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## Segments of Circles:
Suppose you have a segment of a circle, bounded by an arc of the circle
and the chord subtending it.
Let the length of the arc be |

s, c | s, r | s, h | s, theta | s, d | s, K |

c, r | c, h | c, theta | c, d | c, K | |

r, h | r, theta | r, d | r, K | ||

h, theta | h, d | h, K | |||

theta, d | theta, K | ||||

d, K |

Reader Dan Erb has programmed these cases into an Excel Spreadsheet, which is available for download.

for x, which must be done numerically (see below). Thenc/s = sin(x)/x,

theta = 2x, r = s/theta, d = r cos(x), h = r - d, K = r^{2}[theta-sin(theta)]/2.

theta = s/r, c = 2r sin(theta/2), d = r cos(theta/2), h = r - d, K = r^{2}[theta-sin(theta)]/2.

for x, which must be done numerically (see below). Then2h/s = (1-cos[x])/x,

theta = 2x, r = s/theta, c = 2r sin(x), d = r - h, K = r^{2}[theta-sin(theta)]/2.

r = s/theta, c = 2r sin(theta/2), d = r cos(theta/2), h = r - d, K = r^{2}[theta-sin(theta)]/2.

for x, which must be done numerically (see below). Then2d/s = cos(x)/x,

theta = 2x, r = s/theta, c = 2r sin(x), h = r - d, K = r^{2}[theta-sin(theta)]/2.

for theta, which must be done numerically (see below). Then2K/s^{2}= (theta-sin[theta])/theta^{2}

r = s/theta, c = 2r sin(theta/2), d = r cos(theta/2), h = r - d.

theta = 2 arcsin(c/[2r]), s = r theta, d = r cos(theta/2), h = r - d, K = r^{2}[theta-sin(theta)]/2.

r = (c^{2}+4h^{2})/(8h), theta = 2 arcsin(c/[2r]), s = r theta, d = r - h, K = r^{2}[theta-sin(theta)]/2.

r = c/(2 sin[theta/2]), s = r theta, d = r cos(theta/2), h = d - r, K = r^{2}[theta-sin(theta)]/2.

r = sqrt(c^{2}+4d^{2})/2, h = r - d, theta = 2 arcsin(c/[2r]), s = r theta, K = r^{2}[theta-sin(theta)]/2.

for theta, which must be done numerically (see below). Then4K/c^{2}= (theta-sin[theta])/(1-cos[theta])

r = c/(2 sin[theta/2]), s = r theta, d = r cos(theta/2) h = r - d.

d = r - h, theta = 2 arccos(d/r), c = 2r sin(theta/2), s = r theta, K = r^{2}[theta-sin(theta)]/2.

s = r theta, d = r cos(theta/2), h = r - d, c = 2r sin(theta/2), K = r^{2}[theta-sin(theta)]/2.

h = r - d, theta = 2 arccos(d/r), c = 2r sin(theta/2), s = r theta, K = r^{2}[theta-sin(theta)]/2.

for theta, which must be done numerically (see below). Then2K/r^{2}= theta - sin(theta)

s = r theta, c = 2r sin(theta/2), d = r cos(theta/2), h = r - d.

r = h/(1-cos[theta/2]), d = r - h, c = 2r sin(theta/2), s = r theta, K = r^{2}[theta-sin(theta)]/2.

r = h + d, theta = 2 arccos(d/r), c = 2r sin(theta/2) s = r theta, K = r^{2}[theta-sin(theta)]/2.

for theta, which must be done numerically (see below). Then2K/h^{2}= (theta-sin[theta])/(1-cos[theta/2])^{2}

r = sqrt(2K/[theta-sin(theta)]), d = r - h, c = 2r sin(theta/2) s = r theta.

r = d/(cos[theta/2]), h = r - d, c = 2d tan(theta/2), s = r theta, K = r^{2}[theta-sin(theta)]/2.

r = sqrt(2K/[theta-sin(theta)]), s = r theta, d = r cos(theta/2), h = r - d, c = 2r sin(theta/2).

for theta, which must be done numerically (see below). ThenK/d^{2}= (theta-sin[theta])/(1+cos[theta])

r = d/cos(theta/2), h = r - d, c = 2r sin(theta/2), s = r theta.

