Conditional Trigonometric Identities
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Direct answer
Add the condition A + B + C = π and a batch of identities switches on that are false without it: tan A + tan B + tan C = tan A tan B tan C and cos²A + cos²B + cos²C + 2 cos A cos B cos C = 1. Every proof is one substitution — C = π − (A + B), making sin C = sin(A + B) and cos C = −cos(A + B) — then compound formulas. The π/2-set has its own list, headed by tan A tan B + tan B tan C + tan C tan A = 1; using the wrong set for the wrong sum is the chapter's central trap.
What you must remember
- Tan identity: for A + B + C = π, no angle equal to π/2: tan A + tan B + tan C = tan A tan B tan C.
- Cot dual: for the same condition, cot A cot B + cot B cot C + cot C cot A = 1 — valid even when one angle is 90°.
- Double-angle sums: sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C; cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C.
- Cos-squared identity: cos²A + cos²B + cos²C + 2 cos A cos B cos C = 1 for triangle angles; it degenerates to cos²A + cos²B = 1 when C = π/2.
- Half-angle form: tan(A/2)tan(B/2) + tan(B/2)tan(C/2) + tan(C/2)tan(A/2) = 1, safe for obtuse triangles since every half-angle lies in (0, π/2).
- The π/2 set: when A + B + C = π/2, the flagship is tan A tan B + tan B tan C + tan C tan A = 1 — a different list from the π set.
- Proof engine: sin 2A + sin 2B = 2 sin C·cos(A − B) under the condition — most double-angle proofs start here.
Proving the cos-squared identity
Start from the square-sum identity cos²A + cos²B = 1 + cos(A + B)cos(A − B). Since A + B = π − C, cos(A + B) = −cos C, so the pair contributes 1 − cos C·cos(A − B), and the running total is 1 − cos C·cos(A − B) + cos²C. Now simplify what remains: −cos(A − B) + cos C = −cos(A − B) − cos(A + B), because cos C = −cos(A + B); expanding both products gives −(cos A cos B + sin A sin B) − (cos A cos B − sin A sin B) = −2 cos A cos B. Therefore cos²A + cos²B + cos²C = 1 − 2 cos A cos B cos C, which rearranges to the stated identity. Two moves carry the proof — the square-sum identity and one substitution — which is why Advanced can demand it in full.
Read the identity backwards for a bound: for an acute triangle, AM-GM gives cos A cos B cos C ≤ 1/8, so the squared-cosine sum is at least 1 − 2·(1/8) = 3/4, equality at the equilateral triangle.
Condition confusion
Main asks one-line evaluations: plug triangle angles into a listed identity. Advanced embeds the condition — angles from side ratios, or tan A + tan B + tan C = 6 with the product asked, the identity reading backwards to give 6. The traps: using the π-set when the angles sum to π/2 (each sum has its own list; the tan forms differ structurally); applying tan A + tan B + tan C = tan A tan B tan C when one angle is 90°, where tan is undefined but the cot identity still holds; and half-angle sign fears — the half-angle identity is safe because a triangle's half-angles never exceed 45°, keeping every tangent positive.
Frequently asked questions
Which identity links tan A, tan B, tan C for triangle angles?
tan A + tan B + tan C = tan A tan B tan C, provided none of the angles is 90°.
What changes when A + B + C = π/2 instead of π?
A different set applies — tan A tan B + tan B tan C + tan C tan A = 1 replaces the sum-equals-product form.
Can I prove sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C quickly?
Pair sin 2A + sin 2B = 2 sin C·cos(A − B), add sin 2C = 2 sin C cos C, and finish with sum-to-product on the cosines.
What is the minimum of cos²A + cos²B + cos²C for a triangle?
3/4 at the equilateral triangle — from the identity plus cos A cos B cos C ≤ 1/8; obtuse cases push the sum higher.
How do the identities extend to half-angles?
With A/2 + B/2 + C/2 = π/2, the π/2-set gives tan(A/2)tan(B/2) + tan(B/2)tan(C/2) + tan(C/2)tan(A/2) = 1.