If the length of the segment is π / 4, then the value of F (π / 3) is? A.√3/3 B.-√3/3 C.√3 D.-√3

If the length of the segment is π / 4, then the value of F (π / 3) is? A.√3/3 B.-√3/3 C.√3 D.-√3

According to the meaning of the title
tan(ωx)＝tan[ω(x＋π/4)]＝π/3
Thus, (1 + π & sup2 / 9) × Tan (ω π / 4) = 0
So tan (ω π / 4) = 0
Let ω π / 4 = π (or more accurately, let it be equal to π + K π, K ∈ z)
Equal ω = 4
So f (π / 3) = Tan (4 π / 3) = √ 3

If the length of the line segment obtained from two adjacent tangent lines y = π / 6 of the image with function f (x) = tanwx (W > 0) is π / 6, then the value of F (π / 6) is

From the theme
T=π/6=π/w
So w = 6
f(x)=tan6x
f(π/6)=tanπ=0

Deduction of distance formula between two straight lines
That's the C-C divided by the radical a * A-B * B

Take a point from a straight line and use the formula of distance from point to straight line

Derivation of the formula of symmetric point of point about line
Let a known point (m, n), then the symmetric coordinates (x, y) about the line ax + by + C = 0
x=m-2a*[(ma+nb+c)/(a^2+b^2)]
y=n-2b*[(ma+nb+c)/(a^2+b^2)]
The process of reasoning

How to find the symmetrical point of a straight line?
For example, how to find the symmetric point of point (3,1) with respect to X-Y + 9 = 0

The slope of the line X-Y + 9 = 0 is k = 1,
The slope of the line passing through point (3,1) perpendicular to the line X-Y + 9 = 0 is k '= - 1,
The linear equation is: Y-1 = - (x-3), sorted out: x + y-4 = 0,
The solution of two linear equations is: x = - 5 / 2, y = 13 / 2,
So the intersection point (perpendicular foot) of two straight lines is: (- 5 / 2,13 / 2),
Let the symmetric point of point (3,1) with respect to X-Y + 9 = 0 be (x, y), then,
From the formula of midpoint coordinate, we can get that: (the perpendicular foot is the midpoint of two symmetrical points)
x+3=2*（-5/2）=-5,y+1=2*13/2=13,
So x = - 8, y = 12,
So the symmetric point of point (3,1) with respect to X-Y + 9 = 0 is (- 8,12)

The coordinate of a point symmetrical with respect to another point, (how to find it by formula)

Find the symmetric point B (x, y) of point a (m, n) with respect to point P (a, b)
(m+x)/2=a ,x=2a-m
(n+y)/2=b ,y=2b-n

A formula for finding the coordinates of a point with respect to a line

Set the coordinates (a, b) of the point to be solved. According to the set points (a, b) and known points (C, d), the coordinates of the symmetric point (a + C / 2, B + D / 2) can be expressed, and the symmetric point is on a straight line. So substituting this point into a straight line, a, B, the coordinates of the point to be solved can be obtained

The coordinates of the point (4,5) with respect to the line x = 1 are_____ I forgot what I learned today~

The coordinates of the point (a, b) with respect to the line x = m are: (2m-A, b)
Substituting into formula: (2 * 1-4,5) namely: (- 2,5)

Point on the line symmetric point of the solution to say in detail, OK

If any straight line is y = ax + B;
If the known point is (m, n), the required point is (Q, P);
Since the two points are symmetrical with respect to the straight line, the product of the slope of the line connecting the two points and the slope of the known straight line must be - 1;
Then ((p-n) / (q-m)) * a = - 1
And the midpoint of the line between two points is ((M + Q) / 2, (n + P) / 2) on the known line, and then the point is brought into the known line
We have ((p-n) / (q-m)) * a = - 1. Equation 1
(n + P) / 2 = a (M + Q) / 2 + B equation 2
In the two equations, only q p is an unknown number, and then it can be solved. This is the general case

Know how to find the axis of symmetry for two straight line expressions
For example, l1:3x + y + 6 = 0, l2:2x-y-3 = 0, how to find the axis of symmetry of these two lines

1. Solve the intersection point of the two equations, which is on the axis of symmetry
2. The axis of symmetry is the bisector of the angle between two straight lines. The slopes of the two straight lines are - 3 and - 2 respectively
（k+3）/(1-3k)=(2-k)/(1+2k)
Find K, and then substitute it into the intersection