Exercises — Sum-to-product formulas

WHAT: identify the form. We have a difference of two cosines, so we use $\cos A-\cos B=-2\sin \frac{A+B}{2}\sin \frac{A-B}{2}.$ WHY: it is a subtraction of cosines — the only one of the four with a leading minus sign. Here $A=40^{\circ }$, $B=10^{\circ }$:

• half-sum $\frac{40+10}{2}=25^{\circ }$,
• half-difference $\frac{40-10}{2}=15^{\circ }$. So $\cos 40^{\circ }-\cos 10^{\circ }=-2\sin 25^{\circ }\sin 15^{\circ }.$

First, a precise convention. Every product form is written first factor $\times$ second factor, and in all four formulas the first factor always takes the half-sum $\frac{A+B}{2}$ and the second factor always takes the half-difference $\frac{A-B}{2}$. What changes between formulas is only which trig function (sin or cos) sits in each factor, plus the leading sign. Reading straight off the boxed formulas at the top of the page:

• (a) $\sin A+\sin B=2\underset{\text{first: sin, half-sum}}{\underbrace{\sin \frac{A+B}{2}}}\underset{\text{second: cos, half-diff}}{\underbrace{\cos \frac{A-B}{2}}}$ → first factor sin, second factor cos.
• (b) $\sin A-\sin B=2\underset{\text{first: cos, half-sum}}{\underbrace{\cos \frac{A+B}{2}}}\underset{\text{second: sin, half-diff}}{\underbrace{\sin \frac{A-B}{2}}}$ → first factor cos, second factor sin (the two functions swap compared to (a), but the half-sum still lives in the first factor).
• (c) $\cos A+\cos B=2\underset{\text{first: cos, half-sum}}{\underbrace{\cos \frac{A+B}{2}}}\underset{\text{second: cos, half-diff}}{\underbrace{\cos \frac{A-B}{2}}}$ → both factors cos.
• (d) $\cos A-\cos B=-2\underset{\text{first: sin, half-sum}}{\underbrace{\sin \frac{A+B}{2}}}\underset{\text{second: sin, half-diff}}{\underbrace{\sin \frac{A-B}{2}}}$ → both factors sin, leading minus. WHY the swap in (b): subtracting the addition formulas kills the $\sin \cos$ term and leaves $2\cos x\sin y$; after the substitution $x=\frac{A+B}{2},y=\frac{A-B}{2}$, that cosine lands in the first factor (with the half-sum) and the sine in the second (with the half-difference).

WHAT: sine sum $\Rightarrow 2\sin \frac{A+B}{2}\cos \frac{A-B}{2}$. $A=75^{\circ },B=15^{\circ }$: half-sum $=45^{\circ }$, half-difference $=30^{\circ }$. $\sin 75^{\circ }+\sin 15^{\circ }=2\sin 45^{\circ }\cos 30^{\circ }=2\cdot \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{6}}{2}.$ WHY these angles: $45^{\circ }$ and $30^{\circ }$ are known-exact, so the product needs no calculator. (Notice it equals $\cos 75^{\circ }+\cos 15^{\circ }$ from the parent note — coincidence of the special angles, not a general rule.)

WHAT: cosine sum $\Rightarrow 2\cos \frac{A+B}{2}\cos \frac{A-B}{2}$. $A=3x,B=7x$: half-sum $=5x$, half-difference $=\frac{3x-7x}{2}=-2x$. $\cos 3x+\cos 7x=2\cos 5x\cos (-2x)=2\cos 5x\cos 2x.$ WHY $\cos (-2x)=\cos 2x$: cosine is an even function — it doesn't care about the sign of its input, so a negative half-difference is harmless here. (Keep this fact; it protects you when $B>A$.)

Top: $\sin 5x+\sin x=2\sin \frac{6x}{2}\cos \frac{4x}{2}=2\sin 3x\cos 2x.$ Bottom: $\cos 5x+\cos x=2\cos 3x\cos 2x.$ The $2$ and $\cos 2x$ cancel: $\frac{2\sin 3x\cos 2x}{2\cos 3x\cos 2x}=\frac{\sin 3x}{\cos 3x}=\tan 3x.$ WHY it works: both numerator and denominator hide a common factor $2\cos 2x$; factoring makes it visible — exactly the Product-to-Sum Formulas idea run in reverse. This matches the parent note's rule $\frac{\sin A+\sin B}{\cos A+\cos B}=\tan \frac{A+B}{2}$ with $A=5x,B=x$, giving $\tan 3x$.

