)=0
The solution is x = 1 or x = 3
‖ y = 3 or y = 1
The coordinates of point a are (1,3) or (3,1)
∵ the line L is parallel to the line y = X
Let the analytic expression of line l be y = x + B
The line L does not pass through the fourth quadrant
∴b≥0
Substituting the coordinates of point a into the line y = x + B, we get
1 + B = 3, B = 2
Or 3 + B = 1, the solution is b = - 2, not satisfying B ≥ 0, rounding off
∴b=2
The analytic formula of line L is y = x + 2
BCD these points?
Let the line l be y = KX + B, the parallel line y = x, so l k = 1, without passing through the fourth quadrant, so b > 0 of the line.............
The title seems incomplete ~!!???