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Sum-to-Product and Product-to-Sum Formulas

Mathematics Algebra Difficulty 2/5

Beginning of the lesson

Sum-to-Product & Product-to-Sum Formulas

Turning sums into products and products into sums — the bridge between addition and factoring in trigonometry

The Big Idea

You already know the angle-addition formulas. Nothing new is postulated here. Every formula in this lesson is derived by adding or subtracting two addition identities you already have. The payoff is enormous: you gain the ability to factor trigonometric sums (essential for solving equations) and to linearize trigonometric products (essential for integration and simplification).

1. Product-to-Sum Formulas

These formulas rewrite a product of two trigonometric functions as a sum or difference. There are four of them:

The Four Product-to-Sum Identities
  1. $\sin A \cos B = \dfrac{1}{2}\bigl[\sin(A+B) + \sin(A-B)\bigr]$
  2. $\cos A \sin B = \dfrac{1}{2}\bigl[\sin(A+B) - \sin(A-B)\bigr]$
  3. $\cos A \cos B = \dfrac{1}{2}\bigl[\cos(A-B) + \cos(A+B)\bigr]$
  4. $\sin A \sin B = \dfrac{1}{2}\bigl[\cos(A-B) - \cos(A+B)\bigr]$

Derivation (sin A cos B)

Start from the two addition formulas you already know:

$$\sin(A+B) = \sin A\cos B + \cos A\sin B \quad \cdots (i)$$ $$\sin(A-B) = \sin A\cos B - \cos A\sin B \quad \cdots (ii)$$

Add equations (i) and (ii):

$$\sin(A+B) + \sin(A-B) = 2\sin A\cos B$$

Divide both sides by 2:

$$\boxed{\sin A\cos B = \tfrac{1}{2}[\sin(A+B) + \sin(A-B)]}$$

Derivation (sin A sin B)

Start from the cosine addition formulas:

$$\cos(A-B) = \cos A\cos B + \sin A\sin B \quad \cdots (iii)$$ $$\cos(A+B) = \cos A\cos B - \sin A\sin B \quad \cdots (iv)$$

Subtract (iv) from (iii):

$$\cos(A-B) - \cos(A+B) = 2\sin A\sin B$$

Divide by 2:

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