If a is the fourth quadrant angle, Tan (the third school + a) = - 5 / 12, then what is the cos (sixth faction-a)?

If a is the fourth quadrant angle, Tan (the third school + a) = - 5 / 12, then what is the cos (sixth faction-a)?

∵1/cos²(π/3+a)=1+tan²(π/3+a)=1+25/144=169/144
∴cos(π/3+a)=12/13
A is the fourth quadrant angle, Tan (π / 3 + a) = - 5 / 12

cosα*tanα

cosa*tana
=cosa*sina/cosa
=sina

Given that α is an inner angle of triangle sin α + cos α = - 1 / 5, find the value of Tan α

Sin α = - 1 / 5 - cos α sin ^ 2 α + cos ^ 2 α = 1 substitution

Let a be the inner angle of a triangle, tangent sin a + cos a = 1 / 5, and find the value of Tan a

0

0

Solution (1) ∵ sin α + cos α = 1
5，∴cosα=1
5-sinα，
∵sin2α+cos2α=1，
∴25sin2α-5sin α-12=0．
∵ α is the inner angle of the triangle,

Given that ∝ is the inner angle of a triangle, and sin ∝ + cos ∝ = 1 / 5, find Tan ∝. Why is tan ∝ = - 4 / 3? Why is tan ∝ = - 3 / 4 to be omitted?

Sin a + cos a = 1 / 5, so (Sina) ^ 2 + 2sinacosa + (COSA) ^ 2 = 1 / 25 and because (Sina) ^ 2 + (COSA) ^ 2 = 1, so sinacosa = - 12 / 25 because a is the inner angle of a triangle, so a is an obtuse angle

Given α, β∈ (0, π / 4), and 3sin β = sin (2 α + β), 4tan α / 2 = 1-tan α / 2, calculate the value of α + β

3 sin β = sin (2 α + β), that is, 3 sin (α + β) cos α - 3cos (α + β) sin α = sin (α + β) cos α + cos (α + β) sin α sin (α + β) cos α = 2cos (α + β) sin α = 2cos (α + β) sin α, namely Tan (α + β) = 2tan α. 4tan α / 2 = 1-tan α / 2,2 * [2 * Tan α / 2 / (1-tan α / 2)] = 1, that is, 2tan α = 1, Tan α = 1 / 2, so, therefore, therefore, therefore, therefore, this paper, therefore, therefore, therefore, it can be seen that the 2tan α = 1, Tan α = 1, Tan α = 1 / 2, so, therefore, therefore, therefore, therefore, therefore, therefore, therefore, it is Tan (α + β) = 1, because α, β ∈ (0, π / 4), α + β = π / 4

It is known that α is an acute angle, and sin α = 7 / 8sin β, Tan α = 1 / 4tan β

sina=7/8sinb
tana=1/4tanb
sina/cosa=1/4sinb/cosb
cosa=7/2cosb
sin^2a+co^2a=(7/8sinb)^2+(7/2cosb)^2=1
49/64(sinb)^2+49/4(cosb)^2=1
49/64+(15*49)/64(cosb)^2=1
(cosb)^2=1/49
cosb=1/7
cosa=1/2
a=60

Given that Tan α = 1 and 3sin β = sin (2 α + β), find the value of Tan (α + 2 β)

tanα=1
α=kπ+π/4
3sinβ=sin(2α+β)=sin(π/2+β)=cosβ
tanβ=1/3
tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=(1+1/3)/(1-1*1/3)=2
So:
tan(α+2β)=(tan(α+β)+tanβ)/(1-tan(α+β)tanβ)=(1/3+2)/(1-2*1/3)=7

Sin (2 α + β) = 3sin β, Tan α = 1, find Tan β The method is easy to understand

tanα=1
α=kπ+π/4
sin(2α+β)=sin(π/2+β)=cosβ
therefore
tanβ=sinβ/cosβ=sinβ/3sinβ=1/3