WHAT: factor with cosine sum. $A=3x,B=x$: half-sum $=2x$, half-difference $=x$. $\cos 3x+\cos x=2\cos 2x\cos x=0.$ WHY factoring: a product $=0$ splits into "one factor is zero." So $\cos 2x=0$ or $\cos x=0$.

• $\cos 2x=0\Rightarrow 2x=\frac{\pi }{2}+n\pi \Rightarrow x=\frac{\pi }{4}+\frac{n\pi }{2}$.
• $\cos x=0\Rightarrow x=\frac{\pi }{2}+n\pi$. Cover all cases: is the second set already inside the first? Put $n$ odd in the first set: $x=\frac{\pi }{4}+\frac{\pi }{2}=\frac{3\pi }{4}$, $x=\frac{5\pi }{4}$, … these are not $\frac{\pi }{2}+k\pi$. Test $x=\frac{\pi }{2}$: first set gives $\frac{\pi }{4}+\frac{n\pi }{2}=\frac{\pi }{2}\Rightarrow n=\frac{1}{2}$, not an integer. So the second set is genuinely extra. Answer: $x=\frac{\pi }{4}+\frac{n\pi }{2}$ or $x=\frac{\pi }{2}+n\pi ,n\in \mathbb{Z}$. See Solving Trigonometric Equations.

WHAT: sine difference $\Rightarrow 2\cos \frac{A+B}{2}\sin \frac{A-B}{2}$. $A=4x,B=2x$: half-sum $=3x$, half-difference $=x$. $\sin 4x-\sin 2x=2\cos 3x\sin x=0.$ WHY split like this: a product equals zero exactly when one of its factors is zero (the leading $2$ is never zero). So $\cos 3x=0$ or $\sin x=0$ — we solve each piece separately and take the union. (This factored form $2\cos 3x\sin x$ is exactly the "envelope pinches to zero" picture from the opening figure: the equation is zero precisely at those green "soft" points.)

• $\sin x=0$ on $[0,2\pi )$: $x=0,\pi$.
• $\cos 3x=0\Rightarrow 3x=\frac{\pi }{2}+n\pi \Rightarrow x=\frac{\pi }{6}+\frac{n\pi }{3}$. In $[0,2\pi )$, list $n=0,\dots ,5$: $x=\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\frac{7\pi }{6},\frac{3\pi }{2},\frac{11\pi }{6}.$ All cases covered: none of these coincide with $0$ or $\pi$, so we keep all $8$ solutions: $x\in \left\{0,\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\pi ,\frac{7\pi }{6},\frac{3\pi }{2},\frac{11\pi }{6}\right\}.$

WHAT: cosine difference (the minus-sign one!): $\cos 5x-\cos x=-2\sin 3x\sin 2x=0.$ The leading $-2$ never equals zero, so $\sin 3x=0$ or $\sin 2x=0$.

• $\sin 3x=0\Rightarrow 3x=n\pi \Rightarrow x=\frac{n\pi }{3}$. In $[0,\pi ]$: $x=0,\frac{\pi }{3},\frac{2\pi }{3},\pi$.
• $\sin 2x=0\Rightarrow x=\frac{n\pi }{2}$. In $[0,\pi ]$: $x=0,\frac{\pi }{2},\pi$. Merge (remove duplicates $0,\pi$): $x\in \left\{0,\frac{\pi }{3},\frac{\pi }{2},\frac{2\pi }{3},\pi \right\}.$

