## Friday, 3 October 2014

### The Spherical Pythagorean Theorem.

Most high-school students in the USA (OK, include most college math majors) could not give the spherical equivalent of the well known Pythagorean Theorem in the plane. I have managed several times to work it into a diversion in classes for analytic geometry or intro calculus and it always found an appreciative reception from most students. I was recently reminded of the theorem in the wonderful book, Heavenly Mathematics by Glen Van Brummelen, which I highly recommend to any HS teacher or student, or anyone interested in entertaining mathematics. There is even a link below.

At first glance the spherical Pythagorean Theorem , Cos(c) = Cos(a) Cos(b), looks little like the planar version, and it is in bringing them together that the beauty of the Spherical and Planar relationships come alive.

The idea of representing sides as arcs, although it should not be, strikes students a little off center and needs some discussion. If they are well accepting of the idea of using unit circle arc lengths for the (in their mind) angular functions, then they should be able to quickly comprehend the idea of a unit sphere with great circle arcs being the "sides" of a spherical triangle, and be able to visualize using trig measurements for them.

The Taylor expansion of the cosine function which is a key to linking the spherical and planar theorems, although it is a calculus topic, existed well before calculus. It seems it may have been discovered at least as early as the 14th century by an Indian mathematician, Madhava of Sangamagrama. It seems he found a number of special cases of the Taylor series, including those for the trigonometric functions of sine, cosine, tangent, and arctangent.
$cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} \cdots$

For our purposes it would be sufficient to use only two terms of the expansion. The student deserves an explanation about why that is so. Suppose we drew a triangle on the "spherical" earth whose sides were ten miles long or less. With a radius of approximately 4,000 miles, this gives a radian base arc length of 1/400 or about .0025.

At this level, even the second term is very small, but the third term would be $\frac{.0025^4}{24}= 1.67 (10)^{-12}$

Now we take the spherical theorem, and substitute the first two terms of the expansion for each angle.

cos (c) = cos (a) cos(b) becomes $1-\frac{c^2}{2}= (1-\frac{a^2}{2})(1-\frac{b^2}{2})$

Expanding the binomial product on the right side we get

$1-\frac{c^2}{2}= 1-\frac{a^2}{2} -\frac{b^2}{2} + \frac{a^2 b^2}{4}$

Subtracting one from both sides and multiplying by -2 we obtain
$c^2= a^2 + b^2 - \frac{a^2 b^2}{2}$

and that term on the right we are subtracting is less than 2 x 10-10

So even for triangles on the earth at a ten mile side length, the planar version of the
Pythagorean theorem gives a very good approximation, but the spherical is so much easier.

And here is the link to that wonderful book I mentioned above.