Guess a starting value x(0). This will depend heavily on the form of f(x). Then for each n = 0, 1, 2, ..., compute

where f'(x) is the derivative of f(x) with respect to x. Continue this until |x(n+1)-x(n)| is smaller than the accuracy sought. Then x(n+1) agrees with the actual root to at least that level of accuracy.x(n+1) = x(n) - f[x(n)]/f'[x(n)],

If the starting value for x(0) is close enough to a root, this will converge, and very rapidly. If x(0) is rather far from a root, this process may diverge, or exhibit other undesirable behavior.

To solve, for example, sin(x)/x = k for some constant k > 0 (Case 1 above), is the same as finding a root of the equation

This can be done using Newton's Method as follows.f(x) = sin(x) - kx = 0.

Guess a starting value of

Then for each n = 0, 1, 2, ..., computex(0) = sqrt(6-6k).

x(n+1) = x(n) - (sin[x(n)]-kx(n))/(cos[x(n)]-k).

Example: Solve sin(x)/x = 3/4 to five decimal places of accuracy. We carry seven places of accuracy in our calculations:n x(n) sin(x[n]) cos(x[n]) sin(x[n])/x(n) ======================================================== 0 1.2247449 0.9407193 0.3391860 0.7680941 1 1.2786882 0.9576389 0.2879717 0.7489229 2 1.2757074 0.9567763 0.2908250 0.7499967 3 1.2756981 0.9567736 0.2908338 0.7500000 4 1.2756981Since |x(4)-x(3)| < 0.000005, we have the answer correct to five decimal places, x = 1.27570.

There is also an infinite series which can solve sin(x)/x = k, 0 < x <= . Let y = 1 - k. Then 0 < y < 1, and the following infinite series converges in this range:

This converges rapidly for large values of k = 1 - y.x = sqrt(6y)(1 + 3y/20 + 321y^{2}/5600 + 3197y^{3}/112000 + 445617y^{4}/27596800 + 1766784699y^{5}/179379200000 + ...).

Another series is

This converges rapidly for small values of k.x = /(1+k) - (^{3}k^{3}/6)(1 - 4k + [10+9^{2}/20]k^{2}- [20+16^{2}/5]k^{3}+ ...), = /(1+k) - (^{3}k^{3}/[6(1+k)^{4}]) - (3^{5}k^{5}/40)(1 - 64k/9 + ...).

Newton's Method works for (1-cos[x])/x = k (Case 3 above), with the initial start

and iterationx(0) = 2k,

x(n+1) = x(n) - (cos[x(n)]+kx(n)-1)/(-sin[x(n)]+k).

Newton's Method works for cos(x)/x = k (Case 5 above) with the initial start

and iterationx(0) = sqrt(2+k^{2}) - k,

x(n+1) = x(n) - (cos[x(n)]-kx[n])/(-sin[x(n)]-k).

Newton's Method works for k = (x-sin[x])/x

and iterationx(0) = 6k,

x(n+1) = x(n) - (kx(n)^{2}-x(n)+sin[x(n)])/ (2kx(n)-1+cos[x(n)]).

Newton's Method works for k = (t-sin[t])/(1-cos[t]) (Case 11 above), with the initial start

and iterationt(0) = 3k,

t(n+1) = t(n) - (k cos[t(n)]-sin[t(n)]+t(n)-k)/ (1-k sin[t(n)]-cos[t(n)]).

Newton's Method works for k = x - sin(x) (Case 15 above), with the initial start

and iterationx(0) = (6k)^{1/3},

x(n+1) = x(n) - [x(n)-sin(x(n))-k]/[1-cos(x(n))].

Newton's Method works for k = (t-sin[t])/(1-cos[t/2])

and iterationt(0) = 32/(3k),

t(n+1) = t(n) - (k[1-cos(t(n)/2)]^2-t(n)+sin[t(n)])/ (k sin[t(n)/2]-k sin[t(n)]/2-1+cos[t(n)]).

Newton's Method works for k = (t-sin[t])/(1+cos[t]) (Case 21 above), with the initial start

and iterationt(0) = (12k)^{1/3},

t(n+1) = t(n) - (t(n)-sin[t(n)]-k cos[t(n)]-k)/ (1-cos[t(n)]+k sin[t(n)]).

t[0] = (12F)for n = 0, 1, 2, 3, .... The sequence t[n] will converge to theta. Then^{1/3}, t[n+1] = [sin(t[n])-t[n]cos(t[n])+2F]/[1-cos(t[n])],

h/(2r) = [1-cos(theta/2)]/2.

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