Mathematics problem: cos a = - 5 / 13, and a is the value of COS a, Tan a in the fourth quadrant

Mathematics problem: cos a = - 5 / 13, and a is the value of COS a, Tan a in the fourth quadrant

A is the fourth quadrant angle, sin a = - 5 / 13
cosA = √(2-sin^2A) = 12/13
tanA = sinA/cosA = -5/12

If a is the fourth quadrant angle, Tan (π / 3 + a) = - 5 / 12, then cos (π / 6-A) =?

A is the fourth quadrant angle, Tan (π / 3 + a) = - 5 / 12

Cos (3 π / 2 + α) = - 3 / 5, α is the fourth quadrant angle, find Tan (2 π - α)

Cos (3 π / 2 + α) = - 3 / 5, α is the fourth quadrant angle, find Tan (2 π - α)
Cos (3 π / 2 + α) = sin α = - 3 / 5, α is the fourth quadrant angle, so cos α = √ (1-9 / 25) = 4 / 5
∴tan(2π-α)=-tanα=-sinα/cosα=-(-3/5)/(4/5)=3/4

If α is the fourth quadrant angle, sin (α + β) cos β - sin β cos (α + β) = - 12 / 13, then Tan (α - π) / 2 =? Find the detailed process~

According to the trigonometric formula sinacosb sinbcos a = sin (a-b)
Replace α + β and β into the above formula respectively
sin(α+β)cosβ-sinβcos(α+β)=sin(α+β-α)=sinα=-12/13
Because the value of Tan (α - π) / 2 is required
tan(α-π)/2=tan(α/2-π/2)
Because α is the fourth quadrant angle, α / 2 is the second quadrant angle, so tan (α / 2 - π / 2) = - cot (α / 2)
We found that sin α = - 12 / 13
cosα=5/13
According to the formula Tan (α / 2) = sin α / (1 + cos α)
Tan (α / 2) = - 2 / 3 can be calculated
So tan (α / 2 - π / 2) = - cot (α / 2) = - 1 / Tan (α / 2) = 1.5

It is known that afar is the second quadrant angle, Tan AFA = - 1 / 2, cos AFA =? In the triangle ABC, if B = 1, C = follow sign 3, angle c = 2 Pai / 3, then a =?

Cos AFA = negative root 2 a = 1

If α is known to be the angle of the second quadrant, Tan α = 1 / 2, then cos α α is the angle of the second quadrant. Can tan α be greater than 0?

α is the angle of the second quadrant, Tan α = 1 / 2
This is wrong in itself

Given that sin (π / 4 + 2 α) × sin (π / 4-2 α) = 1 / 4, α∈ (π / 4, π / 2), find the value of 2Sin "α + Tan α - cot α - 1 ”Represents square

（π/4-2α）=π/2-（π/4+2α）=
So sin (π / 4 + 2 α) × sin (π / 4-2 α) = sin (π / 4 + 2 α) × cos (π / 4 + 2 α) = 0.5 * sin (π / 2 + 4 α) = 0.5 * Cos4 α = 1 / 4,
So Cos4 α = 1 / 2
If α∈ (π / 4, π / 2), 4 α∈ (π, 2 π), then 4 α = 300 degrees
The original formula = 2Sin "α - 1 + Tan α - cot α
=-cos2α+tanα-cotα
=-cos2α+sinα/cosα-cosα/sinα
=-cos2α-cos2α/0.5sin2α
=(bring in 150 ° and calculate the rest by yourself)
I wish you a happy study!

Using Tan A / 2 to denote sin a, cos a, Tan a

sina=2tana/2/(1+tan^a/2)
cosa=(1-tan^a/2)/(1+tan^a/2)

It is known that Tan α = - 1 / 3 in the first grade of senior high school mathematics. Find the value of 2 (sin α) 2-3 / 2 (sin α * cos α) + 5 (COS α) 2

The original formula is equal to 2 (Tan α squared) - 3 / 2tan α + 1

Senior one mathematics: known Tan α = 1,3 sin β = sin (2 α + β), find (1) Tan β, 2 Tan (α + β), 3 Tan (α + β) / 2, require detailed explanation. Thank you!

tanα=1
sin2α=2tanα/(1+tan^2α) = 2*1/(1+1^2)=1
cos2α=(1-tan^2α)/(1+tan^2α) = (1-1^2)/(1+1^2)=0
3sinβ=sin(2α+β)
3sinβ=sin2αcosβ+cos2αsinβ
3sinβ=1*cosβ+0*sinβ
3sinβ=cosβ
tanβ=1/3
tan(α+β) = (tanα+tanβ)/(1-tanαtanβ) = (1+1/3)/(1-1*1/3) = 2
tan(α+β) = 2tan{((α+β)/2}/ {1-tan((α+β)/2)^2} = 2
2tan{((α+β)/2} = 2- 2{tan((α+β)/2)^2
tan{((α+β)/2} = 1-{tan((α+β)/2)^2
{tan((α+β)/2)} ^2 + tan((α+β)/2) - 1 = 0
Tan ((α + β) / 2) = (- 1 ± Radix 5) / 2