Given that sin (π / 4 + α) * sin (π / 4 - α) = 1 / 4, α∈ (π / 4, π / 2), the value of 2Sin ^ α + Tan α - cot α - 1 is calculated

Given that sin (π / 4 + α) * sin (π / 4 - α) = 1 / 4, α∈ (π / 4, π / 2), the value of 2Sin ^ α + Tan α - cot α - 1 is calculated

sin(π/4+α)*sin(π/4-α)
=sin(π/4+α)*cos(π/4-α)
=1/2sin(π/2+2α)
=1/2cos2α
1/2cos2α=1/4
cos2α=1/2 α∈(π/4,π/2),
SO 2 α = π / 3
2sin^α+tanα-cotα-1
=2sin^α-1+sinα/cosα-cosα/sinα
=-cos2α+(sin^α-cos^α)/sinαcosα
=-cos2α-cos2α/(sin2α/2)
=-(cos2α+2cot2α)
=-(cosπ/3+2cotπ/3)
=-(1/2+2√3/3)
=-2√3/3-1/2

It is known that β is the inner angle of a triangle and sin β + cos β = 1 / 5. Find the value of Tan β

Radix 2cos (β - 45 °) = 1 / 5
Cos (β - 45 °) = root 2 / 5
Sin (β - 45 °) = root 3 / 5
Tan (β - 45 °) = radical 6 / 2 = (Tan β - 1) / (1 + Tan β)
Root 6 + root 6tan β = 2tan β - 2
(2-radical 6) Tan β = 2 + radical 6
Tan β = (2 + radical 6) / (2-radical 6)

Given that Tan α = 1,3 sin β = sin (2 α + β), find (1) Tan α, (2) Tan (α + β), and (3) Tan (α + β) / 2

1) Because Tan α = 1, α = 2 π + π / 4, sin (2 α + β) = sin (π / 2 + β) = cos β. From 3sin β = sin (2 α + β), Tan β = 1 / 3
2）tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=(1+1/3)/(1-1/3)=2
3) Tan (α + β) = 2tan (α + β) / 2 / (1-tan (α + β) / 2 ^ 2), so tan (α + β) / 2 ^ 2 + Tan (α + β) / 2-1 = 0,
Therefore, Tan (α + β) / 2 = (- 1 ±√ 5) / 2

β is an acute angle, and α and β satisfy the requirements of 4tan * α / 2 = 1-tan ^ 2 * α / 2.3sin β = sin (2 α + β)

(1) (2) 3sinb = sin (2a + b) = sin2acosb + cos2asinb = (4 / 5) CoSb + (3 / 5) SINB. = = > tanb = 1 / 3. (3) Tan (a + b) = [Tana + tanb] / (1-tanatanb) = 1. = = > a + B = π / 4

It is known that Tan α = 1.3 sin β = sin (2 α + β) 1.tanβ 2.tan(β+α) 3.tan(α+β)/2

One
sin2α=2tanα/(1+(tanα)^2)=2/2=1
cos2α=[1-(tanα)^2]/[1+(tanα)^2]=0
3sinβ=sin(2α+β)=sin2αcosβ+cos2αsinβ=cosβ
So tan β = sin β / cos β = 1 / 3
2.tan(β+α)=tanβ+tanα/(1-tanβ*tanα)=(4/3)/(1-1/3)=2
3.tan2x=2tanx/[1-(tanx)^2]
Let x = (α + β) / 2, y = TaNx = Tan [(α + β) / 2]
SO 2 = tan2y = 2Y / (1-y ^ 2)
The result shows that Y1 = (radical 5-1) / 2 y2 = (- radical 5-1) / 2

Given that Tan a = 13 sin (a + b) = sin (2a + b), find the value of Tan (a + b)

tana=1
Cosa = √ 2 / 2, Sina = √ 2 / 2 or cosa = - √ 2 / 2, Sina = - √ 2 / 2
cos2a=0,sin2a=1 cos2a=0,sin2a=1
3sin(a+b)=sin(2a+b)
(3√2/2)(sinb+cosb)=cosb
3√2sinb=(2-3√2)cosb
tanb=(2-3√2)/3√2=√2/3-1
tan(a+b)=(tana+tanb)/(1-tanatanb)=(√2/3)/(1-(√2/3-1))=√2/(6-√2)

Given Tan = 1, sin (2 α + β) = 3sin β, find Tan (α + β)

sin(2α+β)=3sinβ
sin(α+β+α)=3sin（α+β-α）
sin(α+β)cosα+cos(α+β)sinα=3sin(α+β)cosα-3cos(α+β)sinα
2sin(α+β)cosα=4cos(α+β)sinα
Divide both sides by cos (α + β) cos α to obtain Tan (α + β) = 2tan α = 2

If 3sinb = sin (2a + b), then Tan (a + b) / Tana=

3sinb=sin(2a+b)
3sin(a+b-a)=sin(a+b+a)
3[sin(a+b)cosa-cos(a+b)sina]=sin(a+b)cosa+cos(a+b)sina
sin(a+b)cosa=2cos(a+b)sina
tan(a+b)=2tana
tan(a+b)/tana=2

Given sin (2a + b) = 3sinb, find Tan (a + b) / Tana

2A + B = (a + b) + A, B = (a + b) - A, so sin (2a + b) = 3sinb is obtained
sin(A+B)cosA＋cos(A+B)sinA＝3sin(A+B)cosA-3cos(A+B)sinA
2sin(A+B)cosA=4cos(A+B)sinA
tan(A+B)=2tanA
tan(A+B)/tanA＝2

Given Tana = 1,3sinb = sin (2a + b), find Tan (a + B / 2)

tana=1 a=kπ+π/4
3sinB=sin(2a+B)=sin(2kπ+π/2+B)=cosB
3sinB=cosB
sin^2B+cos^2B=1
cos^2B=9/10 sin^2B=1/10
sinB=√10/10 cosB=3√10/10
tan(a+b/2)=(1+tamB/2)/(1-tanB/2)=(cosB/2+sinB/2)/(cosB/2-sinB/2)
=(1+sinB)/cosB
=(√10+1)/3
sinB=-√10/10 cosB=-3√10/10
tan(a+b/2)=(1+tamB/2)/(1-tanB/2)=(cosB/2+sinB/2)/(cosB/2-sinB/2)
=(1+sinB)/cosB
=(-√10+1)/3