The last identity family of the chapter, in both directions. Adding and subtracting the sum and difference identities makes the mixed terms cancel, giving three product-to-sum formulas that turn a product of circular functions into a sum of first powers. Reversing them, with the two angles renamed as a half-sum and a half-difference, gives the sum-to-product formulas that turn a sum into a product — the form Section 10.7 needs, because only a product can be set to zero one factor at a time.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.4 Trigonometric Identities, pp. 780-782
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 780-782 — the pages these objectives are drawn from
Warm-up
The last family of identities in this chapter comes from doing something you have not yet tried with the formulas you already have: adding two of them together.
Discussion prompt
Write down the cosine sum identity and the cosine difference identity, one above the other. Now add them. What happens?
Hint: Look at the sine terms specifically.
Answer:
\[ \cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta \]
\[ \cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta \]
Adding them, the two sine terms are exact opposites and cancel completely, leaving twice the product of the cosines. Dividing by two gives a formula for that product as a sum — and subtracting instead would have killed the cosine terms and left the sines.
That is the entire derivation of this lesson. Two additions, one subtraction, and the family is complete.
Concept
The sum and difference identities differ only in one sign. Adding them cancels the terms that carried that sign; subtracting them cancels the others. Either way a single product survives, which is exactly what a product-to-sum formula asserts.
\[ \cos\alpha\cos\beta = \tfrac{1}{2}\big[\cos(\alpha - \beta) + \cos(\alpha + \beta)\big] \]
These are sometimes called the prosthaphaeresis formulas, and they have a real history: before logarithms they were used to turn multiplication into addition, which is precisely what they do.
Figure (svg): A derivation showing the cosine difference and sum identities written one above the other, then added so that the sine terms cancel and only a product of cosines remains
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 780-780
Section
Section 1
Concept
Adding or subtracting the appropriate pair of identities gives three formulas, one for each kind of product. Each turns a product of two circular functions into half a sum or difference of two first powers.
Theorem 10.20 — The product-to-sum formulas, also called the prosthaphaeresis formulas: each product of two circular functions rewritten as half a sum or difference evaluated at the difference and sum of the two angles.
A useful pattern: a matched product — two cosines or two sines — produces cosines, while a mixed product produces sines. That reflects the cosine formulas having matched terms and the sine formulas having mixed ones.
Figure (svg): The three product-to-sum formulas, each turning a product of two circular functions into a sum or difference of cosines or sines at the sum and difference angles
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 780-780
Picture it
The two families in this lesson are inverse to each other, and choosing between them is most of the skill.
Figure (svg): Two columns contrasting when a product is the obstacle and when a sum is the obstacle
A product-to-sum move is the calculus direction: it removes a product that could not be integrated. A sum-to-product move is the equation-solving direction, and Section 10.7 uses it constantly.
Worked example
The same two identities, subtracted rather than added.
\[ \text{Derive a formula for } \sin(\alpha)\sin(\beta) \text{ as a sum or difference.} \]
Write the two cosine identities
Why: They differ only in the sign of the sine term.
Subtract the second from the first
Why: The cosine products cancel and the sine products reinforce.
\[ \cos(a - b) - \cos(a + b) = 2 \sin a \sin b \]
Divide by two
Why: Isolating the product.
\[ \sin a \sin b = (\frac{1}{2}) [\cos(a - b) - \cos(a + b)] \]
Note the order of the angles
Why: The difference angle comes first, and it carries the plus.
Figure (svg): The solution to Worked example derive the sine times sine formula shown as a ladder of expressions, one row per legal move
\[ \sin(\alpha)\sin(\beta) = \tfrac{1}{2}\big[\cos(\alpha - \beta) - \cos(\alpha + \beta)\big] \]
Verify: test at alpha and beta both pi over four
Why: The left side is root two over two squared, which is one half. The right side is half of the cosine of zero minus the cosine of pi over two, which is half of one minus zero, namely one half. They agree.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 780-780
Matching
A matched product gives cosines; a mixed product gives sines.
Match the pairs
Why: The two matched products both produce cosines, because they came from adding or subtracting the two cosine identities. They are distinguished by the sign: adding kept the cosines and gave a sum, subtracting kept the sines and gave a difference. The mixed product came from the sine identities instead and therefore produces sines.
Worked example
Example 10.4.6, part 1. Identify the two angles and substitute.
\[ \text{Write } \cos(2\theta)\cos(6\theta) \text{ as a sum.} \]
Identify the two angles
Why: Alpha is 2 theta and beta is 6 theta.
\[ a = 2 t, b = 6 t \]
Apply the cosine times cosine formula
Why: Half the sum of the cosines at the difference and the sum.
\[ (\frac{1}{2}) [\cos(-4 t) + \cos(8 t)] \]
Clean up the negative angle
Why: Cosine is even, so the minus sign inside disappears.
\[ \cos(-4 t) = \cos(4 t) \]
State the result
Why: Two cosines at first power, each with coefficient one half.
