Sum-to-product formulas: Worked examples — Maths deep dive

The scenario matrix

Before solving anything, let us list every distinct kind of situation these four formulas produce. Each row is a "cell". Every worked example below is tagged with the cell it covers, so you can see the whole territory is filled.

Cell	Situation	What's tricky about it	Example
C1	Solve an equation, sum $= 0$	product $= 0$ → split factors	Ex 1
C2	Exact value, both angles "nice"	pick right formula, know exact values	Ex 2
C3	The minus case ( $cos A - cos B$ )	remembering the $- 2 sin sin$	Ex 3
C4	The swap case ( $sin A - sin B$ )	outer $cos$ , inner $sin$	Ex 4
C5	Degenerate $A = B$	half-difference $= 0$ , collapses to double-angle	Ex 5
C6	Answer comes out negative / sign of factors	tracking sign across quadrants	Ex 6
C7	Word problem / physics (beats)	translate reality → symbols → back	Ex 7
C8	Exam twist: prove an identity (ratio)	common factor cancels	Ex 8
C9	Limiting behaviour: $B \to A$	what the collapse looks like as a limit	Ex 9
C10	Equation $=$ nonzero constant	can't split-to-zero; move to one side first	Ex 10

Every quadrant/sign question is folded into C6; every zero/degenerate case into C5 and C9; the "not equal to zero" trap is C10. Let us fill each cell.

Worked examples

Worked example Ex 1 — Cell C1: solve

$cos 3 x + cos x = 0$ Forecast: guess — how many families of solutions will there be? (Hint: a product of two factors → two families.)

Identify $A = 3 x$ , $B = x$ . Why this step? The formula needs two angles; name them.
Half-sum $\frac{A + B}{2} = \frac{4 x}{2} = 2 x$ ; half-difference $\frac{A - B}{2} = \frac{2 x}{2} = x$ . Why this step? These are the only two arguments the product will use.
Use the cosine-sum formula (both terms are $cos$ , joined by $+$ ): $cos 3 x + cos x = 2 cos 2 x cos x = 0$ Why this step? A product $= 0$ means at least one factor is $0$ — that is the whole point of factoring.
Split: $cos 2 x = 0$ or $cos x = 0$ .
- $cos 2 x = 0 \Rightarrow 2 x = \frac{π}{2} + nπ \Rightarrow x = \frac{π}{4} + \frac{nπ}{2}$ .
- $cos x = 0 \Rightarrow x = \frac{π}{2} + nπ$ . Why this step? Each factor is now a basic equation whose solutions we know.
Answer: $x = \frac{π}{4} + \frac{nπ}{2}$ or $x = \frac{π}{2} + nπ$ , $n \in Z$ .

Verify: try $x = \frac{π}{4}$ : $cos \frac{3 π}{4} + cos \frac{π}{4} = - \frac{2}{2} + \frac{2}{2} = 0$ ✓. Try $x = \frac{π}{2}$ : $cos \frac{3 π}{2} + cos \frac{π}{2} = 0 + 0 = 0$ ✓.

Worked example Ex 2 — Cell C2: exact value of

$sin 7 5^{\circ} + sin 1 5^{\circ}$ Forecast: which formula, and which two "nice" angles will pop out?

$A = 7 5^{\circ}$ , $B = 1 5^{\circ}$ . Half-sum $= 4 5^{\circ}$ , half-difference $= 3 0^{\circ}$ . Why this step? $4 5^{\circ}$ and $3 0^{\circ}$ have exact known values — that is why the problem was designed this way.
Use sine-sum → $sin \cdot cos$ : $sin 7 5^{\circ} + sin 1 5^{\circ} = 2 sin 4 5^{\circ} cos 3 0^{\circ}$ Why this step? Both starting terms are $sin$ , joined by $+$ , so this is the sine-sum row.
Substitute exact values $sin 4 5^{\circ} = \frac{2}{2}$ , $cos 3 0^{\circ} = \frac{3}{2}$ : $= 2 \cdot \frac{2}{2} \cdot \frac{3}{2} = \frac{6}{2}$ Why this step? We replace the two special-angle values with their exact surds so the answer stays exact (no calculator); the $2$ cancels one of the halves and $2 \cdot 3 = 6$ .
Answer: $\frac{6}{2} \approx 1.2247$ .

