Use the product-to-sum and sum-to-product identities to simplify trigonometric expressions and integrals

What this dot point is asking

NESA wants you to convert products like $\sin A \cos B$ into sums (and vice versa) using the standard product-to-sum and sum-to-product identities. The reason this matters is mechanical but important: a product of two trig functions has no antiderivative as it stands, but a sum of single sinusoids integrates term by term, and a sum equals zero is hard to solve, but a product equals zero splits into easy pieces. So the identities are the bridge that turns an intractable integral into an easy one, and an awkward equation into a factored one. Knowing which direction to travel is the real skill.

The answer

The four product-to-sum identities

These come from adding or subtracting the sum and difference identities:

A useful summary: a $\sin\cos$ product gives sines of the sum and difference; the two $\cos\cos$/$\sin\sin$ products give cosines, with $\cos\cos$ keeping the plus and $\sin\sin$ flipping to a minus (difference of cosines).

Where they come from

Add $\sin(A + B) = \sin A \cos B + \cos A \sin B$ and $\sin(A - B) = \sin A \cos B - \cos A \sin B$. The $\cos A \sin B$ terms cancel, leaving $\sin(A + B) + \sin(A - B) = 2\sin A\cos B$; divide by $2$ to get the first identity. Subtracting the same two isolates $2\cos A\sin B$. Adding the two cosine identities gives $2\cos A\cos B$; subtracting gives $-2\sin A\sin B$. Knowing this derivation means you never have to trust a half-remembered sign: you can rebuild any of the four in a few lines.

Sum-to-product (the converses)

Let $A + B = P$ and $A - B = Q$, so $A = \frac{P + Q}{2}$ and $B = \frac{P - Q}{2}$. Substituting turns each product-to-sum identity around:

Every product-to-sum identity has a $\tfrac{1}{2}$ out front; every sum-to-product identity has a $2$.

What sum-to-product looks like: superposition

The sum-to-product identities have a vivid meaning. Add two sine waves of nearby frequencies, say $\sin 5x + \sin 3x$.

Their sum is a single, more complicated-looking wave.

But the identity $\sin 5x + \sin 3x = 2\sin 4x\cos x$ tells you that this wave is really a fast carrier $\sin 4x$ (the average frequency) riding inside a slow envelope $\pm 2\cos x$ (set by the half-difference frequency). The wave hugs the envelope at its crests and pinches to zero where the envelope crosses zero, the "beats" you hear when two close musical notes sound together.

You are not examined on beats, but seeing the product structure makes the identity memorable: a sum of two waves is a product of a carrier and an envelope.

When to convert which way

The single most useful judgement in this dot point is the direction:

• Integration calls for product-to-sum. $\int \sin 3x\cos 5x\,dx$ has no antiderivative as written; rewrite it as $\tfrac12[\sin 8x + \sin(-2x)] = \tfrac12[\sin 8x - \sin 2x]$, which integrates immediately.
• Equation-solving calls for sum-to-product. $\sin x + \sin 3x = 0$ is awkward as a sum, but becomes $2\sin 2x\cos x = 0$, which is zero exactly when $\sin 2x = 0$ or $\cos x = 0$, two easy equations.
• Simplification can go either way; convert to whichever form reveals structure (a factor, a cancellation, a known value).

How exam questions ask about these identities

• "Express $\sin A\cos B$ (a product) as a sum": name the matching product-to-sum identity and substitute $A$, $B$.
• "By first converting to a sum, evaluate $\int (\text{product})\,dx$": product-to-sum, then integrate each single sinusoid with $\int\sin kx\,dx = -\frac1k\cos kx + C$.
• "Simplify $\sin P \pm \sin Q$ / $\cos P \pm \cos Q$": the sum-to-product converse with $P$, $Q$.
• "Solve $\sin mx \pm \sin nx = 0$ (or $= \cos$ something)": sum-to-product to factor, then solve each factor with the general solution.
• "Prove that $\frac{\sin A + \sin B}{\cos A + \cos B} = \tan\frac{A+B}{2}$" and similar: apply sum-to-product to top and bottom and cancel.

Edge cases worth knowing

• $A - B$ can be negative; that is fine. $\sin A\cos B$ uses $A - B$ even when $A < B$. Then $\sin(A - B) = -\sin(B - A)$ tidies the sign cleanly, so do not swap $A$ and $B$ mid-identity.
• Equal arguments recover a double-angle. Setting $A = B$ in $\sin A\cos B$ gives $\sin A\cos A = \tfrac12\sin 2A$, the double-angle identity. The product-to-sum family contains the double-angle formulas as a special case.
• A product equal to zero needs every factor. After sum-to-product, both $\sin\frac{P+Q}{2}=0$ and $\cos\frac{P-Q}{2}=0$ (or the relevant pair) must be solved; the union, not the intersection, is the answer.