Product-to-sum formulas: Worked examples — Maths deep dive

Before we start, one reminder of the toolkit (never used before defined — these came from Angle Addition Formulas):

The scenario matrix

Here is every case class a product-to-sum question can belong to. Each later example is tagged with the cell it fills.

Cell	Case class	What makes it tricky	Example
C1	Same-kind, $cos cos$	must use $+$ sign, keep $\frac{1}{2}$	Ex 1
C2	Same-kind, $sin sin$	must use $-$ sign	Ex 2
C3	Mixed, $sin cos$	output is sines, order matters	Ex 3
C4	Degenerate $A = B$	reduces to a double-angle identity	Ex 4
C5	Zero / collapse ( $A + B$ or $A - B$ lands on $9 0^{\circ}$ , $0^{\circ}$ )	a term vanishes	Ex 5
C6	Negative / large angle (sign of $A - B$ , angles $> 9 0^{\circ}$ )	which term is negative?	Ex 6
C7	Integration application	product $\to$ sum makes it integrable	Ex 7
C8	Real-world word problem (beats)	frequencies, envelope, physical meaning	Ex 8
C9	Exam twist (definite integral / orthogonality) + $cos A sin B$	answer is exactly $0$	Ex 9

Together these fill all sign combinations, both same-kind rules, both mixed rules ( $sin cos$ and $cos sin$ ), the collapse-to-zero edge, the equal-angle edge, and the two big applications.

Worked examples

Worked example Example 1 —

$cos cos$ , both positive angles (Cell C1) Write $2 cos 4 0^{\circ} cos 1 0^{\circ}$ as a sum, then a decimal.

Forecast: Guess — will the two output angles be $3 0^{\circ}$ and $5 0^{\circ}$ , or something else? Jot a guess.

Step 1. Identify the form: two cosines multiplied $\Rightarrow$ use $cos A cos B$ with $A = 4 0^{\circ}$ , $B = 1 0^{\circ}$ . Why this step? "Same kind, both cosine" is exactly the first rule; picking the right rule first prevents sign errors later.

Step 2. Apply the rule. The difference angle is $A - B = 3 0^{\circ}$ , the sum angle is $A + B = 5 0^{\circ}$ , and $cos cos$ uses a plus: $cos 4 0^{\circ} cos 1 0^{\circ} = \frac{1}{2} [cos 3 0^{\circ} + cos 5 0^{\circ}] .$ Why this step? This is the whole point — a product we can't add up becomes a sum of two ordinary cosines.

Step 3. Our expression had a leading $2$ , which cancels the $\frac{1}{2}$ : $2 cos 4 0^{\circ} cos 1 0^{\circ} = cos 3 0^{\circ} + cos 5 0^{\circ} .$ Why this step? Watch the constant out front — the built-in $\frac{1}{2}$ is the single most-forgotten piece.

Verify: $cos 4 0^{\circ} cos 1 0^{\circ} = 0.7660 \times 0.9848 = 0.7544$ ; and $\frac{1}{2} (cos 3 0^{\circ} + cos 5 0^{\circ}) = \frac{1}{2} (0.8660 + 0.6428) = 0.7544$ . ✓ (Your forecast of $3 0^{\circ}, 5 0^{\circ}$ was right!)

Worked example Example 2 —

$sin sin$ , the minus-sign rule (Cell C2) Evaluate $sin 7 5^{\circ} sin 1 5^{\circ}$ exactly.

Forecast: Same kind, but now sines. Will the sign be $+$ like Example 1, or flip to $-$ ? Guess before reading.

Step 1. Two sines multiplied $\Rightarrow$ rule $sin A sin B = \frac{1}{2} [cos (A - B) - cos (A + B)]$ , with $A = 7 5^{\circ}$ , $B = 1 5^{\circ}$ . Why this step? Same-kind but sine flips the inner sign to a minus — that is the one difference from $cos cos$ .

Step 2. $A - B = 6 0^{\circ}$ , $A + B = 9 0^{\circ}$ : $sin 7 5^{\circ} sin 1 5^{\circ} = \frac{1}{2} [cos 6 0^{\circ} - cos 9 0^{\circ}] .$ Why this step? Both output angles are "nice" — this is why the problem was chosen; it becomes exact.

Step 3. $cos 6 0^{\circ} = \frac{1}{2}$ , $cos 9 0^{\circ} = 0$ : $= \frac{1}{2} [\frac{1}{2} - 0] = \frac{1}{4} .$ Why this step? Known unit-circle values finish the job.

Verify: numerically $sin 7 5^{\circ} sin 1 5^{\circ} = 0.9659 \times 0.2588 = 0.2500 = \frac{1}{4}$ . ✓ The sign was a minus — flip caught.

Worked example Example 6 — Large angle, negative output term (Cell C6)

Express $cos 2 0^{\circ} cos 10 0^{\circ}$ as a sum and predict its sign.

Forecast: $A - B = - 8 0^{\circ}$ and $A + B = 12 0^{\circ}$ . Since $cos$ is even, $cos (- 8 0^{\circ}) = cos 8 0^{\circ} > 0$ ; but $cos 12 0^{\circ} < 0$ . Guess: is the product positive or negative?

