Sum-to-product identities · proof ·what it is and when to use it

Proof: sin A + sin B = 2 sin((A + B)/2) cos((A − B)/2)

Sum to product: sin A + sin B — proved by substitute u + v, u − v. Every line below says which rule it uses, so nothing has to be taken on faith.

The proof

Set u = (A + B)/2 and v = (A − B)/2, so A = u + v and B = u − v.

1. sin A + sin B = sin(u + v) + sin(u − v)
   Substitution.
2. = (sin u cos v + cos u sin v) + (sin u cos v − cos u sin v)
   Sine sum and difference. · Sine of a sum, Sine of a difference
3. = 2 sin u cos v
   The cos u sin v terms cancel.
4. = 2 sin((A + B)/2) cos((A − B)/2)
   Return to A and B.

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 + sin B = 2 sin((A + B)/2) cos((A − B)/2) myself? — the derivation, which is a different question from the proof.
• Sum to product: sin A + sin B: when to use it — the practical side.
• All sum-to-product 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