When cos α < 0 and Tan α < 0, what is the quadrant angle of α?

When cos α < 0 and Tan α < 0, what is the quadrant angle of α?

You can use cosa = x / R, Sina = Y / R, Tana = Sina / cosa (where R is the unit circle radius 1)

It is known that α, β are acute angles, α + β ≠ π / 2, and 3sin β = sin (2 α + β) (1) It was proved that Tan (α + β) = 2tan α (2) It is proved that Tan β is less than √ 2 / 4, and the value of Tan α and Tan β when the equal sign is established Reward points can be added

It is proved that 3 [sin (α + β) cos α - cos (α + β) sin α] = sin (α + β) sin α = sin (α + β) cos α + cos (α + β) sin α = sin (α + β) cos α = 4cos (α + β) sin α

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

If sin (2a + b) = sin (2a) CoSb + cos (2a) SINB, sin (2a) = 2tana / (1 + Tana ^ 2) = 1, cos (2a) = (1-tana ^ 2) / (1 + Tana ^ 2) = 0, then from sin (2a) CoSb + cos (2a) SINB = 3sinb, Tan (a + b) = (Tana + tanb) / (1-tanatanb) = 2

It is known that both a and B are greater than zero and less than 90 ° and 3sinb = sin (2a = b), 4tan (A / 2) = 1-tan ^ 2 (A / 2) Find the value of a + B

4tan α/2=1-tan α/2 tan α/2
===>tana = 1/2 .(1)
>tan(α+β) =1
α+β= π/4

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

It is proved that: given 3tana = 2tan (a + b), so 3sinacos (a + b) = 2Sin (a + b) cosa, 5sinacos (a + b) = 2Sin (a + b) cosa + 2sinacos (a + b) = 2Sin (2a + b), sinacos (a + B) = 2Sin (a + b) cosa-2sinacos (a + b) = 2Sin (a + b) = 2Sin (2a + b)] /

It is known that sin (2a + b) = 5sinb, a ≠ K π + (π / 2), a + B ≠ K π + (π / 2), K ∈ Z. it is proved that 2tan (a + b) = 3tana It is known that sin (2a + b) = 5sinb, a ≠ K π + (π / 2), a + B ≠ K π + (π / 2), K ∈ Z. it is proved that 2tan (a + b) = 3tan (a)

2a+b=(a+b)+b
Then, the sum difference product formula is used to expand sin (a + B + b)
Then move the item and merge it

In the triangle ABC, given sin ^ 2A = sin ^ C + sin ^ B + root sign 3sin ^ csin ^ B, then the value of angle a is

By sine theorem
a/sinA=b/cosB=c/sinC
Let a / Sina = B / CoSb = C / sinc = 1 / K
Then Sina = AK
sinB=bk
sinC=ck
Sin ^ 2A = sin ^ C + sin ^ B + Radix 3sincsinb
This is wrong. Finally, sincsinb, there is no square
So a ^ 2K ^ 2 = C ^ 2K ^ 2 + B ^ 2K ^ 2 + √ 3bck ^ 2
So a ^ 2 = C ^ 2 + B ^ 2 + √ 3bC
b^2+c^2-a^2=-√3bc
cosA=(b^2+c^2-a^2)/2bc
=-√3bc/2bc
=-√3/2
So a = 150 degrees

Sn=n+75(5/6)^(n-1)-90 Simplification: SN

Sn=n+75(5/6)^(n-1)-90
S（n+1）=n+75(5/6)^n-89
S（n+1）-Sn=1+75(5/6)^n-75(5/6)^(n-1)=1+75(5/6)^(n-1)(5/6-1)=
=1-15 (5 / 6) ^ n means certificate 1-15 (5 / 6) ^ n > 0 means Certificate (5 / 6) ^ (n-1)

Simplify sin ^ 2 (α - π / 6) + sin ^ 2 (α + π / 6) - Sin ^ 2 α

sin^2(α-π/6)+sin^2(α+π/6)-sin^2α=1/2{1-cos(2α-π/3)+1-cos(2α+π/3)-1+cos2α}=1/2{1+cos2α-2cos2αcosπ/3}=1/2{1+cos2α-cos2α}=1/2

Simplify cot (π / 2 - α)

Cot (π / 2 - α) = Tan α, triangle induction formula