Product-to-sum identities · derivation ·proof
Where cos A cos B = ½[cos(A − B) + cos(A + B)] comes from
Specialise to get the power-reducing form. Put B = A: cos(A − A) = cos 0 = 1 and cos(A + A) = cos 2A.
cos A cos B = ½[cos(A − B) + cos(A + B)]
Building it step by step
- cos A cos A = ½[cos 0 + cos 2A]Set B = A. · Product to sum: cos A cos B
- cos 0 = 1Unit circle.
- cos²A = ½(1 + cos 2A)Simplify — the power-reducing formula. · Power-reducing formula for cos²
What this derivation depends on
Every line above is one of these — nothing else is assumed:
- cos A cos B = ½[cos(A − B) + cos(A + B)] — Product to sum: cos A cos B
- cos²x = (1 + cos 2x)/2 — Power-reducing formula for cos²
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 = ½[cos(A − B) + cos(A + B)] and do the one algebraic move the derivation above makes. If you need the statement itself, it is on the product to sum: cos a cos b page.