Math Core

Sums are usually easier to work with than products. You can add two waves term by term, but multiplying them is harder to picture. The product-to-sum formulas turn a product like $\sin 5x \cos 2x$ into a sum of single sines or cosines, and the sum-to-product formulas go the other way. Engineers use them to analyze sound and radio signals, and you'll use them to evaluate and simplify.

Deriving the product-to-sum formulas

Write the sum and difference formulas for sine next to each other:

\begin{aligned} \sin(A + B) &= \sin A \cos B + \cos A \sin B \\ \sin(A - B) &= \sin A \cos B - \cos A \sin B \end{aligned}

Add them. The $\cos A\sin B$ terms cancel, leaving $\sin(A + B) + \sin(A - B) = 2\sin A \cos B$ . Dividing by $2$ expresses the product $\sin A \cos B$ as a sum.

Do the same with the cosine formulas:

\begin{aligned} \cos(A - B) &= \cos A \cos B + \sin A \sin B \\ \cos(A + B) &= \cos A \cos B - \sin A \sin B \end{aligned}

Adding gives $2\cos A\cos B$ . Subtracting gives $2\sin A \sin B$ .

Product-to-sum formulas

\begin{aligned} \sin A \cos B &= \tfrac{1}{2}\left[\sin(A + B) + \sin(A - B)\right] \\ \cos A \cos B &= \tfrac{1}{2}\left[\cos(A - B) + \cos(A + B)\right] \\ \sin A \sin B &= \tfrac{1}{2}\left[\cos(A - B) - \cos(A + B)\right] \end{aligned}

You don't need a separate formula for $\cos A \sin B$ . Just write it as $\sin B \cos A$ and use the first formula with the letters swapped.

Worked example: An exact value

Find the exact value of $\sin 75^\circ \cos 15^\circ$ .

Use the first formula with $A = 75^\circ$ and $B = 15^\circ$ :

\begin{aligned} \sin 75^\circ \cos 15^\circ &= \tfrac{1}{2}\left[\sin 90^\circ + \sin 60^\circ\right] \\ &= \tfrac{1}{2}\left[1 + \tfrac{\sqrt{3}}{2}\right] = \frac{2 + \sqrt{3}}{4} \end{aligned}

Neither $75^\circ$ nor $15^\circ$ is a basic special angle, but their sum and difference are.

Worked example: Rewriting a product

Write $\sin 5x \sin 2x$ as a sum or difference.

Use $\sin A \sin B$ with $A = 5x$ and $B = 2x$ :

\sin 5x \sin 2x = \tfrac{1}{2}\left[\cos(5x - 2x) - \cos(5x + 2x)\right] = \tfrac{1}{2}\left[\cos 3x - \cos 7x\right].

Common mistake

In $\sin A \sin B$ , the order inside the bracket is $\cos(A - B) - \cos(A + B)$ : the difference comes first. Reversing it flips the sign of the answer. A quick check: at $A = B = 90^\circ$ the product is $1$ , and $\tfrac{1}{2}[\cos 0^\circ - \cos 180^\circ] = \tfrac{1}{2}[1 + 1] = 1$ .

Sum-to-product formulas

To go from a sum to a product, set $u = A + B$ and $v = A - B$ . Then $A = \dfrac{u + v}{2}$ and $B = \dfrac{u - v}{2}$ . Substituting into the product-to-sum formulas and multiplying by $2$ gives:

Sum-to-product formulas

\begin{aligned} \sin u + \sin v &= 2\sin\frac{u + v}{2}\cos\frac{u - v}{2} &\qquad \cos u + \cos v &= 2\cos\frac{u + v}{2}\cos\frac{u - v}{2} \\ \sin u - \sin v &= 2\cos\frac{u + v}{2}\sin\frac{u - v}{2} &\qquad \cos u - \cos v &= -2\sin\frac{u + v}{2}\sin\frac{u - v}{2} \end{aligned}

In words: take the average of the two angles and half the difference.

Worked example: Sum of two cosines

Find the exact value of $\cos 75^\circ + \cos 15^\circ$ .

The average of the angles is $45^\circ$ and half the difference is $30^\circ$ :

\cos 75^\circ + \cos 15^\circ = 2\cos 45^\circ \cos 30^\circ = 2\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2}.

Worked example: Simplify a quotient

Verify: $\dfrac{\sin 3x + \sin x}{\cos 3x + \cos x} = \tan 2x$ .

Apply sum-to-product to the top and bottom. In both, the average angle is $2x$ and half the difference is $x$ .

\begin{aligned} \frac{\sin 3x + \sin x}{\cos 3x + \cos x} &= \frac{2\sin 2x \cos x}{2\cos 2x \cos x} && \text{sum-to-product} \\ &= \frac{\sin 2x}{\cos 2x} && \text{cancel } 2\cos x \\ &= \tan 2x && \text{quotient identity} \end{aligned}

Tip

When a product-to-sum answer involves a negative angle, use the even-odd identities to clean it up: $\sin(-\theta) = -\sin\theta$ and $\cos(-\theta) = \cos\theta$ . Choosing $A$ as the larger angle avoids most negatives.

Why this matters: beats

When two musical notes with nearly equal frequencies play together, you hear a single tone whose loudness swells and fades. Musicians call this a beat, and they use it to tune instruments. Sum-to-product explains it. Model notes of $442$ and $438$ vibrations per second as $\cos u$ and $\cos v$ with $u = 2\pi(442t)$ and $v = 2\pi(438t)$ , where $t$ is in seconds. Their sum is

\cos u + \cos v = 2\cos\big(2\pi(440t)\big) \,\cos\big(2\pi(2t)\big).

The factor $\cos\big(2\pi(440t)\big)$ is a fast wave at the average pitch, which is the tone you hear. The factor $2\cos\big(2\pi(2t)\big)$ changes slowly and acts like a volume knob, rising and falling a few times each second. As the two notes are tuned closer together, the slow factor slows down, and the beats disappear when the notes match exactly.

To decide which direction to go on a problem, ask what you need. To evaluate a product of trig values at odd angles, or to integrate a product later in calculus, go product-to-sum. To simplify a fraction or solve an equation that contains a sum like $\sin 3x + \sin x$ , go sum-to-product, because factors can cancel or be set equal to zero.

Practice

Practice 1

Find the exact value of $\cos 75^\circ \cos 15^\circ$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Find the exact value of $\sin 75^\circ \sin 15^\circ$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Which expression is equal to $2\sin 4x \cos x$ ?

Practice 4

Which expression is equal to $\cos 3x \cos x$ ?

Practice 5

Find the exact value of $\sin 105^\circ + \sin 15^\circ$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

Find the exact value of $\cos 15^\circ - \cos 75^\circ$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Which expression is equal to $\sin 5x - \sin 3x$ ?

Practice 8

Which expression is equal to $\dfrac{\cos x - \cos 3x}{\sin 3x - \sin x}$ ?