Derives the product-to-sum formulas by adding and subtracting the sum and difference identities, then inverts them to get the sum-to-product formulas with their half-sum and half-difference arguments. Applies both directions to verification, exact values, and the acoustic phenomenon of beats.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 7 — Trigonometric Identities and Equations
§7.4 Sum-to-Product and Product-to-Sum Formulas, pp. 835-843
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 835-843 — the pages these objectives are drawn from
Warm-up
The sine of a sum and the sine of a difference are almost the same expression.
Discussion prompt
Write both out and add them. What survives?
Hint: Compare the two second terms.
Answer:
Both expansions have sine A cosine B as their first term, so adding gives twice that.
Their second terms are cosine A sine B with opposite signs, so adding cancels them completely.
So the sum of the two left sides equals twice a single product. Rearranged, that is a formula turning a product into a sum — which is the whole content of this section, obtained in one line.
Concept
Adding or subtracting a matched pair of the sum and difference identities converts between a product of two trigonometric functions and a sum of two.
\[ \sin A\cos B=\tfrac{1}{2}\bigl[\sin(A+B)+\sin(A-B)\bigr] \]
Which direction is useful depends on what you have. A product is easier to differentiate and a sum is easier to integrate, so calculus uses one direction; acoustics uses the other, because a product is what the ear actually perceives.
Figure (svg): A diagram showing that adding two sum and difference identities cancels one pair of terms and doubles the other
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 835-838
Section
Section 1
Concept
Each product-to-sum formula comes from adding or subtracting the sum and difference identities for the same function, so that one pair of terms cancels.
Doing this once removes any need to memorise which formula produces cosines and which produces sines. Adding the two cosine identities leaves the cosine-cosine product, and that fact is visible in the derivation rather than something to recall.
Figure (svg): A diagram showing that adding two sum and difference identities cancels one pair of terms and doubles the other
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 835-838
Picture it
The mixed products have opposite signs, so they vanish.
Figure (svg): A diagram showing that adding two sum and difference identities cancels one pair of terms and doubles the other
Subtracting the same pair instead cancels the other terms and produces the formula for the cosine-times-sine product. Four such combinations give all four formulas.
Worked example
Add the two sine identities and divide.
\[ \text{Derive } \sin A\cos B=\tfrac{1}{2}\bigl[\sin(A+B)+\sin(A-B)\bigr]. \]
Write both sine identities
Why: Sum and difference.
Add them
Why: The mixed products cancel.
\[ 2 \sin A \cos B \]
Write the equality
Why: The sum of the left sides.
\[ \sin(A + B) + \sin(A - B) = 2 \sin A \cos B \]
Divide by two
Why: Isolate the product.
Figure (svg): A diagram showing that adding two sum and difference identities cancels one pair of terms and doubles the other
\[ \sin A\cos B=\tfrac{1}{2}\bigl[\sin(A+B)+\sin(A-B)\bigr] \]
Verify: test at convenient angles
Why: At A sixty and B thirty degrees the left side is root three over two times root three over two, which is three quarters. The right side is half of the sine of ninety plus the sine of thirty, which is half of one and a half — also three quarters.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 835-837
Prediction
You add the sine of a sum and the sine of a difference.
Predict first
Which pair cancels?
Correct: The cosA sinB terms, which have opposite signs.
Why: The sine's expansions differ only in the sign of the second product, so adding annihilates it and doubles the first. That single cancellation is the entire derivation of the formula.
Worked example
The cosine identities behave differently because of their sign reversal.
\[ \text{Derive } \cos A\cos B=\tfrac{1}{2}\bigl[\cos(A-B)+\cos(A+B)\bigr]. \]
Write both cosine identities
Why: Their signs are opposite.
Add them
Why: The sine-sine products cancel.
\[ 2 \cos A \cos B \]
Write the equality
Why: The sum of the left sides.
\[ \cos(A - B) + \cos(A + B) \]
Divide by two
Why: Isolate the product.
