Let f (x) = sin (2x + β) (- π ∠ β ∠ 0), y = f (x), and the equation of a symmetry axis of the image is x = π / 8 (1) Find the value of β (2) Find monotone increasing interval of function

Let f (x) = sin (2x + β) (- π ∠ β ∠ 0), y = f (x), and the equation of a symmetry axis of the image is x = π / 8 (1) Find the value of β (2) Find monotone increasing interval of function

(1) Let 2x + β = π / 2 + K π (K ∈ z) β = π / 2 + K π - 2x (K ∈ z). Because a symmetry axis equation is x = π / 8, so β = k π + π / 4 (K ∈ z) because - π ∈ β ∈ 0, so - 1.25 ＜ K ＜ - 0.25, so k = - 1
∴β=-3π/4
(2) F (x) = sin (2x-3 π / 4)
Let - π / 2 + 2K π ≤ 2x-3 π / 4 ≤ π / 2 + 2K π (K ∈ z)
Just solve X and write it in interval form
As shown in the figure, in the equilateral triangle ABC, the bisectors of ∠ B and ∠ C intersect at O, and the vertical bisectors of OB and OC intersect BC at e and F. try to explore the size relationship of be, EF and FC, and explain the reasons
Conclusion: be = EF = FC (1 point) the reason is: be = EF = FC (1 point) the reason is: ∵ ABC is equilateral triangle, ∵ ABC = ∠ ACB = 60 ° (2 points), ∵ OC, OB bisection ∵ ACB, ∵ ABC, ∵ OBE = ∠ OCF = 30 ° (3 points), ∵ eg, HF vertical bisection ob, OC, ∵ OE = be, of = FC (5 points), ∵ BOE = ∠ OBE = 30 °, ∵ C
If the maximum value of derivative function f '(x) of F (x) = f (x) = sin (Wx - π / 6) - 1 is 3, then an equation of symmetry axis of the image of function f (x) is?
(A)x=2π/9
(B)x=π/6
(C)x=π/3
(D)x=π/2
From F (x) = sin (Wx - π / 6) - 1
F '(x) = w * cos (Wx - π / 6)
The maximum f '(x) is 3
So w = 3
So f (x) = sin (3x - π / 6) - 1
From the equation 3x - π / 6 = k π + π / 2 (k is an integer)
The equation x = k π / 3 + 2 π / 9 is obtained
So when k = 0, x = 2 π / 9
...ftg
It is known that, as shown in the figure, △ ABC is an equilateral triangle, BD is the height of AC side, extend BC to e, so that CE = CD
It is proved that the equilateral points of the triangles are ∩ ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩, ∩
Finding the period of function y = sin (2x + Π / 4) + 2cos (3x - Π / 6)
The period of sin (2x + Π / 4) is Π, and that of 2cos (3x - Π / 6) is Π
2 Π / 3; the minimum positive period is their least common multiple, 2 Π / 3
Theorem: the minimum positive period of the sum of several periodic functions is the least common multiple of their respective minimum positive periods
The period of sin (2x + Π / 4) is k Π, and the period of 2cos (3x - Π / 6) is 2 / 3K Π. The intersection of the two sets is 2K Π. The period can be selected as the minimum positive real number, that is 2 Π
In the triangle ABC, D is the midpoint on the side of BC, and ad = AC, De is perpendicular to BC, de intersects AB at point E, EC intersects ad at point F
1. ABC is similar to FCD. 2. Is f the midpoint of ad
1. From the question, we can see that, because ED is perpendicular to BC and D is the midpoint of BC, ED is the vertical bisector of BC
So EC = EB (the distance from the point on the vertical bisector to both ends of the line segment is equal), so ∠ ECB = ∠ EBC
So there are two equal angles in triangle ABC and triangle FCD, so ABC is similar to FCD
2. It's the midpoint. From the first question, we know that the two triangles are similar, so FD: AC = DC: CB = 1:2 (D is the midpoint of BC)
So 2DF = AC and AC = ad, so 2DF = ad, so f is the midpoint of AD
I drew my own picture·
Do you have a picture
What is the minimum positive period of the function y = sin (3x + π / 4)
The minimum positive period 2 π / 3 of function y = sin (3x + π / 4)
In the triangle ABC, D is the midpoint on the side of BC, and ad = AC, De is perpendicular to BC, de and ab intersect at point E, EC and ad intersect at point F
(1) Because De is perpendicular to BC and D bisects BC, De is the perpendicular of BC, and angle B = angle ECD
And AC = ad, the angle ADC = angle ACD
The triangle ABC is similar to the triangle FCD
(2) Because triangle ABC is similar to triangle FCD
So s triangle ABC = s triangle FCD ^ 2 = 25
S triangle ABC = de * BC / 2
De = 5
Finding the period of y = - sin (3x - Π - 4) function
W=3
Period T = 2 π / | w | = 2 π / 3
As shown in the figure, AB is equal to AC in triangle ABC, and point D is the midpoint of edge BC. Through point a D, make parallel lines of BC and ab respectively, and intersect at point e to connect EC and AD
Proving quadrilateral adceshi rectangle
Because AE / / BC, AB / / De, the quadrilateral ABDE is a parallelogram. Then, because AB = AC, when D is the midpoint of BC, the triangle ABC is an equilateral triangle, and BD = CD = AE, and the angle ADC is a right angle. Because AE = CD, AE / / BC, the quadrilateral adce is a parallelogram