Verify: $sin 7 5^{\circ} \approx 0.9659$ , $sin 1 5^{\circ} \approx 0.2588$ ; sum $\approx 1.2247$ ✓.

Worked example Ex 3 — Cell C3: the minus case,

$cos 2 0^{\circ} - cos 10 0^{\circ}$ Forecast: will the answer be positive or negative? ( $cos 2 0^{\circ}$ is big, $cos 10 0^{\circ}$ is negative — so the difference is a big positive... watch the formula's own minus sign.)

$A = 2 0^{\circ}$ , $B = 10 0^{\circ}$ . Half-sum $= 6 0^{\circ}$ , half-difference $= - 4 0^{\circ}$ . Why this step? Order matters here: $A$ is the first term ( $2 0^{\circ}$ ), so the half-difference is genuinely negative.
Use cosine-difference — the one with the minus: $cos 2 0^{\circ} - cos 10 0^{\circ} = - 2 sin 6 0^{\circ} sin (- 4 0^{\circ})$ Why this step? This is the "odd one out" row; forgetting the leading $-$ is the classic error.
$sin (- 4 0^{\circ}) = - sin 4 0^{\circ}$ , so two minuses make a plus: $= - 2 sin 6 0^{\circ} \cdot (- sin 4 0^{\circ}) = 2 sin 6 0^{\circ} sin 4 0^{\circ}$ Why this step? The odd function $sin$ flips sign, cancelling the formula's minus — this is exactly why the final answer ends up positive, matching our forecast.
$2 \cdot \frac{3}{2} \cdot sin 4 0^{\circ} = 3 sin 4 0^{\circ}$ . Why this step? We substitute the exact value $sin 6 0^{\circ} = \frac{3}{2}$ ; the $2$ cancels the denominator $2$ , leaving the clean form $3 sin 4 0^{\circ}$ (there is no special surd for $4 0^{\circ}$ , so we leave it as $sin 4 0^{\circ}$ ).
Answer: $3 sin 4 0^{\circ} \approx 1.1131$ .

Verify: $cos 2 0^{\circ} \approx 0.9397$ , $cos 10 0^{\circ} \approx - 0.1736$ ; difference $\approx 1.1133$ ✓ (rounding).

Worked example Ex 4 — Cell C4: the swap case, factor

$sin 5 x - sin 3 x$ Forecast: in $sin A + sin B$ the outer factor is $sin$ . In the difference, does it stay $sin$ ? (No — it swaps.)

$A = 5 x$ , $B = 3 x$ . Half-sum $\frac{A + B}{2} = \frac{8 x}{2} = 4 x$ ; half-difference $\frac{A - B}{2} = \frac{2 x}{2} = x$ . Why this step? The formula only ever uses these two arguments, so we compute them first; note $A$ (the $+$ term) is $5 x$ , fixing the order of the difference.
Use sine-difference — outer $cos$ , inner $sin$ (the swap): $sin 5 x - sin 3 x = 2 cos 4 x sin x$ Why this step? Compared with the sine-sum, the $sin$ and $cos$ trade places. Getting this backwards is Ex-mistake #3 in the parent note.
Answer: $2 cos 4 x sin x$ .

Verify: at $x = \frac{π}{6}$ : LHS $= sin \frac{5 π}{6} - sin \frac{3 π}{6} = sin \frac{5 π}{6} - sin \frac{π}{2} = 0.5 - 1 = - 0.5$ . RHS $= 2 cos \frac{4 π}{6} sin \frac{π}{6} = 2 cos \frac{2 π}{3} sin \frac{π}{6} = 2 (- 0.5) (0.5) = - 0.5$ ✓.

Worked example Ex 5 — Cell C5: degenerate

$A = B$ , evaluate $sin A + sin A$ Forecast: if the two angles are identical, the half-difference is $0$ . What does $cos 0$ do to the formula?

