Question bank — Product-to-sum formulas

True. For a product of two first-power sines/cosines the identities always yield exactly two terms, one at angle $A-B$ and one at $A+B$. Higher powers like ${\sin }^{2}A\cos B$ break the rule and need repeated application.

True. Same-kind products (cos·cos, sin·sin) always output cosines; the only difference is the sign in front of the $\cos (A+B)$ term (plus for cos·cos, minus for sin·sin).

False. It is forced: adding or subtracting two addition formulas produces $2\times (\text{product})$, so dividing by 2 is required to isolate the product exactly — not a cosmetic choice.

False. Both output sines, but $\sin A\cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)]$ (a plus) while $\cos A\sin B=\frac{1}{2}[\sin (A+B)-\sin (A-B)]$ (a minus). Swapping the roles of the sine and cosine flips that inner sign.

True here, because $\cos$ is even: $\cos (A-B)=\cos (B-A)$, so swapping $A$ and $B$ leaves both terms unchanged. Be careful — this evenness rescue does not apply to the sine outputs.

False. $\sin (A-B)=-\sin (B-A)$ is odd, so swapping flips that term's sign and turns $\sin A\cos B$ into $\cos A\sin B$ — a genuinely different product. Order matters whenever a sine of a difference appears.

False — they are inverses. Product-to-sum turns a multiplication into an addition; Sum-to-Product Formulas does the reverse. You choose based on whether your starting expression is a product or a sum.

The sign is flipped: cos·cos uses a plus. This is the sin·sin formula in disguise. It comes from adding equations $(1)$ and $(2)$ above, which keeps the $+$; only subtracting them to isolate sines produces the minus.

They kept the formula's $\frac{1}{2}$ but forgot the leading $2$ cancels it. Correct: $2\sin 5x\cos 3x=\sin 8x+\sin 2x$, with no fraction left.

The two cosine terms are in the wrong order, giving an overall sign error. The difference angle always comes first with a plus: $\sin A\sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]$.

The application is correct but they should simplify $\cos 0=1$, giving ${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$ — a genuine double-angle identity, showing $A=B$ is a legal special case.

You cannot integrate a product by integrating each factor — that's not how integration works. You must first turn it into a sum $\frac{1}{2}[\cos 3x+\cos 5x]$, then integrate term by term.

Both the order and the sign are off. The right form is $\cos A\sin B=\frac{1}{2}[\sin (A+B)-\sin (A-B)]$ — $(A+B)$ first with a plus, $(A-B)$ subtracted.

Because we build them by combining $\cos (A\pm B)$ formulas (which only contain $\cos A\cos B$ and $\sin A\sin B$) versus $\sin (A\pm B)$ formulas (which contain the mixed $\sin A\cos B$ and $\cos A\sin B$). The output type is inherited from which pair of addition formulas we started with.

In $\cos (A\pm B)$ the $\cos A\cos B$ term keeps the same sign while $\sin A\sin B$ flips sign. Adding cancels the flipping term (sines) and keeps cosines; subtracting cancels cosines and keeps sines.

Because $\int (\text{product})$ has no term-by-term rule, but $\int (\text{sum of cosines})$ does — each cosine integrates independently. This is why Integration of Trig Functions leans on them constantly.

Both rely on how products and sums of sines/cosines interconvert. In Fourier analysis, integrating a product of two different-frequency waves gives zero (orthogonality) precisely because product-to-sum turns it into cosines that integrate to zero over a full period.

Setting $A=B$ collapses the difference angle to $0$, and $\cos 0=1$, $\sin 0=0$. This turns each product-to-sum identity into a double-angle formula — a legitimate limiting member of the same family, not an exception.

Take $\cos A\cos B$: adding $(1)+(2)$ gives $\cos (A+B)+\cos (A-B)=2\cos A\cos B$, because the $\sin A\sin B$ terms have opposite signs in $(1)$ and $(2)$ and cancel, while the $\cos A\cos B$ terms have matching signs and double. The surviving $\cos (A-B)$ therefore always carries a $+$. For sin·sin you instead compute $(2)-(1)$, which flips the sign of $\cos (A+B)$ but leaves $\cos (A-B)$ positive. So "difference angle with a plus" is a literal consequence of which terms cancel. (Note: for the sine outputs the terms simply appear in $(A+B),(A-B)$ order — check the restated formulas above rather than trusting a single mnemonic.)

With $B=0$: $\cos A\cos 0=\cos A$, and the formula gives $\frac{1}{2}[\cos A+\cos A]=\cos A$. Consistent — the identity gracefully reduces to a triviality.

Then $\cos (A+B)=\cos 90^{\circ }=0$, so $\sin A\sin B=\frac{1}{2}\cos (A-B)$ — one term vanishes. This is exactly why $\sin 75^{\circ }\sin 15^{\circ }$ simplifies so cleanly (their sum is $90^{\circ }$).

It gives $\frac{1}{2}[\sin 2A+\sin 0]=\frac{1}{2}\sin 2A$, so $\sin A\cos A=\frac{1}{2}\sin 2A$ — precisely the sine double-angle formula, recovered as a special case.

The difference term $\cos (\pi (f_1-f_2)t)\to \cos 0=1$, so the envelope stops wobbling and you hear one steady tone at frequency $f_1$ — the beat period stretches to infinity. See Wave Interference and Beats.

Yes, whenever one of the resulting angles makes its trig term vanish or the two terms coincide — e.g. $B=0$ merges both cosines into one, and $A+B=90^{\circ }$ kills a cosine term. These are the same edge cases above, viewed as term-collapse.