\[ (\frac{1}{2}) \cos 4 t + (\frac{1}{2}) \cos 8 t \]
Figure (svg): The solution to Worked example write a product as a sum shown as a ladder of expressions, one row per legal move
\[ \cos(2\theta)\cos(6\theta) = \tfrac{1}{2}\cos(4\theta) + \tfrac{1}{2}\cos(8\theta) \]
Verify: test at theta equal to pi over eight
Why: The left side is the cosine of pi over four times the cosine of three pi over four, which is root two over two times negative root two over two, namely negative one half. The right side is half the cosine of pi over two plus half the cosine of pi, which is zero plus negative one half. They agree.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 781-781
Trap
\[ \sin\alpha\sin\beta = \tfrac{1}{2}\big[\cos(\alpha + \beta) - \cos(\alpha - \beta)\big] \]
Write the sum angle first, since sums usually come before differences
Why: Reading order habit, and the formula is symmetrical enough to look right either way.
But this is the negative of the correct formula. Testing at alpha and beta both pi over four gives negative one half where the product is plainly positive one half.
\[ \sin\alpha\sin\beta = \tfrac{1}{2}\big[\cos(\alpha - \beta) - \cos(\alpha + \beta)\big] \]
Put the difference angle first in every product-to-sum formula
Why: This is what the derivation produces, since the difference identity was the one written first and added or subtracted from.
The fastest safeguard: test at alpha equal to beta. Then the left side is a square, which is non-negative, and the right side becomes half of one minus the cosine of twice the angle — also non-negative. Getting the order backwards makes the right side non-positive, which the test catches immediately.
Fill the middle
Write the product of the sine of 5 theta and the sine of 3 theta as a difference.
Fill in the blanks
\sin 5\theta \sin 3\theta = \tfrac2 theta8 theta\big[\cos(___) - \cos(___)\big]
Why: The difference of the angles is 5 theta minus 3 theta, which is 2 theta, and it comes first with a plus. The sum is 8 theta and comes second with a minus. Note that the resulting angles, 2 theta and 8 theta, are further apart than the originals were — a product of nearby frequencies always produces one very low and one high frequency, which is exactly the beats phenomenon.
Prediction
You rewrite the product of the sine of theta and the cosine of theta using a product-to-sum formula.
Predict first
What do you get?
Correct: Half the sine of 2 theta.
Why: With both angles equal to theta, the formula gives half the sine of zero plus the sine of 2 theta, and the sine of zero is zero, leaving half the sine of 2 theta. That is exactly the double angle formula for sine rearranged, which is a good sign: the product-to-sum formulas must agree with the double angle formulas when the two angles coincide, since both descend from the same sum identities.
Socratic
The name comes from Greek words for addition and subtraction, and the formulas were used by astronomers before logarithms were invented.
Discussion prompt
Explain how these formulas could be used to multiply two numbers using only addition, subtraction and a table.
Hint: To multiply two numbers between negative one and one, what could you take them to be?
Answer:
To multiply two numbers between negative one and one, look each one up in a table of cosines to find angles alpha and beta whose cosines they are. The product of the two numbers is then the product of the two cosines.
\[ \cos\alpha\cos\beta = \tfrac{1}{2}\big[\cos(\alpha - \beta) + \cos(\alpha + \beta)\big] \]
The right-hand side needs only an addition, a subtraction, two more table lookups and a halving — no multiplication at all. Since multiplying long decimals by hand was the dominant cost in sixteenth-century astronomy, this was a genuine breakthrough, and it was superseded only when logarithms did the same job more directly. The formulas survived because calculus turned out to need them for a different reason.
Section
Section 2
Concept
Running the product-to-sum formulas in reverse gives a way to turn a sum into a product. The angles change: if the two output angles are to be alpha and beta, the two input angles must be their half-sum and their half-difference.
Theorem 10.21 — The sum-to-product formulas: each sum or difference of two circular functions rewritten as twice a product evaluated at the half-sum and half-difference of the two angles.
\[ \cos\alpha + \cos\beta = 2\cos\left(\tfrac{\alpha + \beta}{2}\right)\cos\left(\tfrac{\alpha - \beta}{2}\right) \]
The half-sum and half-difference appear because the product-to-sum formulas output the sum and the difference of their inputs. Running that backwards means the inputs must be halved.
Figure (svg): The sum-to-product formulas, each turning a sum or difference of two circular functions into twice a product evaluated at the half-sum and half-difference angles
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 781-781
Picture it
There is one thing you can do with a product that you cannot do with a sum, and it is the reason this family exists in this course.
Figure (svg): A comparison showing that an equation with a sum of two sines equal to zero cannot be split, while the same equation rewritten as a product can be split into two simpler equations
A product equals zero exactly when one of its factors does. A sum equals zero for no such simple reason. Converting a sum into a product turns an unsolvable equation into two easy ones.
