(sin1 / X (SiNx + cosx)), when x approaches infinity, what is the limit of function!

(sin1 / X (SiNx + cosx)), when x approaches infinity, what is the limit of function!

SiNx + cosx = radical 2 sin (x + quarter)
Therefore, the formula is reduced to the radical 2 · sin1 / X · sin (x + quarter). When x approaches infinity, one term always approaches 0, so the limit of the function is 0

Limit: when x approaches a, (cosx COSA) / (x-a) thank you, masters

By definition, this is the derivative of the function y = cosx at x = a, of course - Sina. If we have to prove it, we should use the sum difference product formula: cosx cosa = - 2Sin [(x + a) / 2] sin [(x-a) / 2], and then according to the important limit sinu / U - > 1 (U - > 0)

LIM (sin (x-1)) / (x ^ 2 + X-2), X tends to 1, find the limit

Let a = X-1
Then a tends to zero
x²+x-2=(x-1)(x+2)=a(a+3)
So the original formula = LIM (a → 0) Sina / a (a + 3)
=lim(a→0)sina/a*lim(a→0)1/(a+3)
=1*1/(0+3)
=1/3

Given cos (α - π / 60 + sin α) = (4 √ ~ 3) / 5, then the value of sin [α + (7 π) / 6] is
Given cos (α - π / 6 + sin α) = (4 √ ~ 3) / 5, then the value of sin [α + (7 π) / 6] is
A.-（2√￣5）/5 B.（2√￣3）/5 C.-4/5 D.4/5

Analysis:
Because cos (α - π / 6) + sin α = (4 √￣ 3) / 5
So cos α cos (π / 6) + sin α sin (π / 6) + sin α = (4 √￣ 3) / 5
That is cos α * (√ 3) / 2 + sin α * 3 / 2 = (4 √ 3) / 5
cosα*1/2+sinα*(√￣3）/2=4/5
sin(α+π/6)=4/5
So sin [α + (7 π) / 6]
＝sin(π+α+π/6）
=-sin(α+π/6)
=-4/5

1. Simplification of (tan2 π - θ) sin (- 2 π - θ) cos (6 π - θ) / cos (θ - π) sin (5 π + θ)
2. Simplification of sin (15 π / 2 + α) cos (α - π / 2) / sin (9 π / 2 - α) cos (3 π / 2 + α)
3. Prove sin (π / 2 + θ) - cos (π - θ) / sin (π / 2 - θ) - sin (π - θ) = 2 / 1-tan θ

(tan2π-θ)sin(-2π-θ)cos(6π-θ)／cos（θ-π）sin（5π+θ）=(-tanθ)(-sinθ)cosθ／(-cosθ）(-sinθ）=tanθsinθcosθ／cosθsinθ=tanθsin（15π／2+α）cos（α-π／2）／sin（9π／2-α）cos（3π／2+...

How to simplify sin (- A-3 / 2 π) cos (3 / 2 π - a) tan2 (π - a),

sin(-a-3/2π)=cosa
cos(3/2π-a)=-sina
tan2(π-a)=-tan2a
sin(-a-3/2π)cos(3/2π-a)tan2(π-a)
=cosa(-sina)*(-tan2a)
=1/2sin2a*tan2a

(step) simplification: [sin (- α - 3 / 2 π) cos (3 / 2 π - α) · tan2 (π - α)] / [cos (π / 2 - α) · sin (π / 2 + α)]

[sin(-α-3/2π)cos(3/2π-α)·tan2(π-α)]/[cos（π/2-α）·sin（π/2+α）]
=[cosa(-sina)(-tan2a)]/[sina(-cosa)]
=-tan2a

Cos (α - β) = 12 / 13, sin (α - β) = 3 / 5, tan2 α
π/4

If cos (α - β) = 12 / 13, then sin (α - β) = ± 5 / 13
Can add, ask

Ask a question 1 to prove (Tan α, Tan 2 α) / (Tan 2 α - Tan α) + 3 (Sin & sup2; α - cos & sup2; α) = 2 (2 α - π / 3)

Left = (Sina / cosa * sin2a / cos2a) / (sin2a / cos2a Sina / COSA) - √ 3 (COS & sup2; a-SiN & sup2; a) multiply cosas2a = (sinasin2a / (sin2acosa-sinacos2a) - √ 3 (COS & sup2; a-SiN & sup2; a) = (sinasin2a) / sin (2a-a) - √ 3cos2a = sin2a - √ 3cos2a

Solving problems Sin & sup2; 2 α + sin 2 α × cos α - cos 2 α = 1
The domain of definition is between zero and π / 2, and the values of sin α and Tan α are obtained

：∵sin²(2α)+sin(2α)cosα-cos(2α)=1 ∴sin²(2α)+sin(2α)cosα-cos(2α)-1=0==>sin²(2α)+sin(2α)cosα-2cos²α=0==>[sin(2α)+2cosα][sin(2α)-cosα]=0==>(2sinαcosα+2cosα)(2sin...