From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Let's talk about secs (was: A call for curious trigonometric tidbits...)
Date: 9 Jul 1999 17:13:28 -0400
Newsgroups: sci.math
Keywords: unusual elliptic transformation

[My other tidbit comes later on in the reply, but first the calculation
example:]

In article <37863956.84825@news.globalnet.co.uk>,
Andy Spragg <sparge@globalnet.co.uk> wrote:
[...]
:On the subject of Spivak and his fantastic book, who can forget his
:quote, in the context of evaluating integrals by changing variable,
:that "...the world's sneakiest substitution is undoubtedly t =
:tan(x/2)...", which claim he then backs up with a variety of
:astonishing feats of integration. I had reason to recall this equation
:recently when I was amazed to find it cropping up in a completely
:different context. I was messing around with a geometrical problem and
:was led to have to solve the equation:
:
:x.sin(x) + cos(x) = 1
:
[ Here I offer a shortcut: ]

 x * sin(x) = 1 - cos(x)

When can I divide by sin(x)? If x is an odd multiple of pi then it is not
a solution (LHS=0, RHS=2). If x is an even multiple of pi, we have a
solution (a boring one at that).

Other than that, sin(x) is not 0, so

 x = (1 - cos(x)) / sin(x) 

 x = 2 * (sin(x/2))^2 / (2 * sin(x/2) * cos(x/2))

 x = tan(x/2)

and for a full set of real "non-boring" solutions,

 x = (2*k*pi) + 2 * arctan(x)

where k is an integer.

It is true that the right side is contractive near the non-zero solutions,
so that iterations converge, as noted:

[...]
:
:The best way to solve this is by iteration on the rearranged form:
:
:x = 2.atan(x)
:
:which converges like buggery no matter where you start from, 

 ... including from  x=0 (couldn't resist...:-)=

:and the principal value is:
:
:x = 2.331 122 370 414 ...
:
[...]

Now more on tan(x/2):

Contrary to traditions of story-telling, I will give the punchline first:

There is a transformation, useful in the study of elliptical orbits (and
astronomers will know the traditional names of the angles involved):

With a parameter q strictly between -1 and 1, define

 y = x - 2 * arctan (q * sin(x) / (1 + q * cos(x)))

The amazing fact (to me) is the formula expressing x in terms of y:

 x = y + 2 * arctan (q * sin(y) / (1 - q * cos(y)))

(a neat exercise: note that the inverse is obtained when you replace 
q with (-q)).

The transformation x -> y is continuous (real analytic at that) on the
whole real line. Its derivative is

 dy/dx = (1 - q^2) / (1 + q^2 + 2 * q * cos(x))

and if you play with change of parameter:

 2 * q / (1 + q^2) = eps  

where 

 q = eps / (1 + sqrt(1 - eps^2))

(yes, eps is the relative excentricity of a suitable ellipse!)

then 

 dy/dx = sqrt(1 - eps^2) / (1 + eps * cos(x))

This gives an indefinite integral

 Int (1 + eps * cos(x)) dx

 = (1 - eps^2)^(-1/2) * (x - arctan(q * sin(x) / (1 + q * cos(x))))

 + const.

Now I confess there are more punchlines:

 If you use tan(x/2) = t in the above integration problem, you obtain

 Int (1 + eps * cos(x)) dx

 = 2 * (1-eps^2)^(-1/2) * arctan (r * tan(x/2)) + const

where r = sqrt((1-eps)/(1+eps)).

Ouch! 

 The answer is undefined at certain multiples of pi, and will have jumps
no matter how it is extended. That's not what a differentiable function is
supposed to do.

 Well, between -pi and pi, this solution and "my" solution (not really
mine, it's centuries old, unlike myself) are the same (for suitable
const). It is a nice exercise to show it without integral calculus.

And the relationship between eps and q (punchline #3):

 Suppose eps = sin(alpha), then q = tan(alpha/2).

Have fun, ZVK(Slavek).