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Miscellaneous Exercise · Q5

Q.Prove that sin⁡x+sin⁡3x+sin⁡5x+sin⁡7x=4cos⁡x cos⁡2x sin⁡4x\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\, \cos 2x\, \sin 4x.

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Use the sum-to-product identity twice to combine the four sine terms into a single product. The left-hand side simplifies to 4cos⁡xcos⁡2xsin⁡4x4\cos x \cos 2x \sin 4x.

This is a classic trigonometric identity proof. The left side has four sine terms with arguments in arithmetic progression: x,3x,5x,7xx, 3x, 5x, 7x. When you see a sum of sines (or cosines) with equally spaced angles, the most natural tool is the sum-to-product formula:

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2 \sin\frac{A+B}{2} \cos\frac{A-B}{2}

The trick is to pair the terms cleverly. Pair the first with the last, and the middle two together — that way the averages become nice multiples of xx.

  1. Pair sin⁡x\sin x with sin⁡7x\sin 7x Here A=xA = x, B=7xB = 7x.

sin⁡x+sin⁡7x=2sin⁡x+7x2cos⁡x−7x2=2sin⁡4xcos⁡(−3x)\sin x + \sin 7x = 2 \sin\frac{x+7x}{2} \cos\frac{x-7x}{2} = 2 \sin 4x \cos(-3x)

Since cos⁡\cos is even, cos⁡(−3x)=cos⁡3x\cos(-3x) = \cos 3x. So this pair becomes:

sin⁡x+sin⁡7x=2sin⁡4xcos⁡3x\sin x + \sin 7x = 2 \sin 4x \cos 3x

  1. Pair sin⁡3x\sin 3x with sin⁡5x\sin 5x A=3xA = 3x, B=5xB = 5x:

sin⁡3x+sin⁡5x=2sin⁡3x+5x2cos⁡3x−5x2=2sin⁡4xcos⁡(−x)=2sin⁡4xcos⁡x\sin 3x + \sin 5x = 2 \sin\frac{3x+5x}{2} \cos\frac{3x-5x}{2} = 2 \sin 4x \cos(-x) = 2 \sin 4x \cos x

  1. Add the two results The whole left-hand side is now:

LHS=2sin⁡4xcos⁡3x+2sin⁡4xcos⁡xLHS = 2 \sin 4x \cos 3x + 2 \sin 4x \cos x

Factor out the common 2sin⁡4x2 \sin 4x:

LHS=2sin⁡4x (cos⁡3x+cos⁡x)LHS = 2 \sin 4x \, (\cos 3x + \cos x)

  1. Simplify cos⁡3x+cos⁡x\cos 3x + \cos x using sum-to-product again The formula for cosines is:

cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\cos C + \cos D = 2 \cos\frac{C+D}{2} \cos\frac{C-D}{2}

With C=3xC = 3x, D=xD = x: …

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