Figure (svg): The solution to Worked example derive the cosine version shown as a ladder of expressions, one row per legal move
\[ \cos A\cos B=\tfrac{1}{2}\bigl[\cos(A-B)+\cos(A+B)\bigr] \]
Verify: check which terms cancelled
Why: The cosine of a sum has a minus before the sine-sine product and the cosine of a difference has a plus, so adding cancels them — the same mechanism as for the sine, using the cosine's reversed signs. Subtracting instead would isolate the sine-sine product, giving the second formula.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 837-838
Trap
\[ \text{four formulas, each with its own sign and function pattern} \]
Commit all four to memory separately
Why: Each is learned as an independent fact with its own arrangement of signs.
Under pressure the patterns blur and the wrong one gets used.
All four come from one move: add or subtract a matched pair of §7.2 identities so that one pair of terms cancels.
Two functions times two operations gives exactly four combinations, which is why there are four formulas and not more.
The derivation takes two lines and cannot come out wrong. Recalling which of four sign patterns belongs to which product is far less reliable.
Sorting
Same-named products give cosines; mixed products give sines.
Sort into buckets
Sort each product.
Faded example
Subtracting the two cosine identities.
Fill in the blanks
\cos(A-B)-\cos(A+B)=2\sin A\sin B \;\Longrightarrow\; \sin A\sin B=\tfrac2___}[\cos(A-B)-\cos(A+B)]
Why: Subtracting cancels the cosine-cosine products and doubles the sine-sine ones, and dividing by two isolates the product. The subtraction order matters: reversing it would flip the sign of the whole expression.
Explain it to yourself
The section gives four product-to-sum formulas.
Discussion prompt
Explain why four and not some other number.
Hint: What choices are being made?
Answer:
There are two matched pairs of identities to work with — the sine pair and the cosine pair — and each pair can be added or subtracted.
Two pairs times two operations gives exactly four results, and each isolates a different product: cosine-cosine, sine-sine, sine-cosine and cosine-sine.
So the count is not arbitrary; it is forced by the structure. Knowing why there are four makes it obvious when one is missing from your working, which is a better check than trying to recall a list.
Section
Section 2
Concept
A product of two same-named functions becomes a sum or difference of cosines; a mixed product becomes a sum or difference of sines. Each result carries a factor of one half.
The last two differ only in the order of the factors, and that order decides which sine is subtracted. It is the one place where a product's order matters, which is worth noticing since multiplication is otherwise commutative here.
Figure (svg): Four cards giving the product-to-sum formulas for products of sines and cosines
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 838-840
Picture it
Note which produce cosines and which produce sines.
Figure (svg): Four cards giving the product-to-sum formulas for products of sines and cosines
Same-named products give cosines and mixed products give sines. That single observation covers which family to expect before any signs are considered.
Worked example
Identify the pattern, then substitute.
\[ \text{Rewrite } \sin 5x\cos 3x \text{ as a sum.} \]
Identify the pattern
Why: Sine times cosine, a mixed product.
Write the formula
Why: Half the sum of two sines.
\[ (\frac{1}{2}) [\sin(A + B) + \sin(A - B)] \]
Substitute the angles
Why: A is 5x and B is 3x.
\[ \sum 8 x,\text{ difference } 2 x \]
Write the result
Why: Half the sum.
\[ (\frac{1}{2}) [\sin 8 x + \sin 2 x] \]
Figure (svg): Four cards giving the product-to-sum formulas for products of sines and cosines
\[ \sin 5x\cos 3x=\tfrac{1}{2}\bigl[\sin 8x+\sin 2x\bigr] \]
Verify: test at a convenient value
Why: At x equal to fifteen degrees the left side is the sine of seventy-five times the cosine of forty-five, about 0.683. The right side is half of the sine of a hundred and twenty plus the sine of thirty, which is half of 0.866 plus 0.5 — also 0.683.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 838-839
Faded example
Two sines this time.
Fill in the blanks
\sin 7x\sin 3x=\tfrac410[\cos ___x-\cos ___x]
Why: The sine-sine product gives half the difference of two cosines, with the difference of the angles first and the sum second. Getting the order right matters here, since reversing it flips the sign of the whole expression.
