How I Impressed My Wife: Index

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

\displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}

Yes, I married well indeed.

In this post, I collect the posts that I wrote last summer regarding various ways of computing this integral.

Part 1: Introduction
Part 2a, 2b, 2c, 2d, 2e, 2f: Changing the endpoints of integration, multiplying top and bottom by \sec^2 x, and the substitution u = \tan x.
Part 3a, 3b, 3c, 3d, 3e, 3f, 3g, 3h, 3i: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and the magic substitution u = \tan \theta/2.
Part 4a, 4b, 4c, 4d, 4e, 4f, 4g, 4h: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and contour integration using the unit circle
Part 5a, 5b, 5c, 5d, 5e, 5f, 5g, 5h, 5i, 5j: Independence of the parameter a, the magic substitution u = \tan \theta/2, and partial fractions.
Part 6a, 6b, 6c, 6d, 6e, 6f, 6g:Independence of the parameter a, the magic substitution u = \tan \theta/2, and contour integration using the real line and an expanding semicircle.
Part 7: Concluding thoughts… and ways that should work that I haven’t completely figured out yet.

Proving theorems and special cases (Part 12): The sum and difference formulas for sine

In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:

Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?

Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first 10^{316} cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.

However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.

3. Theorem 1. \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha + \sin \beta

Theorem 2. \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta

For angles that are not acute, these theorems can be proven using a unit circle and the following four lemmas:

Lemma 1. \cos(x - y) = \cos x \cos y + \sin x \sin y

Lemma 2. \cos(x + y) = \cos x \cos y - \sin x \sin y

Lemma 3. \sin(\pi/2 - x) = \cos x

Lemma 4. \cos(\pi/2 - x) = \sin x

Specifically, assuming Lemmas 1-4, then:

\sin(\alpha + \beta) = \cos(\pi/2 - [\alpha + \beta]) by Lemma 4

= \cos([\pi/2 - \alpha] - \beta)

= \cos(\pi/2 - \alpha) \cos \beta + \sin(\pi/2 - \alpha) \sin \beta by Lemma 1

= \sin \alpha \cos \beta + \cos \alpha \sin \beta by Lemmas 3 and 4.

Also,

\sin(\alpha - \beta) = \cos(\pi/2 - [\alpha - \beta]) by Lemma 4

= \cos([\pi/2 - \alpha] + \beta)

= \cos(\pi/2 - \alpha) \cos \beta - \sin(\pi/2 - \alpha) \sin \beta by Lemma 2

= \sin \alpha \cos \beta - \cos \alpha \sin \beta by Lemmas 3 and 4.

However, we see that what I’ve called Lemma 3, often called a cofunction identity, can be considered a special case of Theorem 2. However, this is not circular logic since the cofunction identities can be proven without appealing to Theorems 1 and 2.