Chapter 6: Product-to-Sum and Sum-to-Product Formulas
Overview
Product-to-sum and sum-to-product formulas (quick reference)
Product → Sum
\begin{align*} \sin\alpha\cos\beta &= \tfrac{1}{2}\{\sin(\alpha+\beta) + \sin(\alpha-\beta)\} \\ \cos\alpha\sin\beta &= \tfrac{1}{2}\{\sin(\alpha+\beta) - \sin(\alpha-\beta)\} \\ \cos\alpha\cos\beta &= \tfrac{1}{2}\{\cos(\alpha+\beta) + \cos(\alpha-\beta)\} \\ \sin\alpha\sin\beta &= \tfrac{1}{2}\{\cos(\alpha-\beta) - \cos(\alpha+\beta)\} \end{align*}Sum → Product
\begin{align*} \sin A + \sin B &= 2\sin\tfrac{A+B}{2}\cos\tfrac{A-B}{2} \\ \sin A - \sin B &= 2\cos\tfrac{A+B}{2}\sin\tfrac{A-B}{2} \\ \cos A + \cos B &= 2\cos\tfrac{A+B}{2}\cos\tfrac{A-B}{2} \\ \cos A - \cos B &= -2\sin\tfrac{A+B}{2}\sin\tfrac{A-B}{2} \end{align*}The product-to-sum formulas convert a product of two trigonometric functions into a sum or difference, and the sum-to-product formulas do the reverse. Both follow directly from the angle addition theorems.
These identities appear frequently in integration, in simplification of trigonometric expressions, and in Fourier analysis. This page derives each identity step by step and works through three examples and four exercises.
Prerequisite: angle addition theorems
Angle addition theorems (proof here)
$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$$ $$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$$ $$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$ $$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$$The derivations below take these four identities as the starting point.
Derivation of the product-to-sum formulas
The product-to-sum formulas are obtained by adding or subtracting pairs of addition theorems.
Formula for $\sin\alpha\cos\beta$
Step 1: Take two addition theorems
$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta \quad \cdots (1)$$ $$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta \quad \cdots (2)$$Step 2: Compute (1) + (2)
$$\sin(\alpha + \beta) + \sin(\alpha - \beta) = 2\sin\alpha\cos\beta$$Dividing both sides by 2:
Product-to-sum formula for $\sin\alpha\cos\beta$
$$\sin\alpha\cos\beta = \dfrac{1}{2}\{\sin(\alpha + \beta) + \sin(\alpha - \beta)\} \quad \blacksquare$$Formula for $\cos\alpha\sin\beta$
Step 3: Compute (1) − (2)
$$\sin(\alpha + \beta) - \sin(\alpha - \beta) = 2\cos\alpha\sin\beta$$Dividing both sides by 2:
Product-to-sum formula for $\cos\alpha\sin\beta$
$$\cos\alpha\sin\beta = \dfrac{1}{2}\{\sin(\alpha + \beta) - \sin(\alpha - \beta)\} \quad \blacksquare$$Formula for $\cos\alpha\cos\beta$
Step 4: Take the two cos addition theorems
$$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta \quad \cdots (3)$$ $$\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta \quad \cdots (4)$$Step 5: Compute (3) + (4)
$$\cos(\alpha + \beta) + \cos(\alpha - \beta) = 2\cos\alpha\cos\beta$$Dividing both sides by 2:
Product-to-sum formula for $\cos\alpha\cos\beta$
$$\cos\alpha\cos\beta = \dfrac{1}{2}\{\cos(\alpha + \beta) + \cos(\alpha - \beta)\} \quad \blacksquare$$Formula for $\sin\alpha\sin\beta$
Step 6: Compute (4) − (3)
$$\cos(\alpha - \beta) - \cos(\alpha + \beta) = 2\sin\alpha\sin\beta$$Dividing both sides by 2:
Product-to-sum formula for $\sin\alpha\sin\beta$
$$\sin\alpha\sin\beta = \dfrac{1}{2}\{\cos(\alpha - \beta) - \cos(\alpha + \beta)\} \quad \blacksquare$$Derivation of the sum-to-product formulas
The sum-to-product formulas are the inverse transformations of the product-to-sum formulas, obtained by a change of variables.
