Q.If $A + B + C = 2s$, then prove that $\sin(s - A)\sin(s - B) + \sin s \sin(s - C) = \sin A \sin B$.

The problem this concept solves. Trigonometric functions naturally arise as products of angle expressions in some settings (e.g. amplitude modulation, or three angles of a triangle multiplied together) and as sums in others (e.g. combining two waves). Converting cleanly between the two forms is one of the most-used trigonometric skills, and this concept covers every tool needed to do it.

1. Product-to-sum (from the addition formulas). Adding/subtracting the four expansions of $\sin(A\pm B)$ and $\cos(A\pm B)$ in pairs isolates a pure product on one side and a sum/difference on the other:

Use these whenever you are handed a product of two sines/cosines and need a sum.

2. Sum-to-product (the reverse substitution). Setting $C=A+B,\ D=A-B$ (so $A=\frac{C+D}2,\ B=\frac{C-D}2$) and substituting back into the four identities above inverts the process:

Use these whenever you are handed a sum or difference of two sines/cosines and need a product — which is usually the move that lets a numerator and denominator share a cancelling factor, or that shows an expression equals zero (a product is zero the moment one factor is).

3. The 60°±A triple-product family. Applying the product-to-sum idea twice in a row to three factors spaced $60^\circ$ apart gives three compact identities:

These are worth recognising on sight: any time three factors in a product are centred on some angle $A$ and spread $\pm60^\circ$ around it (e.g. $10^\circ,30^\circ$-adjacent-triples like $10^\circ,50^\circ,70^\circ$, or $12^\circ,48^\circ$-style pairs alongside a third term), one of these three identities collapses the triple product to a single term in $3A$ immediately.

4. Conditional identities for a triangle ($A+B+C=\pi$). When the three angles are constrained to sum to a fixed value — above all, the interior angles of a triangle — the sum-to-product identities become the engine for proving relations that are otherwise false. The recipe is always: eliminate one angle via the condition (e.g. $C=\pi-A-B$, so $\cos C=-\cos(A+B)$, $\sin C=\sin(A+B)$, or at the half-angle level $\sin\frac C2=\cos\frac{A+B}2$), apply a sum-to-product step, and repeat until a single compact product remains. This is exactly how the standard triangle identities are built:

together with the bound $1<\cos A+\cos B+\cos C\le\tfrac32$ that follows from the first identity plus the fact $\sin\frac A2\sin\frac B2\sin\frac C2\le\tfrac18$ (itself proved by treating that product as a quadratic in $\cos\frac{A-B}2$ and demanding a non-negative discriminant).

Worked illustration (mixing both directions). To show $\cos36^\circ\cos72^\circ\cos108^\circ\cos144^\circ=\tfrac1{16}$: rewrite $\cos108^\circ=\cos(90^\circ+18^\circ)=-\sin18^\circ$ and $\cos144^\circ=\cos(180^\circ-36^\circ)=-\cos36^\circ$, and $\cos72^\circ=\sin18^\circ$, so the product becomes $\cos36^\circ\cdot\sin18^\circ\cdot(-\sin18^\circ)\cdot(-\cos36^\circ)=\sin^218^\circ\cos^236^\circ$. Substituting the standard surd values $\sin18^\circ=\tfrac{\sqrt5-1}4$ and $\cos36^\circ=\tfrac{\sqrt5+1}4$ gives $\big(\tfrac{\sqrt5-1}4\big)^2\big(\tfrac{\sqrt5+1}4\big)^2=\big[\tfrac{(\sqrt5-1)(\sqrt5+1)}{16}\big]^2=\big[\tfrac{4}{16}\big]^2=\big[\tfrac14\big]^2=\tfrac1{16}$. No sum-to-product step was even needed here beyond angle relations — the point is that recognising which identity or angle relation applies (product-to-sum, sum-to-product, complementary/supplementary angle, or a known surd value) is the real skill this concept builds, well beyond memorising the eight boxed formulas.