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

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

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

The proof

Identical to the sum case with the opposite operation.

1. sin A − sin B = sin(u + v) − sin(u − v)
   u = (A+B)/2, v = (A−B)/2.
2. = 2 cos u sin v
   Subtract the two sine expansions. · Sine of a sum, Sine of a difference
3. = 2 cos((A + B)/2) sin((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 cos((A + B)/2) sin((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