Change of variables
In the product-to-sum formulas, set $\alpha + \beta = A$ and $\alpha - \beta = B$. Then
$$\alpha = \dfrac{A + B}{2}, \quad \beta = \dfrac{A - B}{2}.$$Formula for $\sin A + \sin B$
Step 7: From the $\sin\alpha\cos\beta$ formula
From $\sin\alpha\cos\beta = \dfrac{1}{2}\{\sin(\alpha + \beta) + \sin(\alpha - \beta)\}$ we get
$$\sin(\alpha + \beta) + \sin(\alpha - \beta) = 2\sin\alpha\cos\beta.$$Substituting $A = \alpha + \beta$ and $B = \alpha - \beta$:
Sum-to-product formula for $\sin A + \sin B$
$$\sin A + \sin B = 2\sin\dfrac{A + B}{2}\cos\dfrac{A - B}{2} \quad \blacksquare$$Formula for $\sin A - \sin B$
Step 8: From the $\cos\alpha\sin\beta$ formula
From $\cos\alpha\sin\beta = \dfrac{1}{2}\{\sin(\alpha + \beta) - \sin(\alpha - \beta)\}$ we get
$$\sin(\alpha + \beta) - \sin(\alpha - \beta) = 2\cos\alpha\sin\beta.$$Substituting $A = \alpha + \beta$ and $B = \alpha - \beta$:
Sum-to-product formula for $\sin A - \sin B$
$$\sin A - \sin B = 2\cos\dfrac{A + B}{2}\sin\dfrac{A - B}{2} \quad \blacksquare$$Formula for $\cos A + \cos B$
Step 9: From the $\cos\alpha\cos\beta$ formula
From $\cos\alpha\cos\beta = \dfrac{1}{2}\{\cos(\alpha + \beta) + \cos(\alpha - \beta)\}$ we get
$$\cos(\alpha + \beta) + \cos(\alpha - \beta) = 2\cos\alpha\cos\beta.$$Substituting $A = \alpha + \beta$ and $B = \alpha - \beta$:
Sum-to-product formula for $\cos A + \cos B$
$$\cos A + \cos B = 2\cos\dfrac{A + B}{2}\cos\dfrac{A - B}{2} \quad \blacksquare$$Formula for $\cos A - \cos B$
Step 10: From the $\sin\alpha\sin\beta$ formula
From $\sin\alpha\sin\beta = \dfrac{1}{2}\{\cos(\alpha - \beta) - \cos(\alpha + \beta)\}$ we get
$$\cos(\alpha - \beta) - \cos(\alpha + \beta) = 2\sin\alpha\sin\beta,$$i.e., $\cos A - \cos B = -2\sin\alpha\sin\beta$ (with $A = \alpha + \beta$, $B = \alpha - \beta$).
Sum-to-product formula for $\cos A - \cos B$
$$\cos A - \cos B = -2\sin\dfrac{A + B}{2}\sin\dfrac{A - B}{2} \quad \blacksquare$$Worked examples
Example 1: Product-to-sum evaluation
Problem
Evaluate $\sin 75° \cos 15°$ using a product-to-sum formula.
Step 1: Apply the formula
Substitute $\alpha = 75°$, $\beta = 15°$ into $\sin\alpha\cos\beta = \dfrac{1}{2}\{\sin(\alpha + \beta) + \sin(\alpha - \beta)\}$.
$$\sin 75° \cos 15° = \dfrac{1}{2}\{\sin(75° + 15°) + \sin(75° - 15°)\}$$Step 2: Compute the angles
$$= \dfrac{1}{2}(\sin 90° + \sin 60°) = \dfrac{1}{2}\left(1 + \dfrac{\sqrt{3}}{2}\right)$$Answer
$$\sin 75° \cos 15° = \dfrac{2 + \sqrt{3}}{4}$$Example 2: Sum-to-product conversion
Problem
Convert $\cos 5\theta + \cos 3\theta$ to product form using a sum-to-product formula.