Worked example
Example 10.4.6, part 2. Watch the negative angle at the end.
\[ \text{Write } \sin(\theta) - \sin(3\theta) \text{ as a product.} \]
Identify the two angles
Why: Alpha is theta and beta is 3 theta.
\[ a = t, b = 3 t \]
Apply the sine difference formula
Why: Twice the sine of the half-difference times the cosine of the half-sum.
\[ 2 \sin(\frac{t - 3 t}{2}) \cos(\frac{t + 3 t}{2}) \]
Simplify the two angles
Why: The half-difference is negative theta and the half-sum is 2 theta.
\[ 2 \sin(-t) \cos(2 t) \]
Clean up with the odd identity
Why: Sine is odd, so the minus sign comes out front.
\[ -2 \sin(t) \cos(2 t) \]
Figure (svg): The solution to Worked example write a difference as a product shown as a ladder of expressions, one row per legal move
\[ \sin(\theta) - \sin(3\theta) = -2\sin(\theta)\cos(2\theta) \]
Verify: test at theta equal to pi over six
Why: The left side is one half minus the sine of pi over two, which is one half minus one, namely negative one half. The right side is negative two times one half times the cosine of pi over three, which is negative one times one half, namely negative one half. They agree.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 781-781
Faded example
Write the sum of the sine of 5x and the sine of x as a product.
Fill in the blanks
\sin 5x + \sin x = 2\sin\left(\frac3x2x\right)\cos\left(\frac______\right) = 2\sin(___)\cos(___)
Why: The half-sum is six x over two, which is 3x, and it goes inside the sine. The half-difference is four x over two, which is 2x, and it goes inside the cosine. For a sum of sines both signs are positive and no cleanup is needed; a difference would have put the minus inside the half-sum and required the odd identity afterwards.
Worked example
The most-used of the family, and the one with no hidden minus sign.
\[ \text{Write } \cos(7x) + \cos(3x) \text{ as a product.} \]
Identify the two angles
Why: Alpha is 7x and beta is 3x.
\[ a = 7 x, b = 3 x \]
Compute the half-sum and half-difference
Why: Ten x over two is 5x; four x over two is 2x.
\[ \text{half-sum } 5 x,\text{ half-diff } 2 x \]
Apply the cosine plus cosine formula
Why: Twice the product of the two cosines.
\[ 2 \cos(5 x) \cos(2 x) \]
Check no sign cleanup is needed
Why: Both angles came out positive, so the even-odd identities are not required.
Figure (svg): The solution to Worked example a sum of cosines shown as a ladder of expressions, one row per legal move
\[ \cos(7x) + \cos(3x) = 2\cos(5x)\cos(2x) \]
Verify: test at x equal to pi over ten
Why: The left side is the cosine of seven pi over ten plus the cosine of three pi over ten, which is about negative 0.588 plus 0.588, namely 0. The right side is twice the cosine of pi over two times the cosine of pi over five, which is twice zero times something, namely 0. They agree.
Error analysis
A student converts a difference of cosines into a product.
Annotate
On: \( \cos(5x) - \cos(x) = 2\sin(3x)\sin(2x) \)
Only one of the six sum-to-product formulas carries a leading minus sign, and it is this one. That makes it easy to forget and worth flagging: cosine minus cosine is the odd one out.
Sorting
Read what kind of thing you have and what kind you want.
Sort into buckets
Sort each expression by which family converts it.
Prediction
You convert the difference of the cosine of 4x and the cosine of 4x into a product.
Predict first
What do you expect?
Correct: Zero.
Why: The two angles are equal, so their half-difference is zero and the formula gives negative two times the sine of 4x times the sine of zero, which is zero because the sine of zero is zero. That is correct, since a quantity minus itself is zero. This is a useful sanity check on the formula: it must give zero whenever the two angles coincide, and the sine of the half-difference is exactly what guarantees that.
Edge cases
The sum-to-product formula turns a sum of two cosines into twice a product.
Discussion prompt
What happens when the two angles are exactly opposite, say alpha and negative alpha? Work it through and say whether the result makes sense.
Hint: Compute the half-sum and half-difference first.
Answer:
\[ \cos\alpha + \cos(-\alpha) = 2\cos(0)\cos(\alpha) = 2\cos(\alpha) \]
The half-sum is zero and the half-difference is alpha, so the product becomes twice the cosine of zero times the cosine of alpha, which is twice the cosine of alpha.
That is exactly right, because cosine is even, so the cosine of negative alpha equals the cosine of alpha and the left side really is twice that. The formula did not need to know about the even identity — it produced the same answer independently, which is a good sign that the whole system of identities is consistent rather than a list of separate facts.
Section
Section 3
Concept
The two families are exact inverses, so neither is more correct. The question is always which form removes the obstacle in front of you, and there are two standard obstacles.
Neither direction simplifies in any absolute sense. Each trades one kind of complexity for another, and the trade is worth making only when the second kind is the kind you can handle.
Figure (svg): Two columns contrasting when a product is the obstacle and when a sum is the obstacle
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 780-781
Picture it
Here is the single most important reason this lesson exists in a trigonometry course rather than only in a calculus one.
Figure (svg): A comparison showing that an equation with a sum of two sines equal to zero cannot be split, while the same equation rewritten as a product can be split into two simpler equations
Section 10.7 has an entire class of equations that are unsolvable as written and routine after one sum-to-product conversion. That conversion is the whole technique.
