Product to Sum and Sum to Product Identities

Some problems are far easier once a product of trigonometric functions is rewritten as a sum or difference (this is especially useful later when integrating a product of sines/cosines), and other problems go the other way — a sum is easier to factor once it is rewritten as a product. Both directions come from the same four building blocks: the sine and cosine addition/subtraction formulas

From product to sum. Add or subtract these four in pairs and the mixed terms cancel, leaving a pure product on one side:

• Adding the two sine formulas: $\sin(A+B)+\sin(A-B)=2\sin A\cos B$, so

• Subtracting them: $\sin(A+B)-\sin(A-B)=2\cos A\sin B$, so

• Adding the two cosine formulas: $\cos(A+B)+\cos(A-B)=2\cos A\cos B$, so

• Subtracting them (cosine-difference minus cosine-sum): $\cos(A-B)-\cos(A+B)=2\sin A\sin B$, so

These four are the product-to-sum identities. Notice the pattern: whichever pair of functions you're multiplying, you get half the sum/difference of $\sin$ or $\cos$ of the added and subtracted angles — the only thing to get right is the sign in the middle and which combination ($A+B$ or $A-B$ first) matches the product you started with.

From sum to product — the reverse direction. To go back the other way, set $C=A+B$ and $D=A-B$. Solving these two simultaneously gives

Substituting these into the four product-to-sum identities above (and multiplying through by 2) turns every product on the right into a sum/difference $C,D$ on the left, giving the sum-to-product identities:

A quick way to keep the last one straight: $\cos C-\cos D$ is negative of $\cos D - \cos C$, and by symmetry of the third line $\cos D-\cos C=2\sin\frac{C+D}2\sin\frac{D-C}2$ —note the $\sin\frac{D-C}2$, not $\sin\frac{C-D}2$, is what keeps the sign consistent for $C>D$ or $C<D$ alike.

A special family — the 60°±A triple products. Three neat identities follow from the product-to-sum machinery:

The sine one is proved by first pairing the two 60±A factors with the product-to-sum rule for cosine ($\sin(60-A)\sin(60+A)$ can be re-read as $\sin X \sin Y$ with $X=60-A, Y=60+A$, giving $\tfrac12[\cos 2A-\cos120^\circ]$), then multiplying the remaining $\sin A$ back in and converting the resulting $\cos2A\sin A$ product into a sum again — the two intermediate product-to-sum passes eventually collapse everything to a single $\sin3A$ term. The cosine version follows by exactly the same two-step route with cosines throughout, and the tangent one follows by dividing the sine version by the cosine version and simplifying. These three are extremely useful shortcuts whenever a problem presents three factors spaced $60^\circ$ apart (or a triple like $10^\circ,30^\circ,50^\circ,70^\circ$ that can be regrouped that way).

How the textbook examples use all this. The worked examples in this section fall into three families:

• Turning a product into a sum directly (e.g. $\sin40^\circ\cos30^\circ$, $\cos110^\circ\sin55^\circ$, or a half-angle product like $\sin\frac{x}{2}\cos\frac{3x}{2}$): identify which of the four product-to-sum formulas matches the order of the two functions, plug in $A$ and $B$, and simplify the resulting angles.
• Turning a sum/difference into a product (e.g. $\sin50^\circ+\sin20^\circ$, $\cos6\theta+\cos2\theta$, $\cos\frac{3x}{2}-\cos\frac{9x}{2}$): identify $C$ and $D$, compute $\frac{C+D}{2}$ and $\frac{C-D}{2}$, and substitute into the matching sum-to-product formula. …