Set $A = B$ . Half-sum $= \frac{A + A}{2} = A$ ; half-difference $= \frac{A - A}{2} = 0$ . Why this step? This is the boundary case — the formula must still hold, and it should reduce to something obvious.
Sine-sum formula: $sin A + sin A = 2 sin A cos 0 = 2 sin A \cdot 1 = 2 sin A$ Why this step? $cos 0 = 1$ — the half-difference factor disappears, leaving the trivial truth $2 sin A$ . The formula degrades gracefully.
Now run the cosine-sum through the identical collapse: $cos A + cos A = 2 cos A cos 0 = 2 cos A$ ✓. Why this step? We repeat the boundary check on a second formula to confirm the collapse behaviour is a general feature of all four identities, not a lucky coincidence of the sine-sum alone.
And the cosine-difference at $A = B$ : $cos A - cos A = - 2 sin A sin 0 = 0$ . Why this step? This is the sharpest consistency test: the formula must return exactly $0$ (a quantity minus itself), and it does because $sin 0 = 0$ — proving the minus-formula is well-behaved at the degenerate point, not broken.

Verify: at $A = 3 0^{\circ}$ : $sin 3 0^{\circ} + sin 3 0^{\circ} = 1$ and $2 sin 3 0^{\circ} = 1$ ✓.

Worked example Ex 6 — Cell C6: sign / quadrant tracking, evaluate

$sin 20 0^{\circ} + sin 4 0^{\circ}$ Forecast: $sin 20 0^{\circ}$ is negative (third quadrant), $sin 4 0^{\circ}$ positive. Will they cancel to something small, or reinforce? Guess the sign of the answer first.

Figure (Ex 6): the unit circle. Two arrows point to the angles $20 0^{\circ}$ (red) and $4 0^{\circ}$ (black). The dashed vertical drop from each arrow's tip to the horizontal axis is that angle's sine — its height. The red arrow's tip sits below the axis, so $sin 20 0^{\circ} < 0$ ; the black arrow's tip sits above, so $sin 4 0^{\circ} > 0$ . The figure's job is to let you literally see the two opposite signs before we add them, so the small positive total is no surprise.

$A = 20 0^{\circ}$ , $B = 4 0^{\circ}$ . Half-sum $= 12 0^{\circ}$ , half-difference $= 8 0^{\circ}$ . Why this step? Notice the half-sum $12 0^{\circ}$ lands in quadrant II, where $sin$ is positive but $cos$ is negative — the sign lives here.
Sine-sum: $sin 20 0^{\circ} + sin 4 0^{\circ} = 2 sin 12 0^{\circ} cos 8 0^{\circ}$ Why this step? Both starting terms are $sin$ , joined by $+$ , so we are in the sine-sum row; this converts the awkward sum of a negative and a positive sine into a single product whose sign we can read factor-by-factor.
$sin 12 0^{\circ} = \frac{3}{2} > 0$ , $cos 8 0^{\circ} \approx 0.1736 > 0$ , so the product is positive. Why this step? We track each factor's sign through its quadrant instead of guessing the total blindly.
$2 \cdot \frac{3}{2} \cdot 0.1736 \approx 0.3006$ .
Answer: $\approx 0.3006$ (positive — the reinforcement wins over the near-cancellation).

Verify: $sin 20 0^{\circ} \approx - 0.3420$ , $sin 4 0^{\circ} \approx 0.6428$ ; sum $\approx 0.3007$ ✓.

Worked example Ex 7 — Cell C7: word problem (beats)

Statement: Two tuning forks vibrate at $f_{1} = 264$ Hz and $f_{2} = 260$ Hz. Struck together, how many times per second does the loudness rise and fall (the beat rate), and what is the pitch you actually hear (carrier)? Forecast: guess — is the beat rate the sum or the difference of the frequencies?

Model each fork as $sin (2 π f_{1} t)$ and $sin (2 π f_{2} t)$ ; the combined pressure is their sum, so let $A = 2 π f_{1} t$ , $B = 2 π f_{2} t$ . Why this step? Sound pressures add; a sum of sines is exactly what sum-to-product eats.
Sine-sum: $sin A + sin B = 2 envelope cos (2 π \frac{f _{1} - f _{2}}{2} t) carrier sin (2 π \frac{f _{1} + f _{2}}{2} t)$ Why this step? The half-difference rides the slow cosine (envelope); the half-sum rides the fast sine (carrier).
Envelope frequency $= \frac{f _{1} - f _{2}}{2} = \frac{264 - 260}{2} = 2$ Hz. But loudness peaks twice per cosine cycle (once at $+ 1$ , once at $- 1$ ), so the audible beat rate $= 2 \times 2 = 4$ Hz $= ∣ f_{1} - f_{2} ∣$ . Why this step? $cos$ has the same magnitude at its peak and trough, so "loud" happens at both — doubling the perceived rate.
Carrier frequency $= \frac{f _{1} + f _{2}}{2} = \frac{264 + 260}{2} = 262$ Hz — the pitch you hear.
Answer: beats at $4$ per second; pitch $262$ Hz.