Worked example
This is the technique Section 10.7 will use. The conversion is the entire difficulty.
\[ \text{Rewrite } \sin(3\theta) + \sin(\theta) = 0 \text{ in a form that can be solved.} \]
Notice that a sum cannot be split
Why: Setting one term to zero does not solve the equation, since the other need not vanish too.
Convert with the sine sum-to-product formula
Why: The half-sum is 2 theta and the half-difference is theta.
\[ 2 \sin(2 t) \cos(t) = 0 \]
Split into factors
Why: A product is zero exactly when one factor is zero.
\[ \sin 2 t = 0\text{ or } \cos t = 0 \]
State the two simpler equations
Why: Each is a basic equation of the kind solved in Lesson 10.2b.
Figure (svg): The solution to Worked example prepare an equation for solving shown as a ladder of expressions, one row per legal move
\[ \sin(3\theta) + \sin(\theta) = 2\sin(2\theta)\cos(\theta) = 0 \]
Verify: check a solution of one factor
Why: Cosine of theta is zero at theta equal to pi over two. Substituting into the original: the sine of three pi over two is negative one and the sine of pi over two is one, and their sum is zero. The factorisation genuinely produces solutions of the original equation.
Discrimination
The goal, not the expression, decides the direction.
Sort into buckets
Sort each goal by the direction it needs.
Worked example
The other direction, and the reason calculus courses teach these formulas.
\[ \text{Rewrite } \sin(4x)\cos(2x) \text{ so that each term is a first power.} \]
Identify the product as mixed
Why: A sine times a cosine, so the result will be a sum of sines.
Apply the sine times cosine formula
Why: Half the sum of the sines at the difference and the sum angles.
\[ (\frac{1}{2}) [\sin(2 x) + \sin(6 x)] \]
Check the angles
Why: The difference is 4x minus 2x, which is 2x; the sum is 6x.
\[ \text{angles } 2 x\text{ and } 6 x \]
Note why this helps
Why: Each term is now a single sine, which integrates to a cosine over a constant.
Figure (svg): The solution to Worked example prepare an expression for integration shown as a ladder of expressions, one row per legal move
\[ \sin(4x)\cos(2x) = \tfrac{1}{2}\sin(2x) + \tfrac{1}{2}\sin(6x) \]
Verify: test at x equal to pi over twelve
Why: The left side is the sine of pi over three times the cosine of pi over six, which is root three over two squared, namely three quarters. The right side is half the sine of pi over six plus half the sine of pi over two, which is one quarter plus one half, namely three quarters. They agree.
Trap
\[ \sin(3\theta) + \sin(\theta) = 0 \quad\Longrightarrow\quad \sin(3\theta) = 0 \text{ or } \sin(\theta) = 0 \]
Split the sum the way a product would be split
Why: The zero-product property is so familiar that it gets applied to sums by reflex.
But a sum of two numbers can be zero without either being zero — three plus negative three is the obvious case. The step is simply invalid.
\[ \sin(3\theta) + \sin(\theta) = 2\sin(2\theta)\cos(\theta) = 0 \quad\Longrightarrow\quad \sin(2\theta) = 0 \text{ or } \cos(\theta) = 0 \]
Convert to a product first, then split
Why: The zero-product property applies to products and only to products, so the conversion is what earns the right to split.
The wrong method does find some genuine solutions, which is what makes it dangerous — the angles where both sines vanish together do solve the equation. But it misses every solution where the two terms cancel without either being zero, and those are the majority.
Fill the middle
Prepare the equation cosine of 5x minus cosine of x equal to zero for solving.
Fill in the blanks
\cos 5x - \cos x = -2\sin(3x)\sin(2x) = 0 \;\Longrightarrow\; \sin 3x = 0 \textsin 2x ___ = 0
Why: The half-sum is 3x and the half-difference is 2x, and the cosine minus cosine formula carries a leading negative two. Since a nonzero constant factor never makes a product zero, the equation reduces to the two sine factors, each of which is a basic equation. The leading minus sign matters for the identity but not for this particular use of it.
Prediction
You want the average value of the product of the sine of 3x and the sine of 5x over a full period.
Predict first
What do you expect the answer to be?
Correct: Zero.
Why: Converting to a sum gives half the cosine of 2x minus half the cosine of 8x, and each cosine averages to zero over a whole number of its own periods. So the average is zero. This is not a curiosity: it is the orthogonality relation that makes Fourier series work, and the product-to-sum conversion is exactly how it is proved. Note the contrast with sine of 3x times sine of 3x, whose conversion leaves a constant term of one half that does not average away.
Real world
Two guitar strings are plucked, one at 440 hertz and one very slightly out of tune at 444 hertz. The combined sound pressure is the sum of two sines at those frequencies.
Discussion prompt
Convert the sum to a product and describe what a listener actually hears. How many times per second does the loudness rise and fall?
Hint: The half-sum and half-difference frequencies are the two that appear.
