Given that the arc length of a bow is C and the chord length is l, how to find its area? The key of the problem is to find the arc radius and the corresponding angle 2 α, R = L / 2Sin α, (α is half angle) C=R*2α, The results show that csin α = l α The problem is how to find this half angle α, except that McLaughlin formula takes the first three terms (SiNx = x-x ^ 3 / 3! + x ^ 5 / 5! - x ^ 7 / 7! +...) Besides the approximate value, is there an exact solution It's no good to simplify the double integral with the axis coordinate, and finally get the differential equation

Given that the arc length of a bow is C and the chord length is l, how to find its area? The key of the problem is to find the arc radius and the corresponding angle 2 α, R = L / 2Sin α, (α is half angle) C=R*2α, The results show that csin α = l α The problem is how to find this half angle α, except that McLaughlin formula takes the first three terms (SiNx = x-x ^ 3 / 3! + x ^ 5 / 5! - x ^ 7 / 7! +...) Besides the approximate value, is there an exact solution It's no good to simplify the double integral with the axis coordinate, and finally get the differential equation

There is a dichotomy iteration method which can achieve the required accuracy. The solution of α when f (α) = csin α - L α f (α) = 0 is a value close to f (α) = 0. The iteration method α 1 = α 0-F (α 0) / F '(α 0) α 2 = 1-f (α 1) / F' (α 1) F (α n) = α (n-1) - f (α (n-1)) / F '(α (n-1)) every more

Given the chord length and arc length, find the radius,
1. Chord length 20.2, arc length 21.92, radius
2. Chord length 28.7, arc length 32.06, radius
I know the formula, but how to solve the two equations? Please master the data and write out the operation process in detail, so that I can learn something,
1. Chord length 20.2, arc length 21.92, radius and area
2. Chord length 28.7, arc length 32.06, radius and area

Let the arc length be l, the chord length be B, the angle of the center of the circle be α, and the radius be r. because l = 2 α R, so α = L / 2R, and because the triangle formed by connecting the center of the circle and the two ends of the chord is isosceles triangle, the angle bisector bisects the chord and is perpendicular to the chord, according to the sine theorem: B / 2 △ r = sin α, so the radius r = B / 2Sin α is substituted into the solution under known conditions

(360-x)/40=(360+x)/60+0.5
solve equations

(360-x)/40=(360+x)/60+0.5
3*(360-x)=2*(360+x)+60
1080-3x=720+2x+60
x=60

6x-360（x-1)=360(x-1)-0.5x

6x-360（x-1)=360(x-1)-0.5x
6x-360x+360=360x-360-0.5x
360x+360x-0.5x-6x=360+360
713.5x=720
x=720÷713.5
X = 1 and 13 / 1427

360/x-2=360/1.5x

360/x-2=360/1.5x
360/x-2=240/x
120/x=2
2x=120
x=60
Test: x = 60 is the solution of the equation
The solution of the equation is x = 60

The sequence {an} satisfies that A0 is a constant, an = 3 (n-1) - 2A (n-1), find an

Let B {{n} = a {{n} = a {{n} n + 1 / 3, then B {n-1} = a {{n-1} - (n-1-1) + 1 / 3, then B {n-1} = a {{n-1} = a {{n-1} = a {{n-1} - (n-1-1-1) + (n-1-1) + (n-1) + 1 / 3 = a {n-1-1-1} - n-1-1} - N + N + 4 + N + 4-1} = a {{n-1}, so B {{{n {n} B {n} is a sequence of the common ratio of the number 2-2-2, the general item is B {{- 2)

An = 3 ^ (n-1) - 2A (n-1) A0 is a constant. Why add 3 ^ n / 5 to both sides of the general formula for finding an

I don't understand. Maybe it can be equal ratio or something. But there is another way
1. First, divide both sides by (- 2) ^ n, and the equation becomes an / (- 2) ^ n = 3 ^ (n-1) / (- 2) ^ n-a (n-1) / (- 2) ^ (n-1)
2. Replace an / (- 2) ^ n with BN: BN = (- 2) * - 3 / 2) ^ (n-1) + B (n-1)
3. After BN = (- 2) * [(- 3 / 2) ^ (n-1) + +1]+b0
4. After calculating BN, we use BN = an / (- 2) ^ n to get the general formula of an
Add: not add 3 ^ n / 5, but subtract it!

Let A0 be constant, and an = 3 Λ (n – 1) - 2An – 1 (n belongs to natural number). Suppose that for any n greater than or equal to 1, an is greater than an – 1, and then calculate the value range of Ao

Using the known formula,
There is a [n + 1] = 3 ^ n-2a [n] > a [n]
After finishing, we get a [n]

Let A0 be a constant and an = 3n-1-2an-1 (n ∈ n *)
It is proved that for any n ≥ 1,
an=1／5[3n+(-1)n-12n]+(-1)n-12na0
Suppose that for any n ≥ 1, there is an > an-1,
Find the value range of A0
Let A0 be a constant and an = 3 ^ (n-1) - 2A (n-1) (n ∈ n *)
It is proved that for any n ≥ 1,
An=0.2[3^n+(-1)^(n-1)2^n]+(-1)^(n-1)2^nA0
Suppose that for any n ≥ 1, there is an > an-1,
Find the value range of A0

0

Let A0 be a constant, and the general formula of the sequence {an} is an = 1 / 5 {3 ^ n + [(- 1) ^ (n-1)] * 2 ^ n} + [(- 1) ^ n] * (2 ^ n) * A0. If any n belongs to a positive integer, it does not mean that an is greater than a (n-1), then the value range of A0 is obtained

It can be divided into two cases. One is when n is odd, arrange an and get a (n-1). When n is odd, use an-a (n-1) > 0 to hold. Calculate a range of A0
The other is to calculate the range of another A0 when n is even,
Take the intersection again
Typing trouble, their own count, not difficult