Step 1. $cos A cos B$ with $A = 2 0^{\circ}$ , $B = 10 0^{\circ}$ : $cos 2 0^{\circ} cos 10 0^{\circ} = \frac{1}{2} [cos (- 8 0^{\circ}) + cos 12 0^{\circ}] .$ Why this step? We keep the negative difference angle — no need to reorder, because…

Step 2. $cos$ is an even function: $cos (- 8 0^{\circ}) = cos 8 0^{\circ}$ . So order of $A, B$ never matters for $cos cos$ . $= \frac{1}{2} [cos 8 0^{\circ} + cos 12 0^{\circ}] = \frac{1}{2} [0.1736 + (- 0.5)] = \frac{1}{2} (- 0.3264) = - 0.1632.$ Why this step? The $cos 12 0^{\circ}$ (second quadrant) is negative and outweighs the small positive term — hence a negative product.

Verify: $cos 2 0^{\circ} cos 10 0^{\circ} = 0.9397 \times (- 0.1736) = - 0.1632$ . ✓ Negative, as the quadrant analysis warned.

Worked example Example 7 — Making an integral doable (Cell C7)

Find $\int cos 4 x cos x d x$ .

Forecast: You cannot integrate a product of cosines term-by-term. What two frequencies will the sum split into?

Radian warning first. From here on $x$ is measured in radians, not degrees. The integration and differentiation rules $\int cos k x d x = \frac{s i n k x}{k}$ and $\frac{d}{d x} sin k x = k cos k x$ are only true when the angle is in radians — that is the measure where the "steepness" of $sin$ at $0$ is exactly $1$ . In degrees an extra $\frac{π}{180}$ factor sneaks in, so we commit to radians for all of calculus.

Step 1. $cos A cos B = \frac{1}{2} [cos (A - B) + cos (A + B)]$ , $A = 4 x$ , $B = x$ : $cos 4 x cos x = \frac{1}{2} [cos 3 x + cos 5 x] .$ Why this step? This is the whole reason product-to-sum exists for calculus — see Integration of Trig Functions. A sum integrates term-by-term; a product does not.

Step 2. Integrate each cosine using $\int cos k x d x = \frac{s i n k x}{k}$ (radians): $\int cos 4 x cos x d x = \frac{1}{2} [\frac{s i n 3 x}{3} + \frac{s i n 5 x}{5}] + C = \frac{s i n 3 x}{6} + \frac{s i n 5 x}{10} + C .$ Why this step? Each term is now elementary.

Verify: differentiate the answer: $\frac{d}{d x} [\frac{s i n 3 x}{6} + \frac{s i n 5 x}{10}] = \frac{3 c o s 3 x}{6} + \frac{5 c o s 5 x}{10} = \frac{1}{2} cos 3 x + \frac{1}{2} cos 5 x = cos 4 x cos x$ . ✓ Back to the integrand.

Worked example Example 8 — Beats: the word problem (Cell C8)

Two tuning forks play frequencies $f_{1} = 256 Hz$ and $f_{2} = 260 Hz$ (Hz = cycles per second). The pressure signals are $cos (2 π f_{1} t)$ and $cos (2 π f_{2} t)$ , where $t$ is time in seconds. Their sum can be written as one slow modulating factor times one fast tone. Find the rate of loud–soft pulses you actually hear.

Forecast: How many loud–soft pulses per second will you hear — $2$ , $4$ , or $516$ ?

The figure below plots (top) the two individual tones in blue and yellow, and (bottom) their green sum riding inside a red envelope. Look at the red curve: it swells and pinches slowly, and each pinch is a moment of silence. The text steps explain where that red envelope comes from.

Figure: two pure tones (top) add to a fast wave trapped inside a slow red envelope (bottom). Loud where the envelope is fat, soft where it pinches to zero.

Step 1. We need the identity that turns a sum of two cosines into a product — the exact reverse of the rules at the top of this page. Reversing $cos A cos B = \frac{1}{2} [cos (A - B) + cos (A + B)]$ : put $P = A + B$ and $Q = A - B$ so that $A = \frac{P + Q}{2}$ and $B = \frac{P - Q}{2}$ ; then the rule reads $cos P + cos Q = 2 cos \frac{P + Q}{2} cos \frac{P - Q}{2} .$ Apply it with $P = 2 π f_{1} t$ and $Q = 2 π f_{2} t$ . Why this step? We derive the sum-to-product form here rather than borrow it, so nothing on the page is used before it is built. (The general version lives at Sum-to-Product Formulas.)

