Prove: sin⁡3x−sin⁡x=2sin⁡xcos⁡2x\sin 3x - \sin x = 2 \sin x \cos 2x

Mathematics

Problem

Prove: sin⁡3x−sin⁡x=2sin⁡xcos⁡2x\sin 3x - \sin x = 2 \sin x \cos 2x
Intuitive explanation

The key is recognizing that sin⁡3x\sin 3x can be rewritten using the triple-angle formula, or more elegantly, by factoring the left side as a difference of sines. Use the sum-to-product identity sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) to collapse the left side into a product form, then simplify to match the right side.

Intuitive explanation

The key is recognizing that sin⁡3x\sin 3x can be rewritten using the triple-angle formula, or more elegantly, by factoring the left side as a difference of sines. Use the sum-to-product identity sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) to collapse the left side into a product form, then simplify to match the right side.

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