Answer:
\[ \sin(2\pi \cdot 440t) + \sin(2\pi \cdot 444t) = 2\sin(2\pi \cdot 442t)\cos(2\pi \cdot 2t) \]
The listener hears a tone at the average frequency of 442 hertz, whose amplitude is modulated by a cosine at 2 hertz. That slow cosine is the envelope.
The loudness peaks whenever the envelope's magnitude peaks, and a cosine reaches maximum magnitude twice per cycle, so the throbbing occurs at 4 hertz — the difference of the two original frequencies. This is precisely how a musician tunes by ear: play both strings, count the beats, and tighten until they slow to nothing. Sum-to-product is the reason the beat rate equals the frequency difference, and it is one of the few places where an identity is directly audible.
Section
Section 4
Concept
A sum-to-product or product-to-sum conversion often produces a negative angle, because the difference of the two angles may be negative. The even-odd identities remove it, and the answer is not finished until they have.
Ordering the two angles so that the larger comes first avoids negative angles altogether, and it is worth doing as a habit — though the even-odd identities will rescue you either way.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 781-781
Picture it
Every one of these can produce a negative angle in its second factor, so the cleanup applies throughout.
Figure (svg): The sum-to-product formulas, each turning a sum or difference of two circular functions into twice a product evaluated at the half-sum and half-difference angles
Notice that the half-difference is the one that can go negative, and it sits inside a cosine in two of the three formulas — which is why the cleanup is usually painless.
Worked example
The angles are given in the order that produces a negative half-difference.
\[ \text{Write } \sin(2x) - \sin(6x) \text{ as a product, fully simplified.} \]
Apply the sine difference formula
Why: Half-difference inside the sine, half-sum inside the cosine.
\[ 2 \sin(\frac{2 x - 6 x}{2}) \cos(\frac{2 x + 6 x}{2}) \]
Simplify the angles
Why: Negative four x over two is negative 2x; eight x over two is 4x.
\[ 2 \sin(-2 x) \cos(4 x) \]
Apply the odd identity to the sine
Why: Sine is odd, so the minus comes out front.
\[ -2 \sin(2 x) \cos(4 x) \]
Check whether the cosine needs cleanup
Why: Its angle is positive, so nothing to do.
Figure (svg): The solution to Worked example a conversion that needs cleanup shown as a ladder of expressions, one row per legal move
\[ \sin(2x) - \sin(6x) = -2\sin(2x)\cos(4x) \]
Verify: test at x equal to pi over eight
Why: The left side is the sine of pi over four minus the sine of three pi over four, which is root two over two minus root two over two, namely zero. The right side is negative two times the sine of pi over four times the cosine of pi over two, which is something times zero, namely zero. They agree.
Sorting
A negative angle inside a cosine is cosmetic; inside a sine it changes the sign.
Sort into buckets
Sort each expression by whether the negative angle changes the value.
Worked example
The same expression, with the two angles taken in the other order.
\[ \text{Write } -\big(\sin(6x) - \sin(2x)\big) \text{ as a product and compare.} \]
Apply the formula to the reordered difference
Why: Now the half-difference is positive.
\[ 2 \sin(\frac{6 x - 2 x}{2}) \cos(\frac{6 x + 2 x}{2}) \]
Simplify the angles
Why: Four x over two is 2x; eight x over two is 4x.
\[ 2 \sin(2 x) \cos(4 x) \]
Restore the leading minus sign
Why: The original expression was the negative of this difference.
\[ -2 \sin(2 x) \cos(4 x) \]
Compare with the previous answer
Why: Identical, as it must be.
Figure (svg): The solution to Worked example reorder to avoid the cleanup shown as a ladder of expressions, one row per legal move
\[ -\big(\sin(6x) - \sin(2x)\big) = -2\sin(2x)\cos(4x) \]
Verify: confirm the two routes agree
Why: Both give negative two times the sine of 2x times the cosine of 4x. Taking the angles in decreasing order avoided the negative angle inside the formula but introduced a leading minus sign instead, so the bookkeeping moved rather than vanished — which is the honest situation.
Error analysis
A student converts a sum of cosines and stops one step early.
Annotate
On: \( \cos(3x) + \cos(7x) = 2\cos(5x)\cos(-2x) \)
This one is a presentation error rather than a mathematical one, and it costs marks rather than correctness. But the same omission with a sine would be a genuine error, because a lost minus sign changes the value.
Faded example
Simplify the result of a sum-to-product conversion.
Fill in the blanks
2\sin(-3x)\cos(5x) = -2\sin(3x)\cos(5x)
Why: Sine is odd, so the sine of negative 3x equals the negative of the sine of 3x. That minus sign multiplies the leading 2, giving negative two. The cosine factor is untouched because its angle was already positive. Bringing every minus sign to the front is the convention, since it makes the sign of the whole expression visible at a glance.
Prediction
You convert the sum of the cosine of 2x and the cosine of 8x into a product, taking the angles in the order given.
Predict first
Will the answer need cleanup?
Correct: Yes, but only cosmetically.
