# 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.