Step 1: Apply the formula
Substitute $A = 5\theta$, $B = 3\theta$ into $\cos A + \cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}$.
$$\cos 5\theta + \cos 3\theta = 2\cos\dfrac{5\theta + 3\theta}{2}\cos\dfrac{5\theta - 3\theta}{2}$$Answer
$$\cos 5\theta + \cos 3\theta = 2\cos 4\theta \cos \theta$$Example 3: Evaluating a product
Problem
Evaluate $\sin 105° \sin 15°$.
Step 1: Convert the product to a difference
Substitute $\alpha = 105°$, $\beta = 15°$ into $\sin\alpha\sin\beta = \dfrac{1}{2}\{\cos(\alpha-\beta) - \cos(\alpha+\beta)\}$.
$$\sin 105° \sin 15° = \dfrac{1}{2}(\cos 90° - \cos 120°)$$Step 2: Use known values
With $\cos 90° = 0$ and $\cos 120° = -\dfrac{1}{2}$,
$$= \dfrac{1}{2}\left(0 - \left(-\dfrac{1}{2}\right)\right) = \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{4}.$$Answer
$$\sin 105° \sin 15° = \dfrac{1}{4}$$Exercises
Exercise 1
Evaluate $\cos 75° \cos 15°$ using a product-to-sum formula.
Hint
Use $\cos\alpha\cos\beta = \dfrac{1}{2}\{\cos(\alpha+\beta) + \cos(\alpha-\beta)\}$. Recall that $\cos 90° = 0$ and $\cos 60° = \dfrac{1}{2}$.
Solution
Substitute $\alpha = 75°$, $\beta = 15°$ into the product-to-sum formula.
$$\cos 75° \cos 15° = \dfrac{1}{2}(\cos 90° + \cos 60°) = \dfrac{1}{2}\left(0 + \dfrac{1}{2}\right) = \dfrac{1}{4}$$Exercise 2
Convert $\sin 7\theta - \sin 3\theta$ to product form using a sum-to-product formula.
Hint
Use $\sin A - \sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}$ with $A = 7\theta$, $B = 3\theta$.
Solution
Substituting into the sum-to-product formula:
$$\sin 7\theta - \sin 3\theta = 2\cos\dfrac{7\theta + 3\theta}{2}\sin\dfrac{7\theta - 3\theta}{2} = 2\cos 5\theta \sin 2\theta$$Exercise 3
Evaluate $\sin 75° \sin 15°$ using a product-to-sum formula.
Hint
Use $\sin\alpha\sin\beta = \dfrac{1}{2}\{\cos(\alpha-\beta) - \cos(\alpha+\beta)\}$ with $\alpha = 75°$, $\beta = 15°$.
Solution
Exercise 4
Solve $\cos 3x + \cos x = 0$ for $0 \leq x \leq \pi$.
Hint
Apply a sum-to-product formula to factor the left side as $2\cos 2x \cos x = 0$. Then solve $\cos 2x = 0$ or $\cos x = 0$.
Solution
Apply the sum-to-product formula:
$$\cos 3x + \cos x = 2\cos 2x \cos x = 0.$$Case $\cos 2x = 0$: $2x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$, so $x = \dfrac{\pi}{4}, \dfrac{3\pi}{4}$.
Case $\cos x = 0$: $x = \dfrac{\pi}{2}$.
Therefore
$$x = \dfrac{\pi}{4}, \quad \dfrac{\pi}{2}, \quad \dfrac{3\pi}{4}.$$