Triangle identities · derivation ·proof
Where a² = b² + c² − 2bc·cos A comes from
solve the same equation for the angle instead of the side. The useful rearrangement is the one that answers 'what is this angle?' when all three sides are known — start from the proved form and isolate cos A.
Building it step by step
- a² = b² + c² − 2bc·cos AThe proved form. · Law of cosines
- 2bc·cos A = b² + c² − a²Move the cosine term to the left, a² to the right.
- cos A = (b² + c² − a²)/(2bc)Divide by 2bc — now the angle is the subject.
What this derivation depends on
Every line above is one of these — nothing else is assumed:
- a² = b² + c² − 2bc·cos A — Law of cosines
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.
Where this route is used
Two sides and the angle between them are known and the third side is wanted, or all three sides are known and an angle is wanted. It also classifies the triangle: the sign of cos A tells you whether A is acute, right or obtuse.
- solve
- evaluate
Checked on these triangles
Deriving a formula and testing it are different things; this table is the test. The numbers come from the same engine as the verifier, computed when the site was built.
| Triangle (A) | Left side | Right side | Agree |
|---|---|---|---|
| 7.5° | 2.62788 | 2.62788 | yes |
| 15° | 0.605613 | 0.605613 | yes |
| 18° | 1.32987 | 1.32987 | yes |
| 22.5° | 3.99025 | 3.99025 | yes |
| 30° | 10.266201 | 10.266201 | yes |
| 37° | 6.274259 | 6.274259 | yes |
Neighbouring derivations
- a/sin A = b/sin B — Law of sines
- sin²θ + cos²θ = 1 — Pythagorean identity
- a = b·cos C + c·cos B — Projection rule
How to recall it under pressure
Re-derive it from the first line rather than searching memory: start from a² = b² + c² − 2bc·cos A and do the one algebraic move the derivation above makes. If you need the statement itself, it is on the law of cosines page.