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A JEE question gives you cos(π/7) cos(2π/7) cos(4π/7) and four options. Product-to-sum formulas will get there, slowly. When each angle is double the one before, there is a one-line route: multiply by sin A and let the double-angle formula fold the product up. This post derives the result, works five products (two with the doubling hidden), and lists the sign slips that cost marks.
Is this product in NCERT and the JEE syllabus?
The NCERT Class 11 chapter Trigonometric Functions (reprint 2026-27) proves sin 2x = 2 sin x cos x as a double-angle formula, but it does not work this product. The NTA Syllabus for JEE (Main) 2026 lists "Trigonometrical identities and trigonometrical functions" in Unit 14. So the trick is an application of a textbook identity, not a separate topic. Check the syllabus for your own exam year.
Why does multiplying by sin A collapse the product?
Let P = cos A cos 2A cos 4A. Multiply both sides by 2 sin A:
2 sin A · P = (2 sin A cos A) cos 2A cos 4A = sin 2A cos 2A cos 4A.
Multiply by 2 again: 4 sin A · P = (2 sin 2A cos 2A) cos 4A = sin 4A cos 4A.
Once more: 8 sin A · P = sin 8A.
So P = sin 8A / (8 sin A), as long as sin A ≠ 0. With n factors:
cos A cos 2A cos 4A ... cos 2^(n−1)A = sin(2ⁿA) / (2ⁿ sin A).
Each step turns a pair 2 sin x cos x into sin 2x, and the next cosine is already waiting with angle 2x. That is why the angles must double.
Worked examples
1. cos(π/7) cos(2π/7) cos(4π/7). Three factors, A = π/7, so P = sin(8π/7)/(8 sin(π/7)). Since sin(8π/7) = sin(π + π/7) = −sin(π/7), P = −1/8. Sign check: 4π/7 is more than π/2, so cos(4π/7) is negative and the other two are positive. A negative answer is right.
2. cos 12° cos 24° cos 48° cos 96°. Four factors: P = sin 192°/(16 sin 12°). Since sin 192° = −sin 12°, P = −1/16. Only cos 96° is negative, which fits.
3. cos(2π/15) cos(4π/15) cos(8π/15) cos(16π/15). P = sin(32π/15)/(16 sin(2π/15)). Because 32π/15 = 2π + 2π/15, the two sines are equal and P = 1/16. Two factors are negative (96° and 192°), so a positive product is right.
4. cos 36° cos 72°. Two factors: sin 144°/(4 sin 36°). Since sin 144° = sin 36°, the product is 1/4.
5. A hidden chain: cos(π/11) cos(2π/11) cos(3π/11) cos(4π/11) cos(5π/11). These angles go up in steps, not doublings. But cos(3π/11) = −cos(8π/11), and cos(5π/11) = −cos(6π/11) = −cos(16π/11). The two minus signs cancel, so the product equals cos(π/11) cos(2π/11) cos(4π/11) cos(8π/11) cos(16π/11), a doubling chain of five. That is sin(32π/11)/(32 sin(π/11)). Since 32π/11 = 2π + 10π/11 and sin(10π/11) = sin(π/11), the answer is 1/32.
We checked every value above numerically to more than ten decimal places.
Do you get the same answer by another route?
Our trigonometric identities practice plan finds cos 20° cos 40° cos 80° with the identity cos θ cos(60° − θ) cos(60° + θ) = (1/4) cos 3θ. The doubling trick gets there too: A = 20° with three factors gives sin 160°/(8 sin 20°) = 1/8. Same answer, different idea.
Comparing two methods on purpose is worth the time. In Rittle-Johnson and Star (2007), 70 seventh-grade students learning to solve equations either compared alternative solution methods side by side or studied the same methods one at a time. The comparison group made greater gains in procedural knowledge and flexibility, with similar gains in conceptual knowledge. Flexibility is what this topic tests: seeing that a product is a doubling chain in disguise.
Which mistakes cost marks?
Wrong power of 2. n factors give 2ⁿ in the denominator. Count the factors before writing it.
Using it when angles do not double. cos A cos 3A cos 9A triples, so the trick does not apply directly.
Dropping the sign. Count how many angles have a negative cosine before you trust the answer.
Dividing by zero. If sin A = 0, the formula is undefined; each cosine is then ±1, so multiply directly.
Missing a hidden chain. For angles kπ/(2m + 1), try rewriting some cosines as minus the cosine of a doubled angle, as in example 5.
A practice routine
Derive the general formula from a blank page once a day for three days.
Redo examples 1 to 4 without notes, and check each sign by counting negative cosines.
Try two new ones: cos(π/9) cos(2π/9) cos(4π/9) (answer 1/8) and cos(π/17) cos(2π/17) cos(4π/17) cos(8π/17) (answer 1/16).
Find cos(π/7) cos(2π/7) cos(3π/7) using the example 5 rewrite. (Answer: 1/8.)
Put three of these into a mixed set with sum-to-product and allied-angle questions, so you have to spot the doubling yourself.
If the double-angle formula itself feels shaky, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson, so you can revise the double-angle formula before coming back to products.
Tonight, prove the four-factor version on paper, then use it on cos 12° cos 24° cos 48° cos 96° without looking.