Why: The half-difference is 2x minus 8x over two, which is negative 3x, so a negative angle does appear. But for a sum of cosines the formula puts both angles inside cosines, and cosine is even, so replacing negative 3x by 3x changes nothing about the value. The answer is twice the cosine of 5x times the cosine of 3x either way — tidier, but not different.
Explain it to yourself
Taking the two angles in decreasing order avoids negative angles inside the formula.
Discussion prompt
Explain why that works, and say whether it actually saves any work overall.
Hint: Compare the two worked examples above.
Answer:
Taking the larger angle first makes the half-difference positive, so nothing negative ever appears inside a function. The half-sum is positive either way, since both angles are.
Whether it saves work depends on which expression you started with. If the expression genuinely is the larger minus the smaller, it saves a step outright. If it is the smaller minus the larger, reordering introduces a leading minus sign that has to be carried instead — so the bookkeeping moves rather than disappearing. The honest advice is to know both routes and use whichever keeps the sign in a place you will not lose it.
Section
Section 5
Concept
This is the end of the identity chapter. It is worth seeing the whole structure at once, because the individual formulas are far easier to hold once the dependencies are clear.
Ten families, one proof. Every row after the second is a derivation, and the second row is the only place where geometry entered.
| Family | Derived from | Used for |
|---|---|---|
| Even-odd | reflection across the x-axis | clearing minus signs |
| Cosine difference | equal chords and the distance formula | everything below |
| Cofunction | the cosine difference identity | complementary pairs |
| Sine sum and difference | cofunction plus cosine difference | exact values, expansions |
| Tangent sum and difference | sine over cosine, divided | exact values |
| Double angle | sum identities with equal angles | doubling, simplification |
| Power reduction | double angle, solved backwards | integration, averaging |
| Half angle | power reduction, halved and rooted | halving |
| Product to sum | sum identities, added or subtracted | integration, averaging |
| Sum to product | product to sum, reversed | solving equations |
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 770-782
Picture it
These three and their three inverses are the last of the chapter.
Figure (svg): The three product-to-sum formulas, each turning a product of two circular functions into a sum or difference of cosines or sines at the sum and difference angles
Section 10.5 leaves identities behind entirely and starts graphing. But Section 10.7 comes back to these, and the sum-to-product formulas are the reason it can.
Worked example
Real problems combine families. Knowing which to reach for is the skill.
\[ \text{Verify that } \frac{\sin(3\theta) + \sin(\theta)}{\cos(3\theta) + \cos(\theta)} = \tan(2\theta). \]
Convert the numerator with sum-to-product
Why: Half-sum 2 theta, half-difference theta.
\[ 2 \sin(2 t) \cos(t) \]
Convert the denominator the same way
Why: Same two angles, both inside cosines.
\[ 2 \cos(2 t) \cos(t) \]
Cancel the common factors
Why: The 2 and the cosine of theta appear in both.
\[ \sin(2 t) / \cos(2 t) \]
Recognise the quotient
Why: Sine over cosine is the tangent.
\[ = \tan(2 t) \]
Figure (svg): The solution to Worked example a verification using two families shown as a ladder of expressions, one row per legal move
\[ \frac{2\sin(2\theta)\cos(\theta)}{2\cos(2\theta)\cos(\theta)} = \frac{\sin(2\theta)}{\cos(2\theta)} = \tan(2\theta) \]
Verify: test at theta equal to pi over twelve
Why: The numerator is the sine of pi over four plus the sine of pi over twelve, about 0.707 plus 0.259, namely 0.966. The denominator is the cosine of pi over four plus the cosine of pi over twelve, about 0.707 plus 0.966, namely 1.673. Their quotient is about 0.577, and the tangent of pi over six is root three over three, also about 0.577.
Comparison
Fill the blanks from memory. This is the dependency structure of Section 10.4.
Comparison matrix
| Family | Comes from | Direction |
|---|---|---|
| Cosine difference | equal chords | the one proof |
| Cofunction | the cosine difference identity | swaps a function for its cofunction |
| Double angle | sum identities, angles equal | one angle in, twice out |
| Power reduction | double angle, solved for the square | square in, first power out |
| Product to sum | sum identities, added or subtracted | product in, sum out |
| Sum to product | product to sum, reversed | sum in, product out |
Every entry in the middle column names something proved earlier. The chapter is one theorem and nine consequences, and reading it that way makes it a page rather than a book.
