Let f (x) = m vector × n vector, where vector M = (2cosx, 1), n vector = (cosx, √ 3sin2x), X belongs to R In the triangle ABC, a, B, C are the opposite sides of angles a, B and C respectively. We know that f (a) = 2, B = 1, and the area of triangle ABC is √ 3 / 2. We can find the value of B + C / SINB + sinc

F (x) = m vector × n vector = (2cosx, 1) * (cosx, √ 3sin2x) = 2 (COX) ^ 2 + √ 3sin2x = 1 + cos2x + √ 3sin2x
= 1 + 2Sin (2x + 30 degrees)
From F (a) = 2, there is 1 + 2Sin (2a + 30 degrees) = 2, a = 60 degrees
The area of the triangle ABC is √ 3 / 2, 1 / 2 * bcsina = 1 / 2 * 1 * csin, 60 degrees = √ 3 / 2, C = 2
According to the cosine theorem, a ^ 2 = B ^ 2 + C ^ 2-2bccosa = 1 + 4-2 * 1 * 2 * 1 / 2 = 3, a = √ 3, C = 90 degrees, B = 30 degrees
B + C / SINB + sinc = 1 + 2 / sin30 degree + sin90 degree = 1 + 4 + 1 = 6

Given vector M = (Radix 3sin2x + 2, cosx), vector n = (1,2cosx), Let f (x) = vector m * vector n. find f (x) In the triangle ABC, a, B and C are the opposite sides of angles a, B and C respectively. If f (a) = 4, B = 1, the area of the triangle is 3 / 2 of the root sign, then find the value of A

(1,2cosx) = (3) sin2x + 2 + 2 (cosx) = (1,2cosx) = (3) sin2x + 2 + 2 (cosx) ^ 2 = (3) sin2x + 2 * (1 + cos2x) / 2 + 2 + 2 = 3 sin2x + cos2x + 3 = 2Sin (2x + cos2x) + 3 = 2Sin (2x + Π / 6) + 3f (a) = 2Sin (2a + Π / 6) + 3 = 4, sin (2a + ⑈ / 6) = 1 / 2, so, 2A + Π / 6 = 2K Π + / 6 or 2A + / 2A + / 2 or 2A + / 2 + / 6 = 2K Π / 6 = 2K Π / 6 or Π / 6 = 2K Π + 5 Π / 6

Let m = (root sign 3sin2x + 2, cosx), vector n = (1,2cosx), and let f (x) = vector m * vector n First of all, I calculated F (x) = 2Sin (2x + π / 6) + 3 In △ ABC, a, B, C are opposite sides of angles a, B and C respectively. If Tana = radical 3, B = f (5 π / 6) △ ABC's area is root three / 2, find the value of A

(1) Your f (x) = 2Sin (2x + π / 6) + 3 (2) B = f (5 π / 6) = 2Sin (2 π - π / 6) + 3 = 2 * (- 1 / 2) + 3 = 3-1 = 2tana = 2tana = √ 3A = 60 ° △ ABC area = 1 / 2 * bcsina = C * √ 3 / 2 = 3 / 2 = √ 3 / 2 / / C = 1BC = 2 cosine theorem cosa = 1 / 2 = (b ^ 2 + C ^ 2 ^ 2) / (2BC) BC = b = 2 ^ 2 ^ 2 = B ^ 2 + C ^ 2-A ^ 2) / (2BC) BC = B ^ 2 ^ 2 ^ 2 = 2 ^ 2 ^ 2 ^ 2 ^ 2) / (2BC) BC = B ^ 2) B ^ 2) 2) + C ^ 2-A ^ 22 = 4 + 1-A ^ 2A ^ 2

What's the name of SiNx's indefinite integral formula to the nth power? I just need to know what it's called. Thank you

Reduction formula
(Reduction Formula)

What about the indefinite integral of the third power of SiNx?

∫1/sin³x dx
=∫csc³x dx
=∫cscx*csc²x dx
=∫cscx d(-cotx)
=-CSCX * Cotx + ∫ Cotx D (CSCX), integration by parts
=-cscx*cotx + ∫cotx*(-cscxcotx) dx
=-cscx*cotx - ∫cscx*cot²x dx
=-CSCX * Cotx - ∫ CSCX * (CSC? X-1) DX, identity CSC? X = 1 + cot? X
=-CSCX * Cotx - ∫ CSC? X DX + ∫ CSCX DX, shift - ∫ CSC? X DX
∫ 2 ∫ CSC ∫ x DX = - CSCX * Cotx + ∫ CSCX DX, note ∫ CSCX DX = ln | CSCX - Cotx | + C
∴∫csc³x dx = (-1/2)cscx*cotx + (1/2)ln|cscx - cotx| + C

Find the indefinite integral of (SiNx) cubic,

∫ (sinx)^3 dx = ∫ (sinx)^2 sinx dx
= ∫ (1-(cosx)^2) (-1) d(cosx)
= - cosx +1/3 (cosx)^3 + C
There are also other calculation methods. The results may be different in appearance, but they are all right (because trigonometric functions add or subtract constants in different forms)

To find the indefinite integral of (SiNx) cubic power ∫ (sinx)^3 dx = ∫ (sinx)^2 sinx dx = ∫ (1-(cosx)^2) (-1) d(cosx) = - cosx +1/3 (cosx)^3 + C In the first step, how does the independent variable become cosx?

d(cosx)=-sinx
So ∫ - SiNx) DX = ∫ dcosx
So ∫ sinxdx = ∫ - D (cosx)

F (x) = (sinx-1) / [(3-2cosx-2sinx)], the value range of this function is - () Please give a brief idea

Reference 1:
Reference 2:

Given SiNx = 2cosx, find three trigonometric functions of angle X

sinx=2cosx
tanx=sinx/cosx=2
sin²x+cos²x=1
When x is the first quadrant angle,
sinx=2√5/5
cosx=√5/5
When x is the third quadrant angle,
sinx=-2√5/5
cosx=-√5/5

The range of y = SiNx + 1 / 2sinx-1

y=(sinx＋1)/(2sinx－1)
=[(sinx－1/2)＋3/2]/(2sinx－1)
=1/2＋(3/2)/[2sinx－1]
Because - 1 ≤ SiNx ≤ 1, so - 3 ≤ 2sinx-1 ≤ 1
Then 1 / [2sinx-1] ≥ 1 or 1 / [2sinx-1] ≤ - 1 / 3
(3 / 2) / [2sinx-1] ≥ 3 / 2 or (3 / 2) / [2sinx-1] ≤ - 1 / 2
Or (2) / [2] / 1 / 2
That is to say, the range of function value is (- ∞, 0] ∪ [2, + ∞)