Given that α and β are acute angles, sin α = 3 / 5, cos (α + β) = - 4 / 5, find the value of 2 α + β

Given that α and β are acute angles, sin α = 3 / 5, cos (α + β) = - 4 / 5, find the value of 2 α + β

Because α, β are acute angles, sin α = 3 / 5, cos (α + β) = - 4 / 5, so cos α = √ (1-sin & # 178; α) = 4 / 5sin (α + β) = √ [1-cos & # 178; (α + β)] = 3 / 5, so cos (2 α + β) = cos [α + (α + β)] = cos α cos (α + β) - sin α sin (α + β) = 4 / 5 × (- 4 / 5) - (3 / 5) × (3 / 5) =

If sin + cos = m, Sin &; Cos = n, then what is the relationship between M and N?

n/sin=cos+sin=m

Let α be an acute angle, M = LG (1-cos α), n = LG (1 / 1 + cos α), then LG (1 / sin α) =?

Let α be an acute angle, M = LG (1-cos α), n = LG (1 / 1 + cos α),
m+n=lg[1/(1-cos^2a)]=lg(1/sin^2a)=2lg(1/sina)
So LG (1 / Sina) = (M + n) / 2

In the RT triangle, the angle a is an acute angle. The proof is sin ^ A + cos ^ B = 1
Mobile phone questions, pay attention to answer the question in the symbol of mobile phone to be able to recognize

A. B is an acute angle and C is a right angle. A, B and C are their corresponding edges
Then (Sina) ^ 2 + (CoSb) ^ 2 = a ^ 2 / C ^ 2 + B ^ 2 / C ^ 2
=(a ^ 2 + B ^ 2) / C ^ 2 (Pythagorean theorem: A ^ 2 + B ^ 2 = C ^ 2)
Then (Sina) ^ 2 + (CoSb) ^ 2 = 1

In the RT triangle ABC, the angle c is 90 degrees, and the angle α is one of the acute angles. If sin α = 3 / 5, the values of cos α and Tan α can be obtained (solved by two methods)

The first kind
∵∠α is an acute angle and sin α = 3 / 5
∴cosα=√(1-sin²α)=√[1-(3/5)²]=4/5
∴tanα=sinα/cosα=(3/5)/(4/5)=3/4
The second kind
Let a = α, ab = 5, BC = 3
∵sinα=3/5
∴AC=√(AB²-BC²)=√(5²-3²)=4
∴cosα=AC/AC=4/5
∴tanα=BC/AC=3/4

What is the relationship between sin a and COS B when the angle c is 90 degrees in RT △ ABC?
emergency

In RT △ ABC, Sina = A / C, CoSb = A / C, that is, Sina = CoSb. Please remind me that after you get the answer, don't forget to adopt it in time. If you adopt it, you will get 2 experience points! Please take time to adopt it, and everything goes well!

As shown in the figure, in RT Δ ABC, e and F are the trisection points of hypotenuse ab. given CE = sin α, CF = cos α (α is an acute angle), what is the length of edge AB?
Online waiting, if wonderful also points, do not use cosine theorem

Through E and F, make eg | BC, FH | BC, intersect AC with G and h. because e and F are the three equal points of hypotenuse AB, they are Ag = GH = HC, AE = EF = FB; 2EG = HF, because CE = sin α, CF = cos α, so CE ^ 2 + CF ^ 2 = 1, that is: (eg ^ 2 + GC ^ 2) + (HF ^ 2 + HC ^ 2) = 1gc = 2ag (eg ^ 2 + 4ag ^ 2) + (4Eg ^ 2 + Ag ^ 2) = 15 (eg

1. Given that a is an acute angle, sin a = cos 50 °, then a is -------- 2. In RT △ ABC, if angle c = 90 degrees, ab = 4bc, then cos a=-----

Let BC be x, then AB be 4xac ^ 2 + x ^ 2 = (4x) ^ 2 (Pythagorean theorem), so: AC = √ 15xcosa = AC / AB = √ 15 / 4 (radical 15:4) sin (Pie / 2 - a) = cosacos (Pie / 2 - a) = sinatg (Pie / 2 - a) = ctgactg (Pie / 2 - a) = TGA

Given that a and B are acute angles, and COS (a + b) = sin (a-b), then a=

Here a = 45 degrees

Given a point P (4T, - 3T), (t ≠ 0) on the terminal edge of angle α, find the value of sin and cos of α

sinα=4t/√(〖(4t)〗^2+〖(-3t)〗^2 )
cosα=-3t/√(〖(4t)〗^2+〖(-3t)〗^2 )
t> At 0, sin α = 4 / 5, cos α = - 3 / 5
t