If α is known to be an acute angle, and there are 2tan (π - α) - 3cos (π 2 + β) + 5 = 0, Tan (π + α) + 6sin (π + β) - 1 = 0, then the value of sin α is______ ．

If α is known to be an acute angle, and there are 2tan (π - α) - 3cos (π 2 + β) + 5 = 0, Tan (π + α) + 6sin (π + β) - 1 = 0, then the value of sin α is______ ．

If α is an acute angle and 2tan (π - α) - 3cos (π 2 + β) + 5 = 0, then - 2tan α + 3sin β + 5 = 0, that is, 2tan α - 3sin β - 5 = 0 ① From Tan (π + α) + 6sin (π + β) - 1 = 0, Tan α - 6sin β - 1 = 0 ② The results show that: 3tan α - 9 = 0, Tan α = 3, Tan α = Si

Given Tan (π - α) = - 3, find √ 3cos α - sin α - -- √ 3cos α + sin α
The long line is a division sign

tan(π-α）=-3
tanα=3
(√3cosα-sinα)/(√3cosα+sinα)
The numerator and denominator are divided by cosa
= (√3-tanα)/(√3+tanα)
= (√3-3)/(√3+3)
= (√3-3)^2/(3-9)
= - 2+√3

It is known that α, β ∈ (0, π / 2) and sin α = sin β cos (α + β). When Tan α takes the maximum value, the value of Tan (α + β) is obtained
I need an answer today. Please help me

sinβ/sinα=cosα*cosβ-sin*αsinβ …… ① (1) * (sin / α cos β) obtains Tan β = sin α cos α - Sin ^ 2 α Tan β, and then divides (1 + sin ^ 2 α) to obtain Tan β = sinacosa / (1 + sin ^ 2 α) because 1 = sin ^ 2 α + cos ^ 2 α, so sin α CoS / α (1 + sin ^ 2 α) =

Given cos α ∈ [1 / 2,1], find the maximum of Tan (α / 2) * (sin α + Tan α)

tan(α/2)*(sinα+tanα)
=sin(α/2)/cos(α/2)*sinα*(1+cosα)/cosα
=sin(α/2)/cos(α/2)*sinα*(2(cos(α/2))^2/cosα
=sin(α/2)*sinα*2cos(α/2)/cosα
=(sinα)^2/cosα
=(1-(cosα)^2)/cosα
=1/cosα-cosα
When cos α is regarded as a variable, 1 / cos α, - cos α are decreasing functions in the domain of definition
The maximum value of the function is at the left end of the domain, that is, cos α = 1 / 2
Original formula = 1 / cos α - cos α
=2-1/2
=3/2

If a is an acute angle and sin (a - π / 6) = 1 / 3, then cosa =?

Sin (a - π / 6) = 1 / 3, sinacos π / 6-cosasin π / 6 = 1 / 3 √ 3sina / 2-cosa / 2 = 1 / 3 √ 3sina cosa = 2 / 3 √ 3sina = CoAA + 2 / 3 square 3 (Sina) ^ 2 = (COSA) ^ 2 + 4cosa / 3 + 4 / 9 3-3 (COSA) ^ 2 = (COSA) ^ 2 + 4cosa / 3 + 4 / 9 let x = cosa 4x ^ 2 + 4x / 3-23 / 9 = 0

Given that a and B are acute angles, cosa = 1 / 7, sin (a + b) = 5 √ 3 / 14, calculate the value of angle B

A is the acute angle sin (A-30 °) = 1 / 3, find cosa

A-30 ° may be an acute angle or a negative angle, but the cosine of the first and fourth quadrants is positive, cos (A-30 °) = √ [1-sin ^ 2 (A-30 °)] = √ 2 / 3, cosa = cos (A-30 ° + 30 °) = cos (A-30 °) cos30 ° - sin (A-30 °) * sin30 ° = (√ 2 / 3) * √ 3 / 2 - (1 / 3) * (1 / 2) = (√ 6-1) / 6

Let a, β ∈ [0, π], if cosa = 3 / 5, sin (a + β) = - 4 / 5, then what is cos β =?

Because cosa = 3 / 5, Sina = 4 / 5
Because sin (a + b) = - 4 / 5
So cos (a + b) = - 3 / 5
sin(a+b)=sinacosb+cosasinb=-4/5
4/5cosb+3/5sinb=-4/5
cos(a+b)=cosacosb-sinasinb=-3/5
3/5cosb-4/5sinb=-3/5
Two equations, eliminate SINB., and solve CoSb = - 1

Given that angle a is an acute angle and cosa = 4 / 5, then sin (90 ° - a) =? Has a process

sin(90°-a)=cosa=4/5

Let Tan α = cos α, then sin α=
How much is it

sina/cosa=cosa
sina=cos²a
sina=1-sin²a
sin²a+sina-1=0
Sina = - 1 / 2 + √ 5 / 2 Sina = - 1 / 2 - √ 5 / 2 rounding off
Given Tan α = cos α, then sin α = - 1 / 2 + √ 5 / 2