Product-to-sum identities · proof ·what it is and when to use it

Proof: sin A cos B = ½[sin(A + B) + sin(A − B)]

Product to sum: sin A cos B — proved by add the two sine expansions. Every line below says which rule it uses, so nothing has to be taken on faith.

The proof

Both expansions contain sin A cos B; adding keeps it and cancels the other term.

1. sin(A + B) = sin A cos B + cos A sin B
   Sine sum. · Sine of a sum
2. sin(A − B) = sin A cos B − cos A sin B
   Sine difference. · Sine of a difference
3. add: sin(A + B) + sin(A − B) = 2 sin A cos B
   The cos A sin B terms cancel.
4. sin A cos B = ½[sin(A + B) + sin(A − B)]
   Divide by 2.

Where the proof stops applying

Both sides are undefined at the same angles — where a denominator in the proof reaches zero (for instance cos x = 0 in a tangent or secant form). Elsewhere the argument above holds for every real angle.

The same claim, checked numerically

A proof is not the same thing as a check, and this page shows both: the reasoning above, and the first few angles fed to the same engine behind the verifier. If they ever disagreed, the data would be broken — the build would fail before publishing (accuracy policy).

Related

• How would I find sin A cos B = ½[sin(A + B) + sin(A − B)] myself? — the derivation, which is a different question from the proof.
• Product to sum: sin A cos B: when to use it — the practical side.
• All product-to-sum identities · proof index

sampled at 800 random anglesSource: OpenStax Precalculus, Ch. 7.2–7.3 · Paul's Online Math Notes · revised 2026-09-26 ·how we check ·accuracy policy ·report an error