Worked example
The sum form may contain special angles the product does not.
\[ \text{Evaluate } \cos 75^\circ\cos 15^\circ \text{ exactly.} \]
Identify the pattern
Why: Cosine times cosine.
Substitute
Why: Difference sixty, sum ninety.
\[ (\frac{1}{2}) [\cos 60 + \cos 90] \]
Evaluate both
Why: Both are special angles.
\[ (\frac{1}{2}) [\frac{1}{2} + 0] \]
Simplify
Why: Half of one half.
\[ \frac{1}{4} \]
Figure (svg): The solution to Worked example evaluate an exact product shown as a ladder of expressions, one row per legal move
\[ \cos 75^\circ\cos 15^\circ=\tfrac{1}{4} \]
Verify: check numerically
Why: The cosine of seventy-five is about 0.2588 and of fifteen about 0.9659, and their product is 0.25 — matching exactly. Neither factor is a special angle but their sum and difference both are, which is precisely when this direction pays off.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 839-840
Error analysis
A student rewrites a product as a sum.
Annotate
On: \( \sin 4x\cos 2x=\sin 6x+\sin 2x \)
The one half is not decorative — it is exactly the 2 that the derivation produced, moved to the other side. Remembering where it comes from makes it much harder to drop.
Prediction
You rewrite a product of two sines.
Predict first
What appears on the right?
Correct: Two cosines.
Why: Products of two same-named functions come from combining the cosine identities, so cosines appear on the right. This is the case most often guessed wrong, since the instinct is to expect sines out of a product of sines.
Matching
Two patterns cover all four cases.
Match the pairs
Why: The function family is set by whether the product is same-named or mixed, and the sum-versus-difference follows the order of the two factors. Both are visible in the derivation rather than needing separate recall.
Step zero
You are asked to rewrite a product of two trigonometric functions.
Discussion prompt
What do you determine before writing a formula?
Hint: Two things about the product.
Answer:
Whether the two functions are same-named or mixed, since that decides whether cosines or sines appear on the right.
And which factor is first, since that decides whether the second term is added or subtracted. Order matters here even though multiplication is commutative.
Both take a glance, and together they pick out one of four formulas. Deciding these before writing anything prevents the common outcome of starting on the right pattern with the wrong signs.
Section
Section 3
Concept
Running the formulas the other way turns a sum of two trigonometric values into a product, with the half-sum and half-difference of the original angles as the new arguments.
The half arguments look mysterious until the derivation is seen. Setting the sum of the two original angles equal to one new variable and the difference to another, then solving, gives exactly the half-sum and half-difference — so they are forced rather than chosen.
Figure (svg): Four cards giving the sum-to-product formulas, each with a half-sum and a half-difference argument
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 840-842
Picture it
The half-sum and half-difference appear in every one.
Figure (svg): Four cards giving the sum-to-product formulas, each with a half-sum and a half-difference argument
The last card's leading minus sign is the one detail worth flagging. Cosine minus cosine is the only formula with a negative in front.
Worked example
Substitute new variables into a product-to-sum formula.
\[ \text{Derive } \sin u+\sin v=2\sin\tfrac{u+v}{2}\cos\tfrac{u-v}{2}. \]
Start from the product-to-sum formula
Why: Twice sinA cosB equals the two sines.
\[ \sin(A + B) + \sin(A - B) \]
Name the two arguments
Why: Set u for the sum and v for the difference.
\[ u = A + B, v = A - B \]
Solve for A and B
Why: Add and subtract the two.
\[ A = \frac{u + v}{2}, B = \frac{u - v}{2} \]
Substitute back
Why: Into the product side.
\[ 2 \sin(\frac{u + v}{2}) \cos(\frac{u - v}{2}) \]
Figure (svg): Four cards giving the sum-to-product formulas, each with a half-sum and a half-difference argument
\[ \sin u+\sin v=2\sin\tfrac{u+v}{2}\cos\tfrac{u-v}{2} \]
Verify: check the substitution
Why: Adding the two defining equations gives twice A equal to u plus v, so A is the half-sum; subtracting gives B as the half-difference. The halves are forced by the algebra, not chosen — which is why they appear in every one of the four formulas.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 840-841
Faded example
For the sine of 6x minus the sine of 2x.
