Using the five point method to make the image of one period of the function y = sin

Using the five point method to make the image of one period of the function y = sin

Let X / 2 - π / 3 = 0, π / 2, π, 3 π / 2,2 π
Then x = 2 π / 3,5 π / 3,8 π / 3,11 π / 3,14 π / 3
So five points are (2 π / 3,0), (5 π / 3,1), (8 π / 3,0), (11 π / 3, - 1), (14 π / 3,0)
Just describe it

The function f (x) = sin (2x + φ) (0

A symmetry axis of y = f (x) image is a straight line x = π / 8. The intersection point between the symmetry axis and the image should be the highest point or the lowest point of the image. Replace it with y = sin (2x + φ), y = sin (2 * π / 8 + φ) = sin (π / 4 + φ) = 1 or - 1,
Zero

Let f (x) = sin (2x + φ) (- π)

Well, this is the third question in our final exam
Let 2x + φ = k π + π / 2, x = (2k π + π - 2 φ) / 4
Because the axis of symmetry is x = π / 8, 4 φ = 4K π + π
Because - π

Let f (x) = sin (2x + φ) (- π)

∵ y = f (x) is a straight line x = π / 8
∴2×π／8+φ＝kπ＋π／2
∴φ＝kπ＋π／4
∵-π

Let f (x) = sin (1 / 2x + φ) (0

Because one symmetry axis of y = f (x) image is a straight line x = Π / 4
So f (π / 4) is equal to 1 or - 1
Because 1 / 2 * π / 4 + φ = π / 8 + φ, and 0

A symmetric axis equation of the image of the function y = sin (2x + 5 π / 2) is A、x=-π/4 B、x=-π/2 C、x=π/8 D、x=5π/4

Observe the image of the function y = sin (2x + 5 π / 2)
We know that its axis of symmetry must be perpendicular to the x-axis and pass through the highest point or the lowest point of the function
While - 1 ≤ sin (2x + 5 π / 2) ≤ 1
So let sin (2x + 5 π / 2) = 1, we get: 2x + 5 π / 2 = 2K π + π / 2, (K ∈ z); X = k π - π, (K ∈ z)
Let sin (2x + 5 π / 2) = - 1, then: 2x + 5 π / 2 = 2K π + 3 π / 2, (K ∈ z); X = k π - π / 2, (K ∈ z)
So the equation of symmetry axis is x = k π / 2, (K ∈ z)
One of them can be: x = - π / 2
Choose B

Given the function f (x) = sin ^ 2 ω x + √ 3sin ω xsin (ω x + π / 2) + 2cos ^ 2 ω x, ω > 0, the abscissa of the first highest point on the right side of the y-axis is π / 6 Find ω

F (x) = √ 3sin ω x * cos ω x + (sin ω x) ^ 2 + (COS ω x) ^ 2 + (COS ω x) ^ 2 = (√ 3 / 2) * sin2 ω x + (1 + Cos2 ω x) / 2 + 1 = (√ 3 / 2) * sin2 ω x + (1 / 2) * Cos2 ω x + 3 / 2 = sin (2 ω x + π / 6) + 3 / 2

The function f (x) = sin (2x + θ) + sin (2x - θ) + 2cos ^ 2x + A is known, where a and θ are constants, and the image of the function passes through the point (5 π / 12, a + 1) (1) Find the minimum positive period and symmetry axis of function f (x); (2) It is pointed out how the image of F (x) can be transformed from the image of y = sin (x + π / 6); (3) If x ∈ [- π / 4, π / 4], the maximum value of F (x) is 1, find the value of A

Expand the formula 2sin2acosq + 2cos ^ 2x + a = 2sin2acosq + cos2x + 1 + A, take the 2sin25 π / 12cosq + cos25 π / 12 = 0 brought in by x = 5 π / 12 to find cosq = radical 3 / 2, and then bring cosq into the
F X = 2Sin (2x + π / 6) + 1 + a period T = π the axis of symmetry is 2x + π / 6 = k π + π / 2
F X has y = sin (x + π / 6) the abscissa is shortened by half and the ordinate is lengthened by two times
If x ∈ [- π / 4, π / 4], then (2x + π / 6) belongs to - 1 / 3 π to 2 / 3 π, and the maximum value is 2sin2 / 3 π + 1 + a = 1

The known function f (x) = sin (ω x + π) 3) If the minimum positive period of (ω > 0) is π, then the graph () A. On the straight line x = π 4 symmetry B. On point (π) 3, 0) symmetry C. On point (π) 4, 0) symmetry D. On the straight line x = π 3 symmetry

∵ function f (x) = sin (ω x + π)
3) The minimum positive period of (ω > 0) is π,
From the periodic formula of trigonometric function, t = 2 π
ω = π, the solution is ω = 2
The function expression is f (x) = sin (2x + π)
3）
Let 2x + π
3 = k π (K ∈ z), x = - π
6+1
2kπ（k∈Z），
The symmetry center of the function image is (- π)
6+1
2kπ，0）（k∈Z）
If k = 1, the center of symmetry is (π)
3, 0), we can get that item B is correct and item C is incorrect
The symmetric axis equation of the function image satisfies x = π
12+1
2kπ（k∈Z），
However, the straight lines of a and D are not consistent, so a and D are not correct
Therefore: B

The known function f (x) = sin (ω x + π) 3) If the minimum positive period of (ω > 0) is π, then the graph () A. On the straight line x = π 4 symmetry B. On point (π) 3, 0) symmetry C. On point (π) 4, 0) symmetry D. On the straight line x = π 3 symmetry

∵ function f (x) = sin (ω x + π)
3) The minimum positive period of (ω > 0) is π,
From the periodic formula of trigonometric function, t = 2 π
ω = π, the solution is ω = 2
The function expression is f (x) = sin (2x + π)
3）
Let 2x + π
3 = k π (K ∈ z), x = - π
6+1
2kπ（k∈Z），
The symmetry center of the function image is (- π)
6+1
2kπ，0）（k∈Z）
If k = 1, the center of symmetry is (π)
3, 0), we can get that item B is correct and item C is incorrect
The symmetric axis equation of the function image satisfies x = π
12+1
2kπ（k∈Z），
However, the straight lines of a and D are not consistent, so a and D are not correct
Therefore: B