Factor the numerator with the sine-difference formula: $\sin A-\sin B=2\cos \frac{A+B}{2}\sin \frac{A-B}{2}.$ Factor the denominator. It is written $\cos B-\cos A$, i.e. the cosine-difference formula with the roles of $A$ and $B$ swapped: $\cos B-\cos A=-2\sin \frac{B+A}{2}\sin \frac{B-A}{2}.$ Since $\frac{B+A}{2}=\frac{A+B}{2}$ and $\sin \frac{B-A}{2}=-\sin \frac{A-B}{2}$ (sine is odd), the two minus signs combine: $\cos B-\cos A=-2\sin \frac{A+B}{2}(-\sin \frac{A-B}{2})=2\sin \frac{A+B}{2}\sin \frac{A-B}{2}.$ Now form the ratio: $\frac{\sin A-\sin B}{\cos B-\cos A}=\frac{2\cos \frac{A+B}{2}\sin \frac{A-B}{2}}{2\sin \frac{A+B}{2}\sin \frac{A-B}{2}}.$ Cancellation — and the domain it requires. We cancel the common factors $2$ and $\sin \frac{A-B}{2}$. Cancelling $\sin \frac{A-B}{2}$ is only legitimate when it is nonzero, i.e. when $\frac{A-B}{2}\ne k\pi$, equivalently $A-B\ne 2k\pi$. Likewise the final $\cot$ requires its denominator $\sin \frac{A+B}{2}\ne 0$, i.e. $A+B\ne 2m\pi$ (otherwise the original left side already has a zero denominator). Under those two conditions: $\frac{2\cos \frac{A+B}{2}\sin \frac{A-B}{2}}{2\sin \frac{A+B}{2}\sin \frac{A-B}{2}}=\frac{\cos \frac{A+B}{2}}{\sin \frac{A+B}{2}}=\cot \frac{A+B}{2}.$ Wait — read carefully. The surviving ratio is $\cot \frac{A+B}{2}$, not $\cot \frac{A-B}{2}$. So the proposed identity as stated is false; the correct simplification of $\frac{\sin A-\sin B}{\cos B-\cos A}$ is $\boxed{\frac{\sin A-\sin B}{\cos B-\cos A}=\cot \frac{A+B}{2}}(A-B\ne 2k\pi ,A+B\ne 2m\pi ).$ Numerical check ($A=80^{\circ },B=20^{\circ }$): LHS $=\frac{\sin 80^{\circ }-\sin 20^{\circ }}{\cos 20^{\circ }-\cos 80^{\circ }}=\frac{0.9848-0.3420}{0.9397-0.1736}=\frac{0.6428}{0.7660}=0.8391$; and $\cot \frac{A+B}{2}=\cot 50^{\circ }=0.8391$ ✓, whereas $\cot \frac{A-B}{2}=\cot 30^{\circ }=1.732$ ✗. So the answer is the half-sum version — a reminder to always verify the surviving argument, not assume it.

WHAT: pair the outer two terms so their half-sum lands on the middle angle $3x$. $\sin x+\sin 5x=2\sin \frac{x+5x}{2}\cos \frac{x-5x}{2}=2\sin 3x\cos (-2x)=2\sin 3x\cos 2x.$ WHY pair 1st and 3rd: their half-sum is exactly $3x$, matching the leftover middle term — that lets us factor $\sin 3x$. Now add back the middle term $\sin 3x$: $\sin x+\sin 3x+\sin 5x=2\sin 3x\cos 2x+\sin 3x=\sin 3x(2\cos 2x+1).$ ∎ (Used $\cos (-2x)=\cos 2x$ again — cosine is even.)

Cosine case (this is where a real double-angle identity appears). Take $\cos A+\cos B=2\cos \frac{A+B}{2}\cos \frac{A-B}{2}.$ Put $B=A$: half-sum $=A$, half-difference $=0$, so $\cos A+\cos A=2\cos A\cos 0=2{\cos }^{2}A.$ The left side is $2\cos A$? No — be careful: $\cos A+\cos A=2\cos A$, but the right side gives $2{\cos }^{2}A$. These can only agree if we do not naively set $B=A$ in a value sense; instead we treat it as a limit statement about the structure. The clean, non-degenerate way to expose the double-angle identity is to instead pick $A=2\theta ,B=0$: $\cos 2\theta +\cos 0=2\cos \frac{2\theta +0}{2}\cos \frac{2\theta -0}{2}=2\cos \theta \cos \theta =2{\cos }^2\theta .$ Since $\cos 0=1$, the left side is $\cos 2\theta +1$, giving $\cos 2\theta +1=2{\cos }^2\theta ⟹\boxed{\cos 2\theta =2{\cos }^2\theta -1},$ which is exactly the cosine double-angle identity from Double Angle Formulas. Sine case — recover $\sin 2\theta$. Use the sine sum with $A=2\theta ,B=0$? That gives $\sin 2\theta +\sin 0=2\sin \theta \cos \theta$, and since $\sin 0=0$ the left side is just $\sin 2\theta$: $\sin 2\theta =2\sin \theta \cos \theta ,$ the sine double-angle identity. WHY this is the honest technique: choosing $B=0$ (rather than $B=A$) makes the half-difference $\theta$ carry real information, so the collapse produces the named double-angle formula the problem promised — not just the trivial $2\cos A$. This is the precise sense in which double-angle formulas are the "$B\to 0$ / equal-angle" special case of sum-to-product.

