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Prove that sin(theta) + sin(3*theta) + sin(5*theta) +... + sin((2n-1)*theta) = sin²(n*theta)/sin(theta). Which standard technique establishes this sum of an arithmetic-progression of sines?

1. Multiply both sides by 2*sin(theta) and telescope using product-to-sum formulae
2. Differentiate a geometric series term by term
3. Apply the binomial theorem to (cos(theta) + i*sin(theta))ⁿ
4. Use mathematical induction on cos rather than sin

Correct answer: Multiply both sides by 2*sin(theta) and telescope using product-to-sum formulae

Solution

Multiplying by 2 sin(theta) converts each product into a difference of cosines that telescopes to cos(0) - cos(2n*theta) = 2 sin²(n theta), giving the result after dividing back.

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