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Miscellaneous Exercise · Q7

Q.Prove that sin⁡3x+sin⁡2x−sin⁡x=4sin⁡x cos⁡x2 cos⁡3x2\sin 3x + \sin 2x - \sin x = 4\sin x\, \cos\frac{x}{2}\, \cos\frac{3x}{2}.

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Pairing sin⁡3x−sin⁡x\sin3x-\sin x via the sum-to-product formula, then factoring out sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x and combining cos⁡x+cos⁡2x\cos x+\cos2x (another sum-to-product step), reduces the left side exactly to 4sin⁡xcos⁡x2cos⁡3x24\sin x\cos\dfrac{x}{2}\cos\dfrac{3x}{2}.

Proof

Step 1 — Combine the first and third terms using sum-to-product.

Using sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2} with A=3x, B=xA=3x,\ B=x:

sin⁡3x−sin⁡x=2cos⁡2x sin⁡x\sin 3x - \sin x = 2\cos 2x\,\sin x

Step 2 — Rewrite the full left-hand side.

LHS=sin⁡3x+sin⁡2x−sin⁡x=sin⁡2x+2sin⁡xcos⁡2x\text{LHS} = \sin 3x+\sin 2x-\sin x = \sin 2x + 2\sin x\cos 2x

Step 3 — Expand sin⁡2x\sin2x and factor out 2sin⁡x2\sin x.

Since sin⁡2x=2sin⁡xcos⁡x\sin2x = 2\sin x\cos x:

sin⁡2x+2sin⁡xcos⁡2x=2sin⁡xcos⁡x+2sin⁡xcos⁡2x=2sin⁡x (cos⁡x+cos⁡2x)\sin2x+2\sin x\cos2x = 2\sin x\cos x + 2\sin x\cos2x = 2\sin x\,(\cos x+\cos2x)

Step 4 — Apply sum-to-product to the cosines. …

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