Step 2. Half-sum and half-difference of the two angles use $\frac{f _{1} + f _{2}}{2} = \frac{256 + 260}{2} = 258$ and $\frac{f _{1} - f _{2}}{2} = \frac{256 - 260}{2} = - 2$ . Using $cos$ even ( $cos (- α) = cos α$ ): $cos (2 π \cdot 256 t) + cos (2 π \cdot 260 t) = slow modulator 2 cos (2 π \cdot 2 t) fast tone, 258 Hz cos (2 π \cdot 258 t) .$ Why this step? This splits the sound into a fast pitch at $258 Hz$ (what note you hear) multiplied by a slowly-varying factor $2 cos (2 π \cdot 2 t)$ — the modulator oscillating at $2 Hz$ . That $2 Hz$ is the frequency of the signed red curve, not yet the beat you hear.

Step 3. Your ear hears loudness, which depends on the amplitude $∣2 cos (2 π \cdot 2 t) ∣$ — the absolute value, because a big negative swing is just as loud as a big positive one. Taking $∣ \cdot ∣$ folds every trough up into a peak, so a modulator of frequency $2 Hz$ produces two amplitude humps per its own cycle. Concretely: $cos (2 π \cdot 2 t)$ reaches magnitude $1$ at $t = 0, \frac{1}{4}, \frac{1}{2}, \dots$ seconds — that is $4$ loudness peaks each second. Hence beat rate $= 2 \times 2 Hz = 4 Hz = ∣ f_{1} - f_{2} ∣$ . Why this step? This is the crucial subtlety: the modulating cosine runs at $\frac{∣ f _{1} - f _{2} ∣}{2} = 2 Hz$ , but the audible beat (loud–soft cycle) runs at twice that, $∣ f_{1} - f_{2} ∣ = 4 Hz$ , because $∣ cos ∣$ has two humps per period. See Wave Interference and Beats.

Verify: beat frequency $= ∣ f_{1} - f_{2} ∣ = ∣256 - 260∣ = 4 Hz$ , and the modulator frequency $= \frac{∣ f _{1} - f _{2} ∣}{2} = 2 Hz$ . Units: Hz $-$ Hz $=$ Hz. You hear $4$ pulses each second. ✓

Worked example Example 9 — Exam twist:

$cos A sin B$ and an orthogonality integral (Cell C9) Show $\int_{0}^{2 π} cos 3 x sin 5 x d x = 0$ , using the fourth rule $cos A sin B$ .

Forecast: This underlies Fourier Series — different-frequency cosine and sine are "orthogonal." Guess the answer before computing: is it $0$ , $π$ , or $2 π$ ?

Radian warning first. As in Example 7, the whole-period integral formula $\int_{0}^{2 π} sin k x d x = 0$ relies on $x$ being in radians, so that $2 π$ really is one full turn. We use radians throughout.

Step 1. This is the other mixed form — cosine times sine — so use $cos A sin B = \frac{1}{2} [sin (A + B) - sin (A - B)]$ , with $A = 3 x$ , $B = 5 x$ : $cos 3 x sin 5 x = \frac{1}{2} [sin 8 x - sin (- 2 x)] = \frac{1}{2} [sin 8 x + sin 2 x] .$ Why this step? Notice the rule has a minus in front of $sin (A - B)$ (unlike $sin A cos B$ 's plus). Then $sin$ is odd, so $sin (- 2 x) = - sin 2 x$ , and the two minus signs make a plus — this is the sign subtlety of the $cos sin$ identity.

Step 2. Integrate over a full period. For any nonzero integer $k$ , $\int_{0}^{2 π} sin k x d x = [- \frac{c o s k x}{k}]_{0}^{2 π} = 0$ because $cos$ returns to its start: $\int_{0}^{2 π} cos 3 x sin 5 x d x = \frac{1}{2} 0 \int_{0}^{2 π} sin 8 x d x + 0 \int_{0}^{2 π} sin 2 x d x = 0.$ Why this step? Each whole-period sine integral cancels its own positive and negative halves — the geometric heart of Fourier orthogonality.

Verify: $\int_{0}^{2 π} sin 8 x d x = 0$ and $\int_{0}^{2 π} sin 2 x d x = 0$ , so the total is $0$ . ✓ Forecast confirmed.

Recall Case-coverage check

Did we hit every cell? ::: Yes — C1 $cos cos$ (+), C2 $sin sin$ (−), C3 mixed $sin cos$ , C4 $A = B$ degenerate, C5 collapse-to-zero, C6 negative/large angle, C7 integral, C8 beats word problem, C9 orthogonality twist using $cos sin$ . In which case does the sign flip if you swap $A$ and $B$ ? ::: The mixed forms $sin A cos B$ and $cos A sin B$ (because $sin$ is odd); never for $cos cos$ (since $cos$ is even). Which output term vanishes in Example 5? ::: The $cos (A + B) = cos 9 0^{\circ} = 0$ term.

Connections

Parent: Product-to-Sum Formulas — the four rules used on every example above.
Angle Addition Formulas — where the rules come from.
Sum-to-Product Formulas — the inverse, used in the beats example.
Integration of Trig Functions — Example 7's whole reason to exist.
Wave Interference and Beats — Example 8's physics.
Fourier Series — Example 9's orthogonality.
Double Angle Formulas — Example 4's degenerate case.

Worked examples — Product-to-sum formulas

The scenario matrix

Worked examples

Connections

Concept Map