Verify: beat rate should equal $∣ f_{1} - f_{2} ∣ = ∣264 - 260∣ = 4$ ✓; carrier $= 262$ ✓.

Worked example Ex 8 — Cell C8: exam identity, prove

$\frac{cos A - cos B}{sin A - sin B} = - tan \frac{A + B}{2}$ Forecast: which common factor will cancel top-and-bottom, and where does the minus survive?

Top (cosine-difference): $cos A - cos B = - 2 sin \frac{A + B}{2} sin \frac{A - B}{2}$ . Why this step? We factor the numerator into a product using the "odd one out" row, because the whole strategy of an identity like this is to expose a factor that the denominator also contains so it can cancel.
Bottom (sine-difference): $sin A - sin B = 2 cos \frac{A + B}{2} sin \frac{A - B}{2}$ . Why this step? Factor the denominator with the swap row so a shared piece appears — the $2$ and $sin \frac{A - B}{2}$ .
Divide; the $2$ and the $sin \frac{A - B}{2}$ cancel, provided $sin \frac{A - B}{2} \neq = 0$ , i.e. $A \neq = B$ (mod $2 π$ ): $\frac{- s i n \frac{A + B}{2}}{c o s \frac{A + B}{2}} = - tan \frac{A + B}{2}$ Why this step? We may only cancel a factor that is nonzero; if $A = B$ the denominator $sin A - sin B$ is itself $0$ and the original fraction is undefined, so excluding $A = B$ costs us nothing.
The leading $-$ from the cosine-difference has nowhere to cancel, so it stays in the answer; the surviving $\frac{s i n}{c o s}$ of the half-sum is exactly $tan \frac{A + B}{2}$ . Why this step? $tan θ = \frac{s i n θ}{c o s θ}$ by definition, so we read the leftover ratio directly as a tangent.
Result: $\frac{cos A - cos B}{sin A - sin B} = - tan \frac{A + B}{2}$ for $A \neq = B$ . Proved.

Verify: at $A = 8 0^{\circ}$ , $B = 2 0^{\circ}$ : LHS $= \frac{c o s 8 0 ^{\circ} - c o s 2 0 ^{\circ}}{s i n 8 0 ^{\circ} - s i n 2 0 ^{\circ}} = \frac{0.1736 - 0.9397}{0.9848 - 0.3420} \approx \frac{- 0.7661}{0.6428} \approx - 1.1918$ . RHS $= - tan 5 0^{\circ} \approx - 1.1918$ ✓.

Worked example Ex 9 — Cell C9: limiting behaviour as

$B \to A$ Statement: What does $\frac{cos A - cos B}{A - B}$ approach as $B \to A$ (angles in radians)? Forecast: guess — does this blow up, vanish, or settle to a finite value? (Recall: a difference over a difference is a slope.)

Figure (Ex 9): the black curve is $y = cos x$ . The red straight segment joins the two points at $x = A$ and $x = B$ — it is the chord, whose slope is exactly $\frac{c o s A - c o s B}{A - B}$ . The black dashed line is the tangent at $x = A$ , whose slope is $- sin A$ . The figure's job is to show that as $B$ slides toward $A$ , the red chord tips over until it lies flat against the dashed tangent — so the ratio we are computing is the slope of $cos$ at $A$ .

