It is proved that sin (a + b) / 2 * cos (a-b) / 2 = COSC / 2

It is proved that sin (a + b) / 2 * cos (a-b) / 2 = COSC / 2

cosC/2=cos(π-A-B)/2
=sin(A+B)/2
If you want to prove that the original form is true, there are two reasons
sin(A+B)/2[cos(A-B)/2-1]=0
sin(A+B)/2≠0
So, for the equation to hold, only cos (a-b) / 2 = 1
A-B=0
As you can see, there is something wrong with your topic
Now I don't know where I got the title. Your time in high school is too precious. It's a pity!

If cos α - cos β = 1 / 2, sin α - sin β = - 1 / 3, then cos (α - β)=

cosα-cosβ=1/2,sinα-sinβ=-1/3
(cosα-cosβ)^2+(sinα-sinβ)^2=1/4+1/9=13/36=2-2cos(α-β)
cos(α-β)= (2-13/36)/2=59/72

Given sin α - cos β = 1 / 3, cos α + cos β = 3 / 4, α, β ∈ (0, π / 2), what is the value of sin [(α + β) / 2]?

Sin α - cos β = 1 / 3, cos α + cos β = 3 / 4. By adding the two formulas, sin α + cos α = 1 and Sina ^ 2 + cosa ^ 2 = 1 can be solved simultaneously. Then sin (A / 2) and COS (A / 2) can be obtained by using the half angle formula. Sin [(α + β) / 2] can be expanded by using the sum of two angles formula and sin (A / 2) and COS (A / 2) can be substituted

Given that α, β, γ ∈ (0, π / 2), sin α + sin γ = sin β, cos β + cos γ = cos α, find the value of α - β

For convenience, a = α, B = β, C = γ
Then sinc = SINB Sina, COSC = CoSb cosa
sin^2c+cos^2c=1
That is, (SINB Sina) ^ 2 + (CoSb COSA) ^ 2 = 1
That is sin & # 178; b-2sinasinb + Sin & # 178; a + cos & # 178; b-2cosacosb + cos & # 178; a = 1
That is, 2-2 (sinasinb + cosacosb) = 1
That is cos (B-A) = 1 / 2, so B-A = π / 6 or - π / 6
Since a, B and C are acute angles, we know b > A from sinc = SINB Sina > 0, so B-A = π / 6

Given α ∈ (0, π), compare the size of 2sin2 α and sin α / 1-cos α

From any angle, we know that 2sin2a > SIA / 1-cosa

Then we can find the process α (sincotans) = α - 2

Tan (π - α) = - 1 / 2, Tan (- α) = - 1 / 2 to get Tan α = 1 / 2 sin α cos α - 2Sin & # 178; α = (sin α cos α - 2Sin & # 178; α) / (Sin & # 178; α + cos & # 178; α) = (Tan α - 2tan & # 178; α) / (Tan & # 178; α + 1) [the molecular denominator of the above formula is divided by cos & # 178; α] = [1

Cos (π + α) + 6cos (- α) / sin (2 π - α) + 4sin (half π - α) + 4sin (half π + α) = 5
Calculation of (1) Tan α (2) sin 2 α

(1) The original formula = - cosa-6cosa / - Sina + 4cosa + 4cosa = - 7cosa / - Sina + 8cosa = 5, - 7cosa = - 5sina + 40cosa, 5sina = 47cosa, so Tana = 47 / 5; (2) because Tana = Sina / cosa = 47 / 5, and Sina + cosa = 1, we can get Sina and cosa

If Tan α = 2, then (4sin α + cos α) / (sin α + cos α)=

(4sinα+cosα)/(sinα+cosα)
The numerator and denominator are simultaneously divided by cos α
=(4tanα+1)/(tanα+1)
=(4*2+1)/(2+1)
=3

Why is the limit of sin ^ x times cos ^ x (when x goes to 0) = 1 / 4sin ^ 2 * 2x?
The limit of sin ^ x times cos ^ x (when x goes to 0)
Why=
1/4sin^2*2x?
When 1 / 4sin ^ 2 * 2x uses the law of Robida,
Also = cos2x * sin2x
The first narrative error should be:
The limit of sin ^ 2 * x multiplied by cos ^ 2 * x (when x goes to 0)
Why=
1/4sin^2*2x?

Double angle formula SiNx times cosx = 1 / 2 * sin2x
So sin ^ 2 * x times cos ^ 2 * x = 1 / 4sin ^ 2 * 2x
The derivation of F (x) = 1 / 4sin ^ 2 * 2x is the derivation of composite function, and the other u = SINV, v = 2x, so f (x) = 1 / 2U ^ 2, the derivation rule of composite function is OK. This is the most basic in the derivation. I suggest reading a book

Using the law of lobida to find the limit LIM (Tan π X / 2x + 1) ^ 1 / x, X tends to infinity
LIM (Tan π X / 2x + 1) ^ 1 / x, X tends to infinity,

If the expression is not clear, please bracket the denominator