Sum-to-product formulas: Question bank — Maths deep dive

Recall the four identities we are stress-testing, so every symbol below is anchored first:

Four pictures to lean on

Before the traps, build the visual intuition the questions keep pointing at. Each figure below is one idea.

1. Where the half-sum and half-difference live. On the unit circle, the two angles $A$ and $B$ are two spokes. Their bisector sits at the half-sum $\frac{A + B}{2}$ (the "average direction"); how far each spoke leans off that bisector is the half-difference $\frac{A - B}{2}$ . Every sum-to-product argument is literally these two angles.

Look at the plum bisector arrow: that is $\frac{A + B}{2}$ . The teal wedge from bisector to each spoke is $\frac{A - B}{2}$ . Swapping $A \leftrightarrow B$ mirrors the two spokes across the bisector — the bisector itself does not move, which is why $cos A + cos B$ is unchanged but $sin A - sin B$ flips sign.

2. Where the factor of $2$ comes from. Adding the two addition formulas cancels one pair of terms and leaves the other pair twice. There is no hidden step: two identical survivors literally stack into $2 \times$ .

The orange bars are the terms that survive; the grey faded bars are the ones that cancel. Two survivors of equal height stack to height $2$ — that is the $2$ in front of every identity.

3. Why cosine differences carry a minus. Plot $cos$ from $0$ to $π$ : it slides downhill. So if $A > B$ , then $cos A$ sits below $cos B$ and the difference $cos A - cos B$ is negative. The factored form must carry a minus to reproduce that downhill sign.

The orange dots mark $cos A$ and $cos B$ with $A > B$ ; the red drop shows $cos A - cos B < 0$ . Sine (teal) climbs uphill over $(0, \frac{π}{2})$ , so its difference stays positive — no leading minus.

4. Beats: slow envelope × fast carrier. Add two sines of nearby frequency and the sum-to-product form splits into a fast wiggle inside a slowly breathing envelope.

The teal curve is the fast carrier $sin (2 π \frac{f _{1} + f _{2}}{2} t)$ ; the plum dashed curves are the slow envelope $\pm 2 cos (2 π \frac{f _{1} - f _{2}}{2} t)$ . The envelope repeats twice per cosine cycle, so the loudness pulses at $∣ f_{1} - f_{2} ∣$ — the audible beat.

True or false — justify

Recall

$sin A + sin B = sin (A + B)$ for all $A, B$ . False ::: $sin$ is not linear, so addition on the outside does not slide inside; the correct form is $2 sin \frac{A + B}{2} cos \frac{A - B}{2}$ . Test $A = B = 9 0^{\circ}$ : LHS $= 2$ but $sin 18 0^{\circ} = 0$ .

Recall Every sum-to-product identity has a factor of

$2$ in front. True ::: Write out, say, $sin (x + y) + sin (x - y)$ : the two $cos x sin y$ terms cancel and the two $sin x cos y$ terms are identical, so they add to $2 sin x cos y$ . Same mechanism (two equal survivors) produces the $2$ in every one of the four — see figure s02.

Recall

$cos A - cos B = 2 sin \frac{A + B}{2} sin \frac{A - B}{2}$ (positive out front). False ::: Cosine decreases on $(0, π)$ , so a difference of cosines runs downhill and is negative for $A > B$ (figure s03); the factored form must carry a minus: $- 2 sin \frac{A + B}{2} sin \frac{A - B}{2}$ . Dropping the minus is the single most common error here.

Recall The outer factor in

$sin A - sin B$ is $sin \frac{A + B}{2}$ , just like in $sin A + sin B$ . False ::: The roles swap for the difference: the outer factor becomes $cos \frac{A + B}{2}$ and the inner becomes $sin \frac{A - B}{2}$ . Sum gives $sin cos$ ; difference gives $cos sin$ .

Recall If

$A$ and $B$ are swapped, $cos A + cos B$ gives the exact same product. True ::: Swapping only flips the sign of the half-difference $\frac{A - B}{2}$ , and $cos$ is even ( $cos (- θ) = cos θ$ ), so $cos A + cos B$ is unchanged — consistent with the fact that addition is commutative. Visually (s01), the bisector direction $\frac{A + B}{2}$ does not move.

Recall Swapping

$A$ and $B$ in $sin A - sin B$ leaves the result unchanged. False ::: $sin A - sin B$ is antisymmetric — swapping flips the whole sign. In the product, $sin \frac{A - B}{2}$ becomes $sin \frac{B - A}{2} = - sin \frac{A - B}{2}$ , matching that the difference itself negates.

Recall Sum-to-product and product-to-sum are inverse operations.

