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Exercise 3.3 · Q14

Q.Prove that sin⁡2x+2sin⁡4x+sin⁡6x=4cos⁡2x sin⁡4x\sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x\, \sin 4x.

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The identity is proved by grouping sin⁡2x+sin⁡6x\sin 2x + \sin 6x using the sum-to-product formula, then factoring sin⁡4x\sin 4x and simplifying 2cos⁡2x+22\cos 2x + 2 to 4cos⁡2x4\cos^2 x, yielding the result.

We need to show that the left-hand side simplifies exactly to the right-hand side. The key is to notice that sin⁡2x\sin 2x and sin⁡6x\sin 6x are symmetric around sin⁡4x\sin 4x — their average angle is 4x4x. This immediately suggests using the sum-to-product identity, which converts a sum of sines into a product involving a cosine of the half-difference.

  1. Group the outer terms Write sin⁡2x+sin⁡6x\sin 2x + \sin 6x. Using the identity

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

with A=2xA = 2x, B=6xB = 6x, we get

sin⁡2x+sin⁡6x=2sin⁡2x+6x2cos⁡2x−6x2=2sin⁡4xcos⁡(−2x).\sin 2x + \sin 6x = 2 \sin\frac{2x+6x}{2} \cos\frac{2x-6x}{2} = 2 \sin 4x \cos(-2x).

Since cos⁡\cos is even, cos⁡(−2x)=cos⁡2x\cos(-2x) = \cos 2x. So

sin⁡2x+sin⁡6x=2sin⁡4xcos⁡2x.\sin 2x + \sin 6x = 2 \sin 4x \cos 2x.

  1. Reassemble the full left-hand side The original expression is

sin⁡2x+2sin⁡4x+sin⁡6x=(2sin⁡4xcos⁡2x)+2sin⁡4x.\sin 2x + 2\sin 4x + \sin 6x = (2 \sin 4x \cos 2x) + 2\sin 4x.

Factor out sin⁡4x\sin 4x:

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

  1. Simplify the bracket Factor the 2:

2cos⁡2x+2=2(cos⁡2x+1).2\cos 2x + 2 = 2(\cos 2x + 1).

Now use the double-angle identity cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1. Then

cos⁡2x+1=(2cos⁡2x−1)+1=2cos⁡2x.\cos 2x + 1 = (2\cos^2 x - 1) + 1 = 2\cos^2 x.

So

2(cos⁡2x+1)=2⋅2cos⁡2x=4cos⁡2x.2(\cos 2x + 1) = 2 \cdot 2\cos^2 x = 4\cos^2 x.

  1. Write the final result …

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