Fill in the blanks
\frac42=___x, \quad \frac______=___x
Why: The half-sum is four x and the half-difference is two x, and those become the arguments of the two factors. Computing both before selecting a formula is the reliable order.
Worked example
Compute the two half arguments first.
\[ \text{Write } \cos 4x+\cos 2x \text{ as a product.} \]
Compute the half-sum
Why: Add and halve.
\[ 3 x \]
Compute the half-difference
Why: Subtract and halve.
Choose the formula
Why: Cosine plus cosine gives two cosines.
\[ 2 \cos \cos \]
Substitute
Why: The two half arguments.
\[ 2 \cos 3 x \cos x \]
Figure (svg): The solution to Worked example apply a sum-to-product formula shown as a ladder of expressions, one row per legal move
\[ \cos 4x+\cos 2x=2\cos 3x\cos x \]
Verify: test at a convenient value
Why: At x equal to fifteen degrees the left side is the cosine of sixty plus the cosine of thirty, about 1.366. The right side is twice the cosine of forty-five times the cosine of fifteen, which is twice 0.7071 times 0.9659 — also 1.366.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 841-842
Trap
\[ \cos 4x+\cos 2x=2\cos 6x\cos 2x \]
Use the sum and difference of the angles directly
Why: Six x and two x are substituted without halving.
A numerical test at fifteen degrees gives the wrong value immediately.
The arguments are the half-sum and the half-difference, so six and two become three and one.
They are halves because the derivation solved a two-equation system, and solving produced a factor of one half in each answer.
Computing both halves before choosing the formula makes the omission impossible, and it is one line of arithmetic.
Prediction
You write a difference of two cosines as a product.
Predict first
What sign appears in front?
Correct: A minus sign.
Why: Cosine minus cosine is the only one of the four formulas with a negative in front, giving negative two times a product of sines. It comes from the cosine's reversed signs in the original identities and is worth flagging separately.
Sorting
Each formula produces a specific pair.
Sort into buckets
Sort each sum or difference.
Explain it
Every sum-to-product formula uses halves.
Discussion prompt
Explain to a classmate where the halves come from.
Hint: What system was solved?
Answer:
The product-to-sum formula has the sum and difference of two angles inside its trigonometric functions. To invert it, those two combinations are given names — call them u and v.
Solving that pair for the original angles requires adding and subtracting the two equations and dividing by two, which is where every half comes from.
So the halves are forced by the algebra, not a convention. A good explanation points out that this is why all four formulas share the same two arguments — they all come from solving the same little system.
Section
Section 4
Concept
Adding two sine waves of nearly equal frequency gives, by the sum-to-product formula, a fast wave at the average frequency multiplied by a slow one at half the difference.
The doubling in the beat rate catches people out. The envelope completes one cycle at half the frequency difference, but loudness peaks at both its maximum and its minimum, so the ear hears twice as many pulses as the envelope has cycles.
Figure (svg): Two close-frequency waves added together, producing a fast oscillation inside a slowly varying envelope
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 842-843
Picture it
The dashed curves are the envelope; the solid one is the sum.
Figure (svg): Two close-frequency waves added together, producing a fast oscillation inside a slowly varying envelope
The picture is the sum-to-product formula drawn. The fast oscillation is the half-sum factor and the envelope is the half-difference factor, exactly as the formula predicts.
Worked example
Two tones a few hertz apart.
\[ \text{Two tones at } 440 \text{ Hz and } 444 \text{ Hz sound together. Find the beat rate.} \]
Apply the sum-to-product formula
Why: The sum becomes a product.
