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Exercise 3.3 · Q19

Q.Prove that sin⁡x+sin⁡3xcos⁡x+cos⁡3x=tan⁡2x\dfrac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x.

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Use sum-to-product identities to convert both numerator and denominator into products, then cancel common factors to reveal tan⁡2x\tan 2x.

The key insight here is that sums of sines and cosines can be rewritten as products using the sum-to-product formulas. Once we express both the numerator and denominator as products, common factors will cancel and the tangent of the middle angle will emerge naturally.

Why does this work? The sum-to-product identities exploit the fact that sin⁡A+sin⁡B\sin A + \sin B can be viewed as twice the product of a sine and cosine of related angles. This transformation is powerful because products often simplify more readily than sums, especially when we're aiming for a ratio that looks like a tangent.