True ::: Product-to-sum expands a product into a sum; sum-to-product runs the same algebra backwards via the substitution $A = x + y, B = x - y$ . They undo each other, so applying one then the other returns the original expression. See Product-to-Sum Formulas.

Recall

$\frac{s i n A + s i n B}{c o s A + c o s B} = tan \frac{A + B}{2}$ holds for every $A, B$ . False — it needs two explicit conditions ::: Factor top and bottom: $\frac{2 sin \frac{A + B}{2} cos \frac{A - B}{2}}{2 cos \frac{A + B}{2} cos \frac{A - B}{2}}$ . You may cancel $2$ and $cos \frac{A - B}{2}$ only when $cos \frac{A - B}{2} \neq = 0$ (else you divided by zero). What is left, $tan \frac{A + B}{2}$ , is itself undefined when $cos \frac{A + B}{2} = 0$ . So the identity holds precisely when both $cos \frac{A - B}{2} \neq = 0$ and $cos \frac{A + B}{2} \neq = 0$ ; at either zero the denominator vanishes and the equation is meaningless.

Spot the error

Recall "

$cos 3 x + cos x = 2 cos 2 x cos 2 x = 2 cos^{2} 2 x$ ." Error: the half-difference was mis-computed ::: Half-sum is $\frac{3 x + x}{2} = 2 x$ ✓ but half-difference is $\frac{3 x - x}{2} = x$ , not $2 x$ . Correct: $2 cos 2 x cos x$ .

Recall "

$sin 5 x - sin x = 2 sin 3 x cos 2 x$ ." Error: sum-form used for a difference ::: A sine difference needs $cos \cdot sin$ , not $sin \cdot cos$ . Correct: $2 cos 3 x sin 2 x$ . The student applied the $sin A + sin B$ pattern by mistake.

Recall "

$cos x - cos 5 x = 2 sin 3 x sin 2 x$ , since sine of a difference is positive." Error: sign reasoning is backwards ::: The formula is $cos A - cos B = - 2 sin \frac{A + B}{2} sin \frac{A - B}{2}$ with $A = x, B = 5 x$ , giving $- 2 sin 3 x sin (- 2 x) = + 2 sin 3 x sin 2 x$ . The answer is right by luck, but the justification ("sine of a difference is positive") is nonsense — the plus came from the odd factor $sin (- 2 x)$ flipping the leading minus.

Recall "To solve

$sin 3 x + sin x = 0$ , I write $sin 3 x = - sin x$ and square both sides." Error: introduces extraneous roots ::: Squaring adds solutions of $sin 3 x = + sin x$ too. Factoring is cleaner: $2 sin 2 x cos x = 0$ , so each factor $= 0$ gives exactly the true roots. See Solving Trigonometric Equations.

Recall "

$cos 4 0^{\circ} + cos 4 0^{\circ} = 2 cos 4 0^{\circ} cos 0^{\circ} = 2 cos 4 0^{\circ}$ — so the formula only works when the angles differ." No error in the result, error in the conclusion ::: With $A = B = 4 0^{\circ}$ the half-difference is $0$ and $cos 0^{\circ} = 1$ , correctly giving $2 cos 4 0^{\circ}$ . Equal angles are a valid limiting case, not a breakdown — this is exactly the Double Angle Formulas connection.

Recall "

$sin A + sin B$ is maximised when both $A, B = 9 0^{\circ}$ , so the product form must peak there too." Error: not a rule, just this case ::: True numerically here ( $2 sin 9 0^{\circ} cos 0^{\circ} = 2$ ), but the reasoning "sum peaks so product peaks" is empty — the product's value is governed by the two new arguments $\frac{A + B}{2}, \frac{A - B}{2}$ , which need separate analysis, not the original angles.

Recall "Solving

$\frac{s i n A + s i n B}{c o s A + c o s B} = 1$ gives $tan \frac{A + B}{2} = 1$ , so $\frac{A + B}{2} = \frac{π}{4}$ and that's the only solution." Error: forgot tangent's period ::: $tan θ = 1$ has solutions $θ = \frac{π}{4} + nπ$ for every integer $n$ , because $tan$ repeats every $π$ . So the general solution is $\frac{A + B}{2} = \frac{π}{4} + nπ$ . Whenever a ratio collapses to a tangent, you must add $nπ$ — see Solving Trigonometric Equations.

Why questions

Recall Why does turning a sum into a product help solve equations?

A product equals zero exactly when one factor is zero ::: So $2 sin 2 x cos x = 0$ splits into the easy sub-problems $sin 2 x = 0$ or $cos x = 0$ . A raw sum offers no such split — that's the entire strategic payoff.