Compute the average
Why: The perceived pitch.
\[ 442 H z \]
Compute the half-difference
Why: The envelope frequency.
\[ 2 H z \]
Double it for the beat rate
Why: Loudness peaks twice per cycle.
\[ 4\text{ beats per second} \]
Figure (svg): Two close-frequency waves added together, producing a fast oscillation inside a slowly varying envelope
\[ \text{4 beats/s at } 442\text{ Hz} \]
Verify: check against the frequency difference
Why: The beat rate equals the difference between the two frequencies, 444 minus 440, which is 4 — matching. That shortcut works because the doubling of the half-difference exactly undoes the halving, which is why musicians quote the difference directly.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 842-843
Prediction
Two tones at 500 and 507 hertz.
Predict first
How many beats per second?
Correct: Seven.
Why: The beat rate is the difference between the two frequencies. The envelope oscillates at half that, but loudness peaks twice per envelope cycle, so the two factors of two cancel and the difference is what is heard.
Worked example
The beat rate measures how far out of tune a string is.
\[ \text{A string beats } 3 \text{ times a second against a } 440 \text{ Hz fork. What is its frequency?} \]
Recall the relationship
Why: Beat rate is the frequency difference.
\[ \text{difference } = 3 \]
Consider both possibilities
Why: The string may be sharp or flat.
\[ 437\text{ or } 443 \]
Note the ambiguity
Why: Beats alone cannot distinguish.
Resolve by tightening
Why: If beats speed up, it was sharp.
Figure (svg): The solution to Worked example tuning by beats shown as a ladder of expressions, one row per legal move
\[ 437\text{ Hz or }443\text{ Hz} \]
Verify: check the resolution method
Why: Tightening the string raises its frequency. If it was already sharp the difference grows and the beats speed up; if flat, the difference shrinks and they slow. One small adjustment settles which, which is exactly how a piano tuner works.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 843-843
Error analysis
A student computes beats from two tones 6 hertz apart.
Annotate
On: \( \text{half-difference}=3\text{ Hz, so } 3 \text{ beats per second} \)
The ear responds to the size of the oscillation rather than its sign, so a negative envelope sounds just as loud as a positive one. That is the whole reason for the factor of two.
Faded example
Two tones at 300 and 306 hertz.
Fill in the blanks
\text303=\frac6___=___\text___, \quad \text___=___\text___
Why: The half-sum gives the perceived pitch and the full difference gives the beat rate. Both come directly from the sum-to-product formula, with the doubling accounting for loudness peaking twice per envelope cycle.
Sorting
The product has a fast factor and a slow one.
Sort into buckets
Sort each description.
Explain it to yourself
As two tones are brought into tune, the beats slow and vanish.
Discussion prompt
Explain why, using the formula.
Hint: What happens to the half-difference?
Answer:
The envelope's frequency is half the difference between the two tones. As they converge, that difference shrinks towards zero.
So the envelope oscillates more and more slowly, and the loudness swells become further apart — the beats slow down.
At exact agreement the difference is zero, the envelope becomes constant, and the loudness stops varying at all. That is why tuning by beats works: the audible signal gets unmistakably slower as the target is approached, which is far easier to hear than a small pitch difference itself.
Section
Section 5
Concept
Neither a product nor a sum is simpler in general. The right direction is decided by what the next step requires.
The second row is the most useful in this course: a product is zero exactly when one factor is, so converting a sum into a product turns an intractable equation into two easy ones. That technique carries straight into §7.5.
| you want to | convert to |
|---|---|
| integrate | a sum, since each term integrates alone |
| find where it vanishes | a product, since a factor being zero suffices |
| describe what is heard | a product, giving pitch and envelope |
| evaluate exactly | whichever form has special angles |
| verify an identity | whichever matches the other side |
Figure (svg): Four cards giving the sum-to-product formulas, each with a half-sum and a half-difference argument
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 835-843
Picture it
This is the direction that makes equations solvable.
Figure (svg): Four cards giving the sum-to-product formulas, each with a half-sum and a half-difference argument
Turning a sum into a product is what allows the zero-product property to be used, which is the same move that makes polynomial equations tractable in chapter 3.
