What is the tan value for a 40 degree angle?

What is the tan value for a 40 degree angle?

zero point eight three nine

If the Tan of an angle is 0.5, what is the angle?

26.6°

Given that sin (π - a) = 1 / 4 of log8 base, tangent a belongs to (- π / 2,0), find the value of Tan (3 π / 2 + a)

sin(π-a)=log8(1/4) = {log2(1/4)}/{log2(8) = (-2)/3 = -2/3
sina=-2/3
A belongs to (- π / 2,0)
tan(3π/2+a)=tan{2π-π/2+a)=tan(-(π/2-a))=-tan(π/2-a)=-cota=-{√(1-sin^2a)/sina}
= -√(1-4/9)/(-2/3) = √5/2

Given that sin (α + β) = 1 / 2, sin (α - β) = 1 / 3, calculate the value of Tan β / Tan α with log base of five

sin(α+β)=sinacosb+cosasinb=1/2
sin(α—β)=sinacosb-cosasinb=1/3
The sum of the two formulas shows that sinacosb = 5 / 12
By subtracting the two formulas, cosasinb = 1 / 12
So,
1/tanα*tanβ=sinbcosa/cosbsina=(1/12) / (5/12)=1/5
Log with five as the base, logarithm of Tan β / Tan α = log5 (1 / 5) = - 1

Log (Tan θ + cot θ) sin θ = - 1 / 4, θ∈ (0, π / 2), then log (Tan θ) sin θ =? It takes process. Thank you

Because (Tan θ + cot θ) = 1 / (sin θ * cos θ),
Therefore, from log (Tan θ + cot θ) sin θ = - 1 / 4, it is concluded that:
log(sinθ) (tanθ+cotθ)=-4,
log(sinθ) [1/(sinθ*cosθ)]=-4,
log(sinθ) (sinθ*cosθ)=4,
1+log(sinθ) (cosθ)=4,
log(sinθ) (cosθ)=3,
log(sinθ) (tanθ)=1-log(sinθ) (cosθ)=-2,
log(tanθ)sinθ=-1/2.

Given that sin (α + β) = 1 / 2, sin (α - β) = 1 / 3, find log (√ 5) (Tan α / Tan β) It is known that sin (α + β) = 1 / 2 sin(α—β)=1/3, Find log (√ 5) [(Tan α / Tan β) ^ 2]

The solution is obtained by sin (α + β) = 1 / 2
That is sin α cos β + cos α sin β = 1 / 2
When sin (α - β) = 1 / 3
That is sin α cos β - cos α sin β = 1 / 3
Adding two formulas
sinαcosβ=5/12
cosαsinβ=1/12
Division of two forms
sinαcosβ/cosαsinβ=5
That is, Tan α / Tan β = 5
Log (√ 5) (Tan α / Tan β)
=log(√5)5
=log(√5)（√5）²
=2log(√5)（√5）
=2

Given sin (π - α) = log8 / 4 and α∈ (- π / 2,0), find Tan (2 / 3 π + α) Given sin (π - α) = log8 quarter and α∈ (- π / 2,0), find Tan (2 / 3 π + α)

The fourth quarter of log 8 = - 2 LG2 / 3lg2 = - 2 / 3 sin (π - α) = log8 (1 / 2 / 3), Sina = - 2 / 3 cosa = √ 5 / 3tan (2 / 3 π + α) = (sin2 / 3 π cosa + Cos2 / 3 π Sina) / (Cos2 / 3 π cosa-sin2 / 3 π Sina) = [(3 / 3 / 2 * (5 / 3-1 / 2 * (- 2 / 3)) / [(- 1 / 2)) * (5 / 3 - / 3-1 / 2 * (- 2 / 3)) / [(- 1 / 2 / 2) * 5 / 3 / 3 / 3 / 2 / 2 * (- 2-2-2-2-2-2-2 / 2 / 2 / 2 / 2 / 2 / / 3

Given Tan θ = - 2 √ 2, find the value of (2cos 2 θ / 2-sin θ - 1) / (√ 2Sin (θ + π / 4))

（2cos²θ/2-sinθ-1)/(√2sin(θ+π/4))={[2cos(θ/2)-1]-sinθ}/{√2[1/√2cosθ+1/√2sinθ}=( cosθ- sinθ)/( cosθ+sinθ)
=(1-tanθ)/(tanθ+1)
=(1+2√2)(1-2√2)
=-9/7-28√2

It is known that sin (π / 42 α) * sin (π / 4-2 α), α∈ (π / 4, π / 2) can be used to find 2 sin 2 α Tan α - 1 / Tan α - 1

The title should be known as sin (π / 4 + 2 α) · sin (π / 4-2 α) = 1 / 4, α∈ (π / 4, π / 2), and find the value of 2Sin ^ 2 α + Tan α - 1 / Tan α - 1
sin(π/4+2α)sin(π/4-2α)
=1/2*(cos(π/4+2α-π/4+2α)-cos(π/4+2α+π/4-2α))
=1/2*(cos(4α)-cos(π/2))
=1/2*cos(4α)
=1/4
So cos (4 α) = 1 / 2
Because α belongs to (π / 4, π / 2)
So π

Given Tan α = 2, find the value of (sin α + cos α) 2

The original formula = 1 + 2Sin α cos α = 1 + 2Sin α cos α / (sin? α + cos? α)
The denominator of sincos α divided by
The original formula = 1 + 2 / (Tan α + 1 / Tan α) = 9 / 5