Recall Why is there a minus sign in

$cos A - cos B$ but not in the other three? Trace it algebraically, then geometrically ::: Algebra: $cos (x + y) = cos x cos y - sin x sin y$ and $cos (x - y) = cos x cos y + sin x sin y$ . Subtract them — the $cos x cos y$ terms cancel and the two $- sin x sin y$ vs $+ sin x sin y$ combine to $- 2 sin x sin y$ ; the minus is inherited directly from the $- sin x sin y$ term inside the addition formula. Substituting $x = \frac{A + B}{2}, y = \frac{A - B}{2}$ keeps that minus. Geometrically (s03), cosine slides downhill on $(0, π)$ , so $cos A - cos B < 0$ for $A > B$ — the minus is exactly what makes the sign come out right. The other three subtract terms that carry a $+$ , so no minus appears.

Recall Why do the two new arguments come out as

$\frac{A + B}{2}$ and $\frac{A - B}{2}$ specifically? They come from the substitution $A = x + y, B = x - y$ ::: Solving gives $x = \frac{A + B}{2}$ (half-sum) and $y = \frac{A - B}{2}$ (half-difference). Since the derivation's products used $x$ and $y$ , those halves are forced — no free choice. Picture s01 shows them as bisector and lean-angle.

Recall Why does the beats formula split into a "slow" and a "fast" factor?

Because the half-difference frequency is tiny and the half-sum is large ::: For close frequencies $f_{1} \approx f_{2}$ , $\frac{f _{1} - f _{2}}{2}$ is small (slow envelope) while $\frac{f _{1} + f _{2}}{2}$ is large (fast carrier). Sum-to-product hands you exactly this envelope×carrier split — figure s04 shows the plum envelope hugging the teal carrier; see Beats and Superposition (Waves).

Recall Why can we treat sum-to-product as "angle-addition run backwards"?

Because the identities were built by adding/subtracting the addition formulas ::: We never invented new math — combining $sin (x + y)$ and $sin (x - y)$ and then renaming is fully reversible, so the whole family lives inside Angle Addition Formulas.

Edge cases

Recall What does

$sin A + sin B$ become when $A = B$ ? The half-difference is $0$ , so $cos 0 = 1$ ::: You get $2 sin A \cdot 1 = 2 sin A$ , which is just "adding a thing to itself." Consistent and correct — equal angles are a valid degenerate input.

Recall What is

$sin A - sin B$ when $A = B$ ? Zero, cleanly ::: The inner factor is $sin \frac{A - B}{2} = sin 0 = 0$ , killing the product. This matches the obvious fact that a quantity minus itself is $0$ .

Recall The formula

$\frac{s i n A + s i n B}{c o s A + c o s B} = tan \frac{A + B}{2}$ — where exactly is it undefined? Where $cos A + cos B = 0$ , i.e. $cos \frac{A + B}{2} = 0$ or $cos \frac{A - B}{2} = 0$ ::: If $cos \frac{A + B}{2} = 0$ the tangent itself blows up (a genuine infinity); if $cos \frac{A - B}{2} = 0$ the "cancellation" was illegal because you divided by zero. Always check the denominator before cancelling.

Recall Is

$cos A + cos B$ ever negative even though it's " $+$ "? Yes ::: The sign is set by $cos \frac{A + B}{2} cos \frac{A - B}{2}$ , either of which can be negative (e.g. $A = B = 17 0^{\circ}$ gives $2 cos 17 0^{\circ} < 0$ ). The " $+$ " refers to combining the two cosines, not to the sign of the result.

Recall In beats, what happens to the "slow envelope" as

$f_{1} \to f_{2}$ ? Its frequency $\frac{∣ f _{1} - f _{2} ∣}{2} \to 0$ ::: The envelope stops wobbling and becomes constant, so the beats disappear and you hear one steady tone at frequency $f_{1}$ . The formula degrades gracefully to a single sine.

Recall Can

$sin A + sin B$ exceed $2$ in magnitude? No ::: Each sine lies in $[- 1, 1]$ , so the sum lies in $[- 2, 2]$ ; the product form $2 sin \frac{A + B}{2} cos \frac{A - B}{2}$ is a product of $2$ with two factors each $\leq 1$ in size, capping it at $2$ . Both viewpoints agree.

Connections

Angle Addition Formulas — the source every trap traces back to.
Product-to-Sum Formulas — the exact inverse; "spot the error" items lean on it.
Double Angle Formulas — the $A = B$ edge cases.
Solving Trigonometric Equations — why factoring beats squaring, and the $nπ$ tangent trap.
Beats and Superposition (Waves) — the envelope-limit edge cases.