Worked example
A sum of two sines is hard; a product is easy.
\[ \text{Solve } \sin 3x+\sin x=0 \text{ for } 0\le x<2\pi. \]
Convert to a product
Why: Half-sum 2x, half-difference x.
\[ 2 \sin 2 x \cos x = 0 \]
Apply the zero-product property
Why: Either factor may vanish.
\[ \sin 2 x = 0\text{ or } \cos x = 0 \]
Solve the first
Why: Doubled angle at multiples of pi.
\[ x = 0, \frac{\pi}{2}, \pi, 3 \pi / 2 \]
Solve the second
Why: Cosine vanishes at quarter turns.
\[ x = \frac{\pi}{2}, 3 \pi / 2 \]
Figure (svg): Four cards giving the sum-to-product formulas, each with a half-sum and a half-difference argument
\[ x=0,\;\tfrac{\pi}{2},\;\pi,\;\tfrac{3\pi}{2} \]
Verify: check one solution
Why: At a quarter turn the sine of three quarters of a turn is negative one and the sine of a quarter turn is 1, so the sum is zero — confirming that solution. Converting to a product was what made the zero-product property available; the original sum offered no such handle.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 841-843
Prediction
You need to solve an equation where a sum of two sines equals zero.
Predict first
Which direction helps?
Correct: Sum to product, so a factor can be set to zero.
Why: A product is zero exactly when one of its factors is, which splits one hard equation into two easy ones. A sum offers no such handle, which is why this direction is the standard first move for equations of this shape.
Worked example
Match the form of the other side.
\[ \text{Verify } \frac{\sin 4x+\sin 2x}{\cos 4x+\cos 2x}=\tan 3x. \]
Convert the numerator
Why: Half-sum 3x, half-difference x.
\[ 2 \sin 3 x \cos x \]
Convert the denominator
Why: The same half arguments.
\[ 2 \cos 3 x \cos x \]
Cancel the common factors
Why: The twos and the cosine of x.
\[ \sin 3 x / \cos 3 x \]
Recognise the quotient
Why: By the quotient identity.
\[ \tan 3 x \]
Figure (svg): The solution to Worked example verify by converting shown as a ladder of expressions, one row per legal move
\[ \frac{\sin 4x+\sin 2x}{\cos 4x+\cos 2x}=\tan 3x \]
Verify: check the shared half arguments
Why: Both conversions produce the same half-sum and half-difference, which is why the cosine of x cancels so cleanly. That shared structure is what makes ratios of this kind collapse, and spotting it is the whole trick.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 842-843
Trap
\[ \text{convert every product or sum encountered, on principle} \]
Apply a formula because it is available
Why: The conversion is performed without asking what the next step needs.
The expression changes form without becoming any more useful.
Convert towards what the next step needs: a product when you want a zero factor, a sum when you want to integrate term by term.
If neither applies, the conversion is motion without progress and may make the expression harder to read.
Ask what the target form looks like before converting. In a verification the other side answers this directly, which is why it is worth reading first.
Sorting
The next step decides.
Sort into buckets
Sort each task.
Faded example
A sum of two cosines set to zero.
Fill in the blanks
\cos 5x+\cos x=2\cos 3x\cos 2x=0
Why: The half-sum is three x and the half-difference is two x, and the product form lets each factor be set to zero separately. That conversion turns an equation with no obvious approach into two routine ones.
Explain it
The formulas work in both directions.
Discussion prompt
Explain to a classmate how to decide which way to go.
Hint: What does the next step need?
Answer:
Look at what you are about to do, not at the expression itself. Neither form is simpler in the abstract.
If you are solving, go to a product, because a product is zero exactly when a factor is — that turns one hard equation into two easy ones.
If you are integrating or verifying against a sum, go to a sum. A good explanation adds that in a verification the other side tells you the target, so the decision is usually already made for you.
Comparison
Fill the blanks from memory. Both directions use the same four relationships.
Comparison matrix
| product to sum | sum to product | |
|---|---|---|
| arguments on the right | the sum and the difference | the half-sum and the half-difference |
| leading coefficient | one half | two |
| useful for | integrating, separating terms | solving, describing sound |
| derived from | adding two identities | the same, with substituted variables |
The first two rows are inverses of each other, which is expected — the two directions are the same relationships read from opposite ends.
