TaNx = 5 / 12, tany = 4 / 3. X belongs to 0,90, y belongs to 180270, find sin (x + y)

TaNx = 5 / 12, tany = 4 / 3. X belongs to 0,90, y belongs to 180270, find sin (x + y)

tan(x+y)={tanx+tany]/(1-tanxtany)=(5/12+4/3)/(1-5/12*4/3)=(7/4)/(4/9)=63/16
Another 180 degrees

What is the product of the maximum and minimum of y = tanx-tanx ^ 3 / (1 + 2tanx ^ 2 + TaNx ^ 4)

y = (tanx - tan³x)/(1 + 2tan²x + tan^4x)= (tanx - tan³x)/(1 + tan²x)²= (tanx - tan³x)/(sec²x)²= (tanx - tan³x) * cos^4x= sinxcos³x - sin³xcosx= (...

The function y = - TaNx ^ 2 + 2tanx, X belongs to [π / 4, π / 3]. The maximum and minimum values of y = - TaNx ^ 2 + 2tanx, X belong to [π / 4, π / 3] are?

Y = - TaNx ^ 2 + 2tanx, let TaNx = B, because x belongs to [π / 4, π / 3], so B belongs to [1, radical 3]
y=-b^2+2b=-(b-1)^2+1
So, the minimum value of Y is - 3 + 1 = - 2
Maximum 1

Given that a function y = KX + B passes through points (1,2) and (- 1,8), ① find the analytic expression of the function, ② judge whether the point (2, - 1) is in the function
It is known that the linear function y = KX + B passes through points (1,2) and (- 1,8)
① Find the analytic expression of the function
② Judge whether the point (2, - 1) is on the function image

1、
Put two points in
Then 2 = K + B
8=-k+b
Add
10=2b
b=5
k=2-b=-3
So y = - 3x + 5
2、
x=2
y=-6+5=-1
So (2, - 1) on the function image

Once the function image passes through point (2,1), (1,3) 1, judge whether point a (2,3) is on the function image. 2, answer when x according to the image=-------

Let the relation of a function be y = KX + B
Set points (2,1), (1,3) to
Substituting
Then: 1 = 2K + B 3 = K + B
The solution is: k = - 2, B = 5
Then y = - 2x + 5
Substituting point (2,3) into relation
-2*2+5≠3
The point a is not on the function image
2. When X0
When x > 2, y

Given the square of y = negative 2x, how to translate the function image to pass through (0,0) (1,6) points

Let y = - 2 (X-H) ^ 2 + K,
Substituting (0,0), (1,6) to get,
-2h^2+k=0,
-2(1-h)^2+k=6,
h=2,k=8,
Move up 8 units and right 2 units

If the image of a function y = 5x + m passes through a point (- 1,0), then M=______ ．

If the image of the first-order function y = 5x + m passes through (- 1,0), substituting (- 1,0) into the analytic formula, we get - 5 + M = 0, and the solution is m = 5

The image of a function of a certain degree intersects the image of the inverse scale function y = 4x at point a (m, 1) and is parallel to the straight line y = - 12x. (1) find the analytic expression of the function of a certain degree; (2) find the coordinates of point m whose abscissa is - 4 on the image of the function of a certain degree in (1); (3) whether there is a point P on the image of the function of a certain degree so that the distance from it to the x-axis is 2? If it exists, find out the coordinates of point p; if not, explain the reason

(1) The image of inverse scale function y = 4x is at point a (m, 1), 1 = 4m, M = 4, and the image of primary function passes through point a (4, 1) and is parallel to the straight line y = - 12x. Let the image of primary function be y = - 12x + B1 = - 12 × 4 + BB = 3, and the analytic expression of primary function be y = - 12x + 3; (2) when x = - 4, y = - 12 × (- 4) + 3 = 5, and the coordinate of m point is (- 4, 5); (3) there exists that the distance from it to X axis is 2, that is | y|= 2-12x + 3 = 2 or - 12x + 3 = - 2, x = 2, or x = 10, the coordinates of point P are (2, 2) (10, - 2)