Sum-to-product identities · derivation ·proof
Where cos A + cos B = 2 cos((A + B)/2) cos((A − B)/2) comes from
Both cosines, both stay cosines. Mnemonic: with a sum of cosines, nothing changes function; with a difference of cosines, everything does.
cos A + cos B = 2 cos((A + B)/2) cos((A − B)/2)
Building it step by step
- cos + cos → cos·cosPattern.
- cos − cos → sin·sin with a minusThe companion formula. · Sum to product: cos A − cos B
What this derivation depends on
Every line above is one of these — nothing else is assumed:
- cos A − cos B = −2 sin((A + B)/2) sin((A − B)/2) — Sum to product: cos A − cos B
That list is the argument for learning fewer formulas: most of this catalogue is a short chain away from the unit circle and the sum formulas.
How to recall it under pressure
Re-derive it from the first line rather than searching memory: start from cos A − cos B = −2 sin((A + B)/2) sin((A − B)/2) and do the one algebraic move the derivation above makes. If you need the statement itself, it is on the sum to product: cos a + cos b page.