Given that sin (2 α + β) = 3sin β, calculate the value of Tan (α + β) / Tan α

Given that sin (2 α + β) = 3sin β, calculate the value of Tan (α + β) / Tan α

Sin (α + β) cos α + cos (α + β) sin α = 3sin (α + β) cos α - 3cos (α + β) sin α = 2Sin (α + β) cos α - 3cos (α + β) sin α = 2Sin (α + β) cos α sin (α + β) cos α / cos (α +...) can be obtained by sum difference product

It is known that 3 sin β = sin (2 α + β), and it is proved that Tan (α + β) = 2tan α

It is proved that: 3sin [(α + β) - α] = sin [(α + β) + α], and the result is: 3sin (α + β) cos α - 3cos (α + β) sin α = sin (α + β) cos α + cos (α + β) sin α, that is, 2Sin (α + β) cos α = 4cos (α + β) sin α

Given sin (2a + b) = 3sin (b), find the value of Tan (a + b) / Tan (a)

cos[(a＋b)＋a]=3sin[(a＋b)－a]
cos(a＋b)cosa－sin(a＋b)sina=3sin(a＋b)cosa－3cos(a＋b)sina
cos(a＋b)[cosa＋3sina]=sin(a＋b)[3cosa＋sina]
tan(a＋b)=sin(a＋b)/cos(a＋b)=[cosa＋3sina]/[sina＋3cosa]=[1＋3tana]/[3＋tana]
Therefore, Tan (a + b) / Tana = [1 + 3tana] / [Tan 2A + 3tana] =
What's the problem of this problem?

3 SINB = sin (2a + b), Tan (a + b) = 4, find Tana ditto

sin(α+β+α)=3sin(α+β-α)
sin(α+β)cosα+cos(α+β)sinα=3sin(α+β)cosα-3cos(α+β)sinα
4cos(α+β)sinα=2sin(α+β)cosα
That is, Tan α = 1 / 2tan (α + β) = 2

In the triangle ABC, 3sinb = sin (2a + b), 4tan (A / 2) = 1-tan 2 (A / 2). It is proved that a + B = π / 4

The proof is as follows:
Because 4tan (A / 2) = 1-tan 2 (A / 2), according to the tangent formula of double angle, we can get
tanA=2tan(A/2)/[1-tan²(A/2)]=2tan(A/2)/4tan(A/2)=1/2
sinA/cosA=1/2
So cosa = 2sina
3sinB=sin(2A+B)
3sinB=sin2AcosB+cos2AsinB
3sinB=2sinAcosAcosB+(cos²A-sin²A)sinB
3sinB=2sinA(2sinA)cosB+[(2sinA)²-sin²A]sinB
3sinB=4sin²AcosB+3sin²AsinB
4sin²AcosB=3sinB-3sin²AsinB
4sin²AcosB=3sinB(1-sin²A)
4sin²AcosB=3sinBcos²A
4sin²AcosB=3sinB(2sinA)²
4sin²AcosB=12sinBsin²A
cosB=3sinB
tanB=1/3
tan(A+B)=(tanA+tanB)/(1-tanAtanB)=(1/2+1/3)/[1-(1/2)*(1/3)]=1
So a + B = π / 4
This is square
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Given that 3sinb = sin (2a + b), Tan (a + b) = 2tan a

3sinB=sin（2A+B）
3sinB=sinAcos(A+B)+cosAsin(A+B)
3sinB=sinA(cosAcosB-sinAsinB)+cosA(sinAcosB+cosAsinB)
3sinB=sinAcosAcosB-sinAsinAsinB+cosAsinAcosB+cosAcosAsinB
3sinB=2cosAsinAcosB+sinB(cosAcosA-sinAsinA)
3(sinAsinA+cosAcosA)sinB=2cosAsinAcosB+sinB(cosAcosA-sinAsinA)
4sinAsinAsinB+2cosAcosAsinB=2cosAsinAcosB
2sinAsinAsinB+cosAcosAsinB=cosAsinAcosB
2sinAsinAsinB+cosAcosAsinB=2cosAsinAcosB-cosAsinAcosB
2sinAcosAcosB-2sinAsinAsinB=cosAsinAcosB+cosAcosAsinB
2sinA(cosAcosB-sinAsinB)=cosA(sinAcosB+cosAsinB)
2sinAcos(A+B)=cosAsin(A+B)
tan（A+B）=2tan A

5sinb = sin (2a + b) prove that 2tan (a + b) = 3tana

The original formula can be written as follows:
5sin[(A + B) - A] = sin[(A + B) + A]
Left = 5sin (a + b) cosa - 5cos (a + b) Sina
Right = sin (a + b) cosa + cos (a + b) Sina
4sin(A + B)cosA = 6cos(A + B)sinA
Divide the two sides by 2cosacos (a + b)
2tan(A+B)=3tanA

Given 5sinb = sin (2a + b), it is proved that 2Sin (a + b) = 3tana

It is proved that 5sin [(a + b) - A] = sin [a + (a + b)]
5sin(A+B)cosA-5cos(A+B)sinA=sin(A+B)cosA+cos(A+B)sinA
4sin(A+B)cosA=6cos(A+B)sinA
2sin（A+B）=3tanA

Given that 3sin β = sin (2a + β), let Tana = x, Tan β = y, y = f (x) (1) The analytic expression of F (x) (2) If angle a is the smallest inner angle of a triangle, try to find the value range of function f (x)

(1) By expanding the known term, 3sin β = sin2acos β + cos2asin β
Divide both sides by cos β to get 3tan β = sin2a + cos2atan β
When the non Tan β is divided to the other side, Tan β = 2sin2a / 3-cos2a is obtained
Using universal formula
sinα=2tan(α/2)/[1+tan^2(α/2)]
cosα=[1-tan^2(α/2)]/[1+tan^2(α/2)]
Tan β = 2tana / 2 + 4tan ^ 2A
That's what you want
(2) It is easy to know that the minimum inner angle of a triangle belongs to the interval of (0, Π / 3]
Tana belongs to (0, radical 3]
So tan β > 0, and Tan β

If Tana = 1, sin (2a + β) = 3sin β, then Tan (a + β) = 3

tana=1 sin2a=2tana/[1+(tana)^2]=2/(1+1)=1 cos2a=[1-(tana)^2]/[1+(tana)^2]=0 sin(2a+b)=sin2acosb+cos2asinb=3sinb sin2acosb=sinb(3-cos2a) tanb=sin2a/(3-cos2a) tan(a+b)=(tana+tanb)/(1-tanatanb)=(1+tanb)/...