TaNx = 1,3sinb = sin (2x + b), find Tan (x + b)

TaNx = 1,3sinb = sin (2x + b), find Tan (x + b)

3sin(x+b-x)=sin(x+b)cosx+cos(x+b)sinx 3sin(x+b)cosx-3cos(x+b)sinx=sin(x+b)cosx+cos(x+b)sinx 2sin(x+b)cosx=4cos(x+b)sinx tan(x+b)=2tanx=2

Proof: Tan square x-sin square x = Tan x

Tan squared x-sin squared x
=tan^2x(1-cos^2x)
=tan^2xsin^2x

sin(30+x)=5/13,60

sin(30°+x)=5/13,60°

We know the image of a function of degree through a (0,2), B (2,4). (1): find the analytic formula of this function; (2) try to judge whether the point P (3, - 5) is on a straight line

Solution
Let the linear function be:
y=kx+b
Substituting (0,2) (2,4) into
b=2
2k+b=4
∴k=1
∴y=x+2
Substitute x = 3 into the equation
We get: y = 3 + 2 = 5 ≠ - 5
P (3, - 5) is not on a straight line

The image of a function of degree is known to pass through two points a (- 2, - 3) and B (1,3). (1) find the analytic expression of the image of the function of degree. (2) try to judge whether the point P (- 1,1) is on the image of the function of degree

(1) Using the point oblique formula y = KX + B, and then substituting the coordinates of a and B, the solution is y = 2x + 1
(2) Substituting P (- 1,1) into y = 2 × (- 1) + 1 = - 1 is not equal to 1
So the point P (- 1,1) is not on the graph of this linear function

It is known that the image of a function of degree passes through points a (- 3,2), B (1,6). ① find the analytic expression of the function. ② find the area of the triangle formed by the image of the function and the coordinate axis

① Let the analytic formula of a function be y = KX + B (K ≠ 0). Substituting points a (- 3,2), B (1,6) into 2 = − 3K + B6 = K + B, the solution is b = 5K = 1. So the analytic formula of the function is y = x + 5 (4 minutes). ② the intersection of the function and X, Y axis is: x = - 5 when y = 0; y = 5 when x = 0. The area of the triangle enclosed by the function image and coordinate axis is 12 × 5 ×| - 5 | = 252

Given that the graph of a function passes through two points (3, 6) and (0, 0), we can find the analytic expression of the function and the value of a if the point (a, 2) is on the graph of the function

(1) Let the analytic formula of a function be y = KX + B, and substitute two points to get 6 = 3K + B0 = B, and the solution is k = 2, B = 0, so the analytic formula of the function is y = 2x; (2) ∵ point (a, 2) is on the function image, and the solution is a = 1

It is known that the image of a function passes through two points (3,5) and (- 4,9). Find the analytic expression of the function. If the point (a, 2) is on the image of the function, find the value of A

Let the analytic expression of the function be y = KX + B
Substituting points (3,5) and (- 4,9), we can get k = - 4 / 7, B = 47 / 7
So the analytic formula is y = - 4 / 7X + 47 / 7
Substituting point (a, 2) into the analytic formula, 2 = - 4 / 7a + 47 / 7, the solution is a = 33 / 4

Given that the image of a function of degree passes through points a (- 1,3) and (2,3), (1) find the analytic expression of the function of degree; (2) judge whether point C (- 2,5) is on the image of the function

(1) Let the analytic expression of the function of the line AB be y = KX + B (k, B are constants and K ≠ 0) ∵ the image of a linear function passes through points a (- 1,3) and points (2, - 3), ∵ 3 = − K + B − 3 = 2K + B, and the solution is k = − 2B = 1. ∵ the analytic expression of the function of the line AB is y = - 2x + 1. (2) substituting x = - 2 into y = - 2x + 1, we get y = - 2 × (- 2) + 1 = 5, so the point C (- 2,5) is on the function image

Given that the image of a function passes through points (- 4,9) and (6,3), the analytic expression of the function is obtained

Let the analytic expression of this function be y = KX + B
-4k＋b=9
6K + B = 3, k = - 3 / 5, B = 33 / 5
The analytic expression of this function is y = - 3x / 5 + 33 / 5