If sin AFA + cos AFA = 1 / 3, then Tan AFA = what?

If sin AFA + cos AFA = 1 / 3, then Tan AFA = what?

The solution (Sina COSA) / (Sina + COSA) = 1 / 3  Sina cosa = 1 / 3 (Sina + COSA), that is, Sina cosa = 1 / 3 Sina + 1 / 3 cosa, and the two sides are multiplied by 3 to get: 3sina-3cosa = Sina + Cosa 3sina Sina = cosa + 3cosa 2sina = 4cosa  Tana = Sina / cosa = 2

Let Tan (a + b) = 1 / 3, cos = - 2 / 2

Tan (a + b) = (Tana + tanb) / (1-tana * tanb), Tana + tanb = 7 / 3; Tana + tanb = (Sina / COSA) + (SINB / CoSb) = (sinacosb + sinbcosa) / (COSA * CoSb) = sin (a + b) / (COSA * CoSb), Tan (a + b) = 7, sin (a + b) = (+) (7 * √ 2) / 10, so cosa * CoSb = (+ -)

If Tan a = 1 / 2, Tan B = 3, what is Tan (a-b) = then?

tan（a—b）＝(tan a -tan b）/（1+tan a*tan b)
=(1/2-3)/(1+3/2)
=-1

It is known that Tan α = 2, a ∈ (π, 3 π / 2)
sin(π＋α)＋2sin((3π／2)＋α)
————————————
cos(3π－α)＋1
(2）sin(－4π－α)
Please do me a favor. I'm in the exam,

The original formula = - sin α - 2cos α. = -- (sin α + cos α) = - sec α (2 + 1) = - (- √ (1 + Tan ^ 2 α) = 3 √ 5. Cos (3 π - α) + 1 = = - cos α + 1. = 1 + √ 5 / 5. Sin (- 4 π - α) = - sin α = - 1 / CSC α = 2 √ 5 / 5

Given that β - a = γ - β = π / 3, find the value of Tan α Tan β + Tan β Tan γ + Tan γ Tan α

Because Tan (β - α) = (Tan β - Tan α) / (1 + Tan α, Tan β), Tan α, Tan β = (Tan β - Tan α) / Tan (β - α) - 1 = (Tan β - Tan α) / √ 3-1
Similarly, Tan β, Tan γ = (Tan γ - Tan β) / 3-1, Tan γ, Tan α = - (Tan γ - Tan α) / 3-1
The sum of Tan α, Tan β, Tan γ is - 3

It is known that Tan α = 3 / 4 α ∈ (π, 3 π / 2)
1) Finding the value of cot2 α
2) Finding the value of 2cos ^ 2 (α - π / 4) - 1

tan2α=2tanα/(1-tan²α）=2*3/4/(1-9/16)=24/7∴cot2α=1/tan2α=7/24（2）2cos²(α-π/4)-1=cos(2α-π/2)=sin2αα∈(π,3π/2)2α∈(2π,3π)sin2α>0∵tan2α>0∴cos2α>0∵tan2α=7/24∴sin2α=7/2...

Given that Tan (a + b) = 7, Tana = 3 / 4, and B belongs to (0, π), then B=

tan(a+b）
=(tana+tanb)/[1-tanatanb]
=(3/4+tanb)/(1-3tanb/4)
=7
tanb=1
And B belongs to (0, π)
So B = π / 4

It is known that cos (15 π / 2 - α) = 1 / 3, | α|

cos(15π/2-α)=-sinα
∴sinα=-1/3
Ψ cos α = 2 / 3 · radical 2
Tan α = - radical 2 / 4
tan(π-α)
=-tanα
=Root 2 / 4

It is known that Tan (α - β) = 1 / 2, Tan β = - 1 / 7, and α and β belong to (- π, 0), so the value of 2 α - β can be obtained
In the analysis, Tan α = 1 / 3, Tan (2 α - β) = 1
But "by - π"

But "by - π"

What is the initial image in trigonometric function

In the trigonometric function y = asin (ω x + φ), the,
A> 0 is called amplitude,
ω> 0 is called angular frequency,
ω x + φ is called phase, where the phase when x = 0 is called initial phase