Worked example
Two families in sequence, each removing a different obstacle.
\[ \text{Rewrite } \sin^2(x)\cos(2x) \text{ as a sum of first powers.} \]
Reduce the square first
Why: Power reduction turns the sine squared into a first power at 2x.
\[ (\frac{1 - \cos 2 x}{2}) \cos 2 x \]
Distribute
Why: Half the cosine of 2x minus half the cosine squared of 2x.
\[ (\frac{1}{2}) \cos 2 x - (\frac{1}{2}) \cos ^{2}(2 x) \]
Reduce the remaining square
Why: Power reduction again, doubling 2x to 4x.
\[ (\frac{1}{2}) \cos 2 x - (\frac{1}{4}) (1 + \cos 4 x) \]
Collect
Why: Gather the constant terms.
\[ -\frac{1}{4} + (\frac{1}{2}) \cos 2 x - (\frac{1}{4}) \cos 4 x \]
Figure (svg): The solution to Worked example combine product-to-sum with power reduction shown as a ladder of expressions, one row per legal move
\[ \sin^2(x)\cos(2x) = -\tfrac{1}{4} + \tfrac{1}{2}\cos(2x) - \tfrac{1}{4}\cos(4x) \]
Verify: test at x equal to zero
Why: The left side is zero squared times one, which is 0. The right side is negative one quarter plus one half minus one quarter, which is 0. They agree.
Trap
Faced with the equation sine of 3x plus sine of x equal to zero, a student applies the sine sum identity to expand the sine of 3x into a mess of products.
\[ \sin(3x) = \sin(2x)\cos(x) + \cos(2x)\sin(x) = \cdots \]
Every step is legal, and the expression gets steadily longer. The identity chosen was a valid one but it was moving in the wrong direction.
\[ \sin(3x) + \sin(x) = 2\sin(2x)\cos(x) = 0 \]
Ask what is blocking you before choosing an identity
Why: The obstacle was that a sum cannot be set to zero factorwise, so the direction needed was toward a product.
With ten families available, the constraint is no longer which identities you know but which one addresses the actual obstacle. Naming the obstacle out loud first — a product I cannot integrate, a sum I cannot factor, a square I cannot integrate — turns the choice from guesswork into a lookup.
Elimination
Solve the equation cosine of 4x minus cosine of 2x equal to zero.
Eliminate the wrong options
Which family gets you started?
Survives elimination: B
Why: The obstacle is that a difference cannot be set to zero one term at a time. Converting to a product gives negative two times the sine of 3x times the sine of x, and then the zero-product property splits it into two basic equations. Identifying the obstacle as a sum blocking factorisation is what selects the family, and it does so before any algebra is attempted.
Ranking
Put the families of Section 10.4 in the order they were derived.
Put in order
Why: The even-odd identities come first because the proof of the cosine sum formula uses them. The cosine identities are the one thing proved from geometry. The cofunction identities are the cosine difference formula with pi over two substituted. The double angle formulas set the two angles equal in the sum identities. The product-to-sum formulas add and subtract those same sum identities, which is why they could equally have come earlier — but the book places them last because their applications are the ones needed later.
Explain it
A classmate has written all ten families on flashcards and is trying to memorise thirty-odd formulas.
Discussion prompt
In four sentences or fewer, tell them what to memorise and what to derive. Which formulas genuinely have to be known cold?
Hint: Which one was actually proved, and how long does each derivation take?
Answer:
A usable answer: memorise the cosine sum and difference formulas and the sine sum and difference formulas — four lines — and derive everything else. Setting the two angles equal gives all the double angle formulas in ten seconds; solving those for the square gives power reduction; halving gives the half angle formulas; adding or subtracting the sum identities gives product-to-sum.
The even-odd identities are worth knowing simply because they are one sentence: cosine and secant are even, the other four are odd. That is six formulas condensed into a fact about two functions. Four lines memorised and one sentence beats thirty flashcards, and the derivations are quick enough to do in an exam margin.
Comparison
Fill the blanks from memory. Each family undoes the other.
Comparison matrix
| Product to sum | Sum to product | |
|---|---|---|
| Input | a product of two functions | a sum or difference of two functions |
| Output | a sum of first powers | twice a product |
| Angles that appear | the difference and the sum | the half-sum and the half-difference |
| Leading constant | one half | two |
| Used when | a product blocks integration | a sum blocks factoring |
The leading constants are reciprocals and the angles are related by doubling and halving, which is exactly what you would expect from two operations that undo one another.
Pattern
Whether the question converts a product or a sum, the same five moves cover it.
Testing at one value is worth doing every time in this lesson, because the formulas differ from each other only in signs and in which angle comes first, and a single substitution catches every one of those slips.
OpenStax Algebra and Trigonometry 2e, §9.4 Sum-to-Product and Product-to-Sum Formulas §9.4
Check
Product to sum. Watch which angle comes first.
Check your understanding
Which expression equals the product of the cosine of 5x and the cosine of 3x?
Answer: A
Why: A product of two cosines gives half the sum of the cosines at the difference and the sum angles. The difference is 2x and comes first; the sum is 8x. Both terms carry a plus, which is the feature distinguishing the cosine-times-cosine formula from the sine-times-sine one.
Check
Sum to product. Halve the sum and the difference.
Check your understanding
Which expression equals the sine of 6x minus the sine of 2x?
Answer: B
Why: For a difference of sines the half-difference goes inside the sine and the half-sum inside the cosine. The half-difference is 6x minus 2x over two, which is 2x, and the half-sum is 8x over two, which is 4x. So the answer is twice the sine of 2x times the cosine of 4x.
Check
Choosing a direction. Name the obstacle first.
