What are sine and cosine functions like? Is the graph axisymmetric and centrosymmetric? Write the axis of symmetry or the point of symmetry

What are sine and cosine functions like? Is the graph axisymmetric and centrosymmetric? Write the axis of symmetry or the point of symmetry

Sine function and y = SiNx symmetry center: [K π, 0] symmetry axis; X = π - 2 + K π cosine function and y = cosx symmetry center [π - 2 + K π, 0] symmetry axis X = k π

Relationship between symmetric axis symmetry center of sine function and cosine function and image key points

For the sine function y = SiNx, the axis of symmetry is x = π / 2 ± K π (k is an integer)
The center of symmetry is x = k π (k is an integer)
For cosine function y = cosx, the axis of symmetry is x = k π (k is an integer)
The center of symmetry is x = π / 2 ± K π (k is an integer)
Keys: Intersections
When x = π / 4 ± K π

Let a be the inclination angle of the straight line y = - root 3x + 2. Then cosa's value is!

According to the title, Tana = 3 and a belongs to (0,90)
So we can get Sina / cosa = 3 and Sina squared + cosa squared = 1
So cosa squared = 1 / 10, because a belongs to (0,90)
So cosa is positive

Through point a (2,1), its inclination angle is half of the inclination angle of the straight line L: 3x4y 5 = 0, and solve the equation of the line

Let the inclination angle of this line be 2n, the new line inclination angle N.2), t denote Tan n.! Tan 2n = 2T / [1-T ^ 2) = - 3 / 4. Cross multiplication, 3T ^ 2-3 = 8t, 3T ^ 2-8t-3 = 0. Cross multiplication, (3T + 1) (T-3) = 0,3t + 1 = 0, t = - 1 / 3

If the inclination angle of the line 3x + 4y-5 = 0 is a, then the inclination angle of the symmetrical line with respect to the line X3 is

Line x = 3 symmetry
Inclination angle = 180 ° - arctan (- 3 / 4)
=arctan(3/4)

If the equation of the second line is 3x-4y = 0, find the equation of the other three lines

The solution of Tan θ = - 3tan3 θ = (Tan θ + tan2 θ) / (1-tan θ, tan2 θ = (Tan θ + tan2 θ) / (1-tan θ tan2 θ) is solved, and the solution of tan3 θ = - 9 / 13tan4 θ = 2tan2 θ / [1 - (tan2 θ) ^ 2] the solution of Tan 4 θ = 3 / 5, so K1 = - 3, K2 = 3 / 4, K3 = - 3 = - 3, K3 = - 3 = - 3, K3 = - 3 = - 3, K3 = - 3 = - 3, K2 = 3 / 4, K3 = - 3 = - 3, K3 = - 3 = - 3, K3 = - 3 = 3, K3 = - 3 = 3, K3 = - 9

Make a straight line L through point a (8,6) and the inclination angle of L is half of the inclination angle of the line 3x-4y-2 = 0

If the inclination angle of the line is a, then tan2a = 3 / 4
The solution is Tana = 1 / 3
The linear equation is y-6 = 1 / 3 (X-8)
It is concluded that x-3y + 10 = 0

The following conditions are solved: passing through the intersection of two lines 2x-3y + 10 = 0 and 3x + 4Y-2 = 0, and perpendicular to the line 3x-2y + 4 = 0; There are two problems in solving the linear equation satisfying the following conditions: (1) Passing through the intersection point of two straight lines 2x-3y + 10 = 0 and 3x + 4Y-2 = 0, and perpendicular to the line 3x-2y + 4 = 0; (2) Through the intersection of two lines 2x + Y-8 = 0 and x-2y + 1 = 0, and parallel to the line 4x-3y-7 = 0

(1) By solving the equations 2x-3y + 10 = 0,3x + 4Y-2 = 0, we can get x = -- 2, y = 2, so the intersection point is (- - 2,2)

(1) The equation of the line parallel to the straight line 3x + 4y-12 = 0 and the distance from it is 7; (2) the equation of the line perpendicular to the line x + 3y-5 = 0 and the distance from point P (- 1,0) is three fifths and root ten

(1) Let the straight line be 3x + 4Y + C = 0. According to the meaning of the title C - (- 12) ︱ / √ (3 ﹢ 4 ﹣ 2) = 7 ︱ C + 12 = 35C + 12 = 35 or C + 12 = - 35C = 23 or - 47, the equation of straight line is 3x + 4Y + 23 = 0 or 3x + 4y-47 = 0 (2) x + 3y-5 = 03y = - x + 5Y = - X / 3 + 5 / 3, and the slope of the line is (- 1) / (-...)

Find the equation of circle C: (x + 2) ^ 2 + (y-6 = 1) on the symmetric circle of the straight line 3x-4y + 5 = 0

The symmetry radius of a circle about a straight line is constant. So, as long as the symmetrical point of the center of the circle is calculated about the line, then C (- 2,6) can be set as C '(a, b), then the vertical straight line CC' 3x-4y + 5 = 0, that is, the midpoint coordinates of the slope multiplied by - 1 cc 'are listed on the line 3x-4y + 5 = 0 according to these two equivalent relations, the equation system can be obtained