Factor the top with the cosine-difference formula: $cos A - cos B = - 2 sin \frac{A + B}{2} sin \frac{A - B}{2}$ Why this step? We want to expose the tiny factor $sin \frac{A - B}{2}$ that will control the limit.
Introduce a name for the small angle: let $θ = \frac{A - B}{2}$ (this is the half-difference; as $B \to A$ it shrinks to $0$ ). Then $A - B = 2 θ$ , so divide top and bottom by $A - B = 2 θ$ : $\frac{c o s A - c o s B}{A - B} = \frac{- 2 s i n \frac{A + B}{2} s i n θ}{2 θ} = - sin \frac{A + B}{2} \cdot \frac{s i n θ}{θ}$ Why this step? Renaming $\frac{A - B}{2}$ as $θ$ lets us see the classic pattern $\frac{s i n θ}{θ}$ cleanly; the $2$ in the numerator's $- 2$ cancels the $2$ in $2 θ$ , leaving one clean $sin θ / θ$ .
Use the small-angle fact: for a tiny angle, $sin θ \approx θ$ , so $\frac{sin θ}{θ} \to 1$ as $θ \to 0$ . Why this step? Geometrically a tiny chord equals its arc, so the sine of a small angle equals the angle itself — this is the one limit we must import, and it kills the $sin θ / θ$ factor down to $1$ .
As $B \to A$ : $θ \to 0$ so $\frac{s i n θ}{θ} \to 1$ , and $\frac{A + B}{2} \to A$ so $sin \frac{A + B}{2} \to sin A$ . Multiplying the two limits: $\frac{c o s A - c o s B}{A - B} ⟶ - sin A \cdot 1 = - sin A$ Why this step? Each factor has its own clean limit, so the product's limit is the product of the limits — collapsing the whole ratio to $- sin A$ .
Answer: the limit is $- sin A$ (which is exactly the derivative of $cos A$ — the formula reveals calculus).

Verify: at $A = 1$ rad, $B = 1.001$ rad: $\frac{c o s 1 - c o s 1.001}{1 - 1.001} \approx - 0.8409$ , while $- sin 1 \approx - 0.8415$ ✓ (close, and tightens as $B \to A$ ).

Worked example Ex 10 — Cell C10: equation equal to a

nonzero constant, solve $cos 5 x - cos x = 1$ (on $[0, 2 π)$ , illustrative) Forecast: the tempting move is to factor and "set each factor to $1$ " — but a product equal to $1$ does not split that way. Guess what must change first.

Factor the left side with the cosine-difference row: $A = 5 x$ , $B = x$ , half-sum $= 3 x$ , half-difference $= 2 x$ : $cos 5 x - cos x = - 2 sin 3 x sin 2 x$ Why this step? Sum-to-product still turns the left side into a single product — but the right side is $1$ , not $0$ .
The equation is now $- 2 sin 3 x sin 2 x = 1$ . Why this step? Crucially you cannot split this into separate factor-equations — a product $= 1$ can happen with neither factor equal to $1$ (e.g. $2 \times \frac{1}{2}$ ). The zero-product trick only works against $0$ .
So the correct route is: keep everything on one side, $- 2 sin 3 x sin 2 x - 1 = 0$ , and solve the resulting product-form-plus-constant either numerically or by a further identity — there is no clean factoring shortcut. Why this step? Recognising when a tool does not apply is itself the lesson of this cell; forcing "each factor $= 1$ " would produce wrong answers.
Answer / takeaway: sum-to-product is for factoring, and factoring only solves an equation when the equation is set to $0$ . Against a nonzero constant, first subtract to make one side $0$ , or bound the product ( $∣ - 2 sin 3 x sin 2 x ∣ \leq 2$ , so a value of $1$ is at least possible), then solve numerically.

Verify: the false shortcut " $sin 3 x = 1$ and $sin 2 x = - \frac{1}{2}$ " would need both simultaneously; at $x = \frac{π}{6}$ , $- 2 sin \frac{π}{2} sin \frac{π}{3} = - 2 (1) (\frac{3}{2}) = - 3 \approx - 1.732 \neq = 1$ , confirming the product form is correct and the naive split is invalid.

Active recall

Recall Which cell is each problem?

" $cos 4 0^{\circ} - cos 8 0^{\circ}$ exactly" ::: C3 (minus case)
" $sin 3 x - sin x = 0$ " ::: C4 (swap) then C1 (solve)
"beat rate of 440 and 443 Hz" ::: C7 (word problem) → $3$ Hz
" $lim_{B \to A} \frac{s i n A - s i n B}{A - B}$ " ::: C9 → $cos A$
" $sin 3 x + sin x = 1$ " ::: C10 (move to one side; no zero-split)

Connections

Parent: Sum-to-product formulas — the four identities and their derivation.
Angle Addition Formulas — where the identities come from.
Product-to-Sum Formulas — the reverse direction, used implicitly in Ex 8.
Solving Trigonometric Equations — Cells C1, C4, C10.
Beats and Superposition (Waves) — Cell C7.
Double Angle Formulas — the degenerate $A = B$ case, Cell C5.