Pattern
Five steps, and the first two are arithmetic rather than trigonometry.
Doing steps 1 and 2 before choosing the formula prevents the commonest error, which is substituting the sum and difference rather than their halves.
OpenStax Algebra and Trigonometry 2e, §9.4 Sum-to-Product and Product-to-Sum Formulas §9.4
Check
Which functions come out.
Check your understanding
A product of two sines converts to what?
Answer: A
Why: Products of same-named functions come from combining the two cosine identities, so cosines appear on the right. Subtracting them isolates the sine-sine product, which is why this case gives a difference.
Check
The arguments in the other direction.
Check your understanding
When writing a sum of two sines as a product, what arguments appear?
Answer: A
Why: Inverting the product-to-sum formula requires solving a two-equation system, and that solution divides by two. The halves are forced by the algebra, which is why all four sum-to-product formulas share them.
Check
Beats.
Check your understanding
Two tones at 300 and 304 hertz are played together. What is the beat rate?
Answer: A
Why: The beat rate is the difference between the two frequencies. The envelope oscillates at half that, but loudness peaks at both its extremes, so the ear hears twice as many pulses as the envelope has cycles.
Real world
Radio transmission is a product-to-sum problem, run deliberately.
Discussion prompt
An AM radio station multiplies an audio signal by a high-frequency carrier. What does the product-to-sum formula say about the result?
Hint: What does a product of two sinusoids equal?
Answer:
The product becomes a sum of two sinusoids, at the carrier frequency plus and minus the audio frequency. Those are the sidebands.
So a station broadcasting on one nominal frequency actually occupies a band whose width is set by the audio it carries — which is why stations must be spaced apart on the dial.
The formula predicts exactly where that energy lands, so it determines how many stations fit in a given range. Radio spectrum allocation is this identity applied at scale, and the same computation underlies every modulation scheme in use.
Commit first
State your confidence along with your answer.
Predict first
Why do the sum-to-product formulas use half arguments?
Correct: Inverting the other direction requires solving a system, which divides by two.
Why: Naming the sum and difference as new variables and solving for the original angles produces the half-sum and half-difference. The halves are forced by that algebra, which is why every one of the four formulas contains them.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why two slightly out-of-tune notes produce a wavering sound.
Hint: What does the sum become?
Answer:
Adding the two waves gives, by the formula, a product: a fast oscillation at the average frequency times a slow one at half the difference.
The ear hears the fast factor as the pitch and the slow factor as a change in loudness, because it varies far too slowly to be a tone.
So one hears a single note that swells and fades. The rate of the swelling is the frequency difference, which is why tuners listen for beats rather than trying to hear a small pitch difference directly — a slow pulsing is much easier to detect.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third is where the errors are, since substituting the sum instead of the half-sum is easy to do. The first makes all eight formulas reconstructible rather than memorised.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the sine sum and difference identities one above the other and show what adding them gives, then what subtracting gives. From one of those results, derive a sum-to-product formula by naming the sum and difference as new variables and solving. Underneath, sketch two close waves adding into an envelope and label the pitch and beat frequencies.
If your sum-to-product formula came out of the substitution rather than from memory, the half arguments will have appeared on their own — which is the point of doing the derivation at all.
Recap
Five things, and the first makes the other four reconstructible.
| if you remember one thing | it should be this |
|---|---|
| about the derivation | add or subtract two §7.2 identities and one pair cancels |
| about which functions appear | same-named products give cosines, mixed give sines |
| about the half arguments | they come from solving a system, so they are forced |
| about beats | the beat rate is the full frequency difference |
Section 7.5 puts the whole chapter to work solving trigonometric equations, where the sum-to-product direction becomes the standard route to a factorable form.
OpenStax, Precalculus, §7.4 Sum-to-Product and Product-to-Sum Formulas §7.4, pp. 835-843 — everything on these slides traces back here
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