The value of log4 (1 + √ 2 + √ 3) + log4 (1 + √ 2 - √ 3) is equal to

The value of log4 (1 + √ 2 + √ 3) + log4 (1 + √ 2 - √ 3) is equal to

log4(1+√2+√3)+log4(1+√2-√3)
=log4[(1+√2+√3)(1+√2-√3)]
=log4[(1+√2)^2-(√3)^2]
=log4[3+2√2-3]
=log4(2√2)
=log4(2^3/2)
=ln(2^3/2)/ln4
=3/2*ln2/(2ln2)
=3/4

It is known that cos α = 1 / 7, cos (α - β) = 13 / 14, 0 < β < α < π / 2, (1) calculate the value of tan2 α and (2) calculate the value of β

(1) Because cosa = 1 / 7, (Sina) ^ 2 + (COSA) ^ 2 = 1, so (Sina) ^ 2 = 48 / 49, because 0 ＜ β ＜ π / 2, so Sina = 4 √ 3 / 7, Tana = Sina / cosa = (4 √ 3 / 7) / (1 / 7) = 4 √ 3, so tan2a = 2tana / [1 - (Tana) ^ 2] = 8 √ 3 / (- 47) = - 8 √ 3 / 47 (2) is determined by cos α = 1 / 7, cos (α - β) =

Given that cos α = 1 / 7, cos (α - β) = 13 / 14, and 0 < β < α < π / 2 (1), find the value of Tan α (2) find the value of tan2 α (3) find β β = 60

(1)∵cosα=1/7,0＜ β ＜α＜π/2
∴sinα=4√3/7
∴tanα=sinα/cosα=4√3.
(2)tan2α=2tanα/(1-tanα*tanα)=-8√3/47.
(3) Cos (α - β) = 13 / 14, 0 ＜ β ＜ α ＜ π / 2, sin (α - β) = 3 √ 3 / 14
∴sinβ=sin(α-(α-β))
=sinα*cos(α-β)-sin(α-β)*cosα
=(4√3/7)*(13/14)-(3√3/14)*(1/7)
=√3/2.
∴β=60°.

Given that cos α = 1 / 7, cos (α - β) = 13 / 14, and 0 < β < α < π / 2, we can find the value of Tan α and cos β

(1) Cos α = 1 / 7, because 0 < α < π / 2,
So sin α = √ (1-cos 2 α) = √ [1 - (1 / 7) 2] = 4 √ 3 / 7
So tan α = sin α / cos α = 4 √ 3
(2) Cos (α - β) = 13 / 14, because - π / 2 ＜ α - β ＜ π / 2,
So sin (α - β) = √ [1 - (COS 2 (α - β)] = √ [1 - (13 / 14) 2] = 3 √ 3 / 14
According to the cosine formula of two angle difference:
cos[α - (α-β)] = cosαcos(α-β) + sinαsin(α-β)
cosβ = (1/7) * (13/14) + (4 √ 3 / 7) * (3√3 /14)
= 1/2

It is known that cos α = 1 / 7, cos (α - β) = 13 / 14, and 0 ＜ β ＜ α ＜ 2 ＜ (1) calculate tan2 α and (2) calculate β

(1) Cos α = 1 / 7, because 0 < α < π / 2,
So sin α = √ (1-cos 2 α) = √ [1 - (1 / 7) 2] = 4 √ 3 / 7
So tan α = sin α / cos α = 4 √ 3
(2) According to the tangent formula of double angle:
tan2α = (2tanα)/(1 - tan²α)
= (2 * 4 √3)/[1 - (4 √3)²]
= - 8√3 / 47
(3) Cos (α - β) = 13 / 14, because - π / 2 ＜ α - β ＜ π / 2,
So sin (α - β) = √ [1 - (COS 2 (α - β)] = √ [1 - (13 / 14) 2] = 3 √ 3 / 14
According to the cosine formula of two angle difference:
cos[α - (α-β)] = cosαcos(α-β) + sinαsin(α-β)
cosβ = (1/7) * (13/14) + (4 √ 3 / 7) * (3√3 /14)
= 1/2
β = 60 degrees

Given Tan α = 2, find the value of sin (π / 4 + α) / cos (π / 4 + α) · tan2 α

Tan2a = [2tana] / [1-tan 2A] = - 4 / 3 sin (π / 4 + a) / cos (π / 4 + a) = [sin (π / 4) cosa + cos (π / 4) Sina] / [cos (π / 4) cosa sin (π / 4) Sina] = [cosa + Sina] / [cosa Sina] = [1 + Tana] / [1-tana] = - 3, then the original formula = [- 3] / [- 4 / 3] = 9 / 4

If (α / π? Is known, then (α / π?) = 3, π - a) = 1

Tan (π / 4 - α) =? Because cos α = - 12 / 13, α∈ (π, 3 π / 2), so sin α = - 5 / 13, so tan α = 5 / 12, Tan (π / 4 - α) = (1-tan α) / (1 + Tan α * 1) = (7 / 12) / (17 / 12) = 7 / 17

Given cos θ = - 12 / 13, θ∈ [π, 3 π / 2], find the value of Tan (θ - π / 4)

Because cos θ = - 12 / 13, θ∈ [π, 3 π / 2],
So, sin = 5
So tan θ = 5 / 12
tan(θ-π/4)=(tanθ-1)/(1+tanθ）=-7/17

Cos (α + β) = 4 / 5, cos (α - β) = 12 / 13

Cos (α + β) = 4 / 5, cos (α - β) = 12 / 13, Tan α Tan β cos (α - β) - cos (α + β) = (COS α cos β + sin α sin β) - (COS α cos β + sin α sin β) = - 2Sin α sin β = 12 / 13-4 / 5 = 8 / 652sin α sin β = - 8 / 65cos (α + β) + cos (α - β) = (COS α cos β - sin α sin)

It is known that α β belongs to (3 π / 4, π), Tan (α + β) = - 3 / 4, sin (β - π / 4) = 12 / 13 Sorry, alpha is known, not alpha beta

It belongs to (3 π / 4, π) a + β belongs to (3 π / 2,2 π) β - π / 4 belongs to (π / 2,3 π / 4) Tan (α + β) = - 3 / 4sin (α + β) = - 3 / 4sin (α + β) = - 3 / 5ccos (α + β) = 4 / 5sin (β - π / 4) = 12 / 13cos (β - π / 4) = 5 / 13cos (β - π / 4) = 5 / 13cos (a + π / 4) = cos (a + β - β + π / 4) = cos (a + β + β + π / 4) = cos [(a + β + β - (β - π / 4) (4) β - π / 4) cos [(a + β + β - π / 4] = cos (a + β)