Apply sine sum with $A=2\pi f_{1}t,B=2\pi f_{2}t$: $\sin A+\sin B=2\underset{\text{slow envelope}}{\underbrace{\cos \left(2\pi \frac{f_1-f_2}{2}t\right)}}\underset{\text{fast carrier}}{\underbrace{\sin \left(2\pi \frac{f_1+f_2}{2}t\right)}}.$

• (a) Carrier $=\frac{f_1+f_2}{2}=\frac{440+436}{2}=438\text{ Hz}$ — the pitch you hear.
• (b) Envelope $=\frac{|f_1-f_2|}{2}=\frac{4}{2}=2\text{ Hz}$ — the cosine's own frequency.
• (c) Beat $=|f_1-f_2|=4\text{ Hz}$. WHY twice the envelope: loudness depends on the envelope's magnitude $|\cos (\dots )|$, and $|\cos |$ peaks twice per full cosine cycle — so the ear counts $4$ loud swells per second, not $2$. See Beats and Superposition (Waves).

Read the figure below — it is the opening figure redrawn with these real numbers: the thin lavender curve is the $438\text{ Hz}$ carrier (too fast to resolve individual wiggles), the coral curves are the $2\text{ Hz}$ envelope, and the green arrows mark the $4$ loudness swells per second — that is the beat you actually hear.

In the figure, count the coral envelope's tall bulges: there are 4 within one second (the green arrows), confirming beat $=4\text{ Hz}$ while the envelope curve itself completes only $2$ full cosine cycles.

WHAT / WHY the tool: the numerator is a difference of cosines, so sum-to-product turns it into a product with a $\sin \frac{A-B}{2}$ factor that we can pair with $h$ using the small-angle limit ${\lim }_{u\to 0}\frac{\sin u}{u}=1$. $A=x+h,B=x$: half-sum $=x+\frac{h}{2}$, half-difference $=\frac{h}{2}$. $\cos (x+h)-\cos x=-2\sin (x+\frac{h}{2})\sin \frac{h}{2}.$ Divide by $h$ and split the $\sin \frac{h}{2}$: $\frac{-2\sin (x+\frac{h}{2})\sin \frac{h}{2}}{h}=-\sin (x+\frac{h}{2})\cdot \frac{\sin \frac{h}{2}}{\frac{h}{2}}.$ As $h\to 0$: $\frac{h}{2}\to 0$ so $\frac{\sin \frac{h}{2}}{\frac{h}{2}}\to 1$, and $\sin (x+\frac{h}{2})\to \sin x$. Therefore ${\lim }_{h\to 0}\frac{\cos (x+h)-\cos x}{h}=-\sin x.$ This is the derivative of $\cos$ — sum-to-product is the classic route to it.

$A=\theta ,B=180^{\circ }-\theta$. Half-sum $=\frac{\theta +180^{\circ }-\theta }{2}=90^{\circ }$; half-difference $=\frac{\theta -(180^{\circ }-\theta )}{2}=\theta -90^{\circ }$. $\cos \theta -\cos (180^{\circ }-\theta )=-2\sin 90^{\circ }\sin (\theta -90^{\circ })=-2\sin (\theta -90^{\circ }).$ Now $\sin (\theta -90^{\circ })=-\cos \theta$ (co-function shift), so $-2\sin (\theta -90^{\circ })=-2(-\cos \theta )=2\cos \theta .$ Sanity across quadrants: directly, $\cos (180^{\circ }-\theta )=-\cos \theta$, so $\cos \theta -(-\cos \theta )=2\cos \theta$ ✓ — matches for every $\theta$. Largest value: $2\cos \theta$ is maximised at $\theta =0^{\circ }$, giving $2$.

• L1 ::: Read the operator first; the half-sum always lives in the first factor, half-difference in the second.
• L2 ::: Compute half-sum & half-difference; exploit even $\cos$ / odd $\sin$.
• L3 ::: Product $=0$ → set each factor to zero; never divide out a factor.
• L4 ::: Pair terms so a shared factor appears; after cancelling, re-read which argument survives.
• L5 ::: Sum-to-product powers beats, derivatives, and clean sign checks.