Check your understanding
You need to solve the equation cosine of 5x plus cosine of x equal to zero. What is the right first move?
Answer: C
Why: A sum cannot be set to zero term by term, but a product can be set to zero factor by factor. Converting gives twice the cosine of 3x times the cosine of 2x, and each factor then yields a basic equation of the kind solved in Lesson 10.2b.
Real world
An AM radio transmitter multiplies a carrier wave at 1000 kilohertz by an audio signal. For a pure audio tone at 5 kilohertz, the transmitted signal is the product of the cosine of the carrier and the cosine of the tone.
Discussion prompt
Convert the product to a sum and identify the frequencies actually transmitted. Then explain why a radio station is allocated a band of frequencies rather than a single one.
Hint: The two output angles are the difference and the sum.
Answer:
\[ \cos(2\pi \cdot 1000t)\cos(2\pi \cdot 5t) = \tfrac{1}{2}\cos(2\pi \cdot 995t) + \tfrac{1}{2}\cos(2\pi \cdot 1005t) \]
The transmitter puts out two frequencies, at 995 and 1005 kilohertz, and no energy at the carrier frequency itself. These are the sidebands.
A real audio signal contains many tones, not one, and each produces its own pair of sidebands spaced by its own frequency. Audio up to 5 kilohertz therefore fills the whole band from 995 to 1005 kilohertz — which is why stations are allocated a band rather than a frequency, and why the allocation's width is set by the audio bandwidth being carried. The product-to-sum identity is what converts a statement about multiplying signals into a statement about which frequencies are occupied, and the entire design of the radio spectrum rests on it.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Is the sum-to-product conversion a simplification?
Correct: Neither; it trades one form for another, and which is better depends on the goal.
\[ \sin(3\theta) + \sin(\theta) = 2\sin(2\theta)\cos(\theta) \]
For solving, the right-hand side. For integrating, the left.
Why: The two forms are exactly equivalent, so neither is objectively simpler. A product is better when you want to set the expression to zero, because only a product factors. A sum is better when you want to integrate or average, because those operate term by term. The same expression is therefore best written differently for different purposes, and the pair of identities exists precisely so that you can move between them. Calling either direction a simplification hides the fact that the choice depends entirely on the obstacle in front of you.
Explain it
They can see that these formulas are true but cannot see why anyone would want them.
Discussion prompt
In no more than five sentences, give them one concrete reason for each direction. Use an equation for one and an integral for the other.
Hint: What can you do with a product that you cannot do with a sum, and vice versa?
Answer:
A usable answer: suppose you have to solve the sine of 3x plus the sine of x equals zero. You cannot set each term to zero, because two numbers can add to zero without either being zero. Rewrite it as a product — twice the sine of 2x times the cosine of x — and now you can, because a product is zero exactly when a factor is.
Going the other way: to integrate the sine of 4x times the cosine of 2x, the product is hopeless, but rewriting it as half the sine of 2x plus half the sine of 6x makes each piece a one-line integral. A product blocks integration and a sum blocks factoring, so you convert toward whichever one is not blocking you.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Product-to-sum angles are fixed by remembering that the difference always comes first. The stray minus sign is fixed by flagging cosine minus cosine as the one exception in the family. Cleanup is fixed by asking whether the function is even or odd, since a cosine absorbs the sign and a sine releases it. The direction choice is fixed by naming the obstacle out loud before reaching for any formula. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page, write the cosine sum and difference identities one above the other, and show explicitly what happens when you add them and when you subtract them, deriving two of the product-to-sum formulas from scratch. Below that, write all three product-to-sum formulas and all three sum-to-product formulas in two columns, and draw arrows between the matching pairs to show that they are inverse. Circle the one formula that carries a leading minus sign. On the right, draw two boxes: label one integrate and write which direction it needs, and label the other solve and do the same. In the bottom left, convert sine of 5x plus sine of 3x to a product and check your answer at x equal to pi over sixteen. In the bottom right, draw the dependency diagram for the whole of Section 10.4, with the cosine difference identity at the top and arrows down to every family that descends from it. Finally, write one sentence saying which four formulas in the whole chapter you would memorise if you could keep only four.
The four worth keeping are the cosine and sine sum and difference identities. Every other family in the chapter can be derived from those four in under a minute, and doing so is more reliable than recalling thirty separate lines.
Recap
Five things, and the last one is what actually decides which formula you reach for.
| If the question says | Your first move is |
|---|---|
| Write this product as a sum | Difference angle first, leading constant one half |
| Write this sum as a product | Half-sum and half-difference, leading constant two |
| Solve an equation that is a sum of two terms | Convert to a product, then split |
| Integrate or average a product | Convert to a sum of first powers |
| A negative angle appeared | Cosine absorbs it; sine releases a minus sign |
That is the end of Section 10.4 and of the identities. Section 10.5 turns to something entirely different: what these six functions look like when they are graphed. The identities do not go away — the graphs will make several of them visible — but for three lessons the tools are pictures rather than algebra.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 780-782 — everything on these slides traces back here
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