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Prove that the product sin(pi/n) sin(2pi/n)... sin((n-1)pi/n) equals n / 2^(n-1).

1. Product = n / 2^(n-1) (proved)
2. Product = n / 2ⁿ
3. Product = (n-1) / 2^(n-1)
4. Product = n² / 2^(n-1)

Correct answer: Product = n / 2^(n-1) (proved)

Solution

The nth roots of unity give product of (1 - omega^k) = n for k=1..n-1. Pairing |1 - e^(i theta)| = 2 sin(theta/2) yields the product of sines equal to n / 2^(n-1).

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