Maxima and Minima of Trigonometric Products
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Direct answer
Three numbers are worth carrying into the hall: 3√3/2, 3√3/8 and 1/8 — the maxima of sin A + sin B + sin C, of sin A sin B sin C, and of cos A cos B cos C when A, B, C are angles of a triangle. Symmetry forces each extreme to the equilateral case A = B = C = π/3, and the linking identity sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C converts product questions into sum questions that known bounds finish. Equality conditions are part of the answer, not an afterthought.
What you must remember
- Sum bounds: 1 < cos A + cos B + cos C ≤ 3/2 and sin A + sin B + sin C ≤ 3√3/2 for triangle angles — both peak at the equilateral triangle.
- Product bounds: sin A sin B sin C ≤ 3√3/8 and cos A cos B cos C ≤ 1/8, equality only at π/3, π/3, π/3; an obtuse angle makes the cosine product negative, so the 1/8 bound survives automatically.
- Linking identities: sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C and cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C — products become sums of known range.
- Pairwise squeeze: sin B sin C = [cos(B − C) + cos A]/2 ≤ (1 + cos A)/2, equality iff B = C — the balance-the-two lemma behind every asymmetric version.
- Square identity: sin²A + sin²B + sin²C = 2 + 2 cos A cos B cos C, giving 9/4 at the equilateral triangle.
- Sides bridge: a = 2R sin A turns the product into abc/(8R³), connecting angle bounds to side data.
- Unconstrained basics: |sin x cos x| ≤ 1/2 by the double angle, and for fixed x + y, sin x sin y is maximised at x = y by the same squeeze.
Maximising sin A sin B sin C
Fix A and balance the free angles: by the squeeze lemma, sin B sin C ≤ (1 + cos A)/2 with equality when B = C. So the product is at most sin A(1 + cos A)/2. Convert with half-angles: sin A = 2 sin(A/2)cos(A/2) and 1 + cos A = 2cos²(A/2), giving the bound 2 sin(A/2)·cos³(A/2). Let u = sin(A/2) ∈ (0, 1); the square of the bound is 4u²(1 − u²)³, and differentiating shows the maximum at u² = 1/4, i.e. A = π/3. The value is 2·(1/2)·(3/4)^(3/2) = 3√3/8, attained when B = C = π/3 as well. Every step is a standard tool, and the same skeleton handles "maximise the product given one angle" questions verbatim.
The cosine version follows from the same balance argument, giving 1/8 at the equilateral triangle; for an obtuse triangle the product is negative and approaches −1 as one angle nears π, so minimum questions have a different answer.
The acute-triangle catch
Main asks the three quotable maxima as single-correct or numerical answers; a bound without its equality condition loses the discriminating option. Advanced pushes to asymmetric products (one angle fixed — the squeeze lemma is exactly that tool), to sides-based data through abc/(8R³), and to true/false items on the 1 < cos A + cos B + cos C ≤ 3/2 range. The traps: claiming the cos-product minimum is −1/8 or 1/8 (the infimum is −1, approached in the degenerate limit); applying AM-GM to cos A cos B cos C without checking the angles are acute — for an obtuse triangle the cosines are not all positive and AM-GM is void; and forgetting that sin A sin B sin C has no attained minimum over genuine triangles, only an infimum.
Frequently asked questions
What is the maximum of sin A sin B sin C for a triangle?
3√3/8, attained only at A = B = C = π/3 — provable via the identity sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.
Why does cos A cos B cos C ≤ 1/8 hold for obtuse triangles too?
An obtuse angle makes the product negative, automatically below 1/8; the positive maximum still sits at the equilateral case.
What are the bounds of cos A + cos B + cos C?
1 < sum ≤ 3/2 for triangle angles — the upper end at the equilateral triangle, the lower approached as the triangle degenerates.
How do I maximise sin B sin C for a fixed A?
Use sin B sin C = [cos(B − C) + cos A]/2 ≤ (1 + cos A)/2, with equality when B = C — balance the two free angles.
How does the product connect to the triangle's sides?
Since a = 2R sin A, the product equals abc/(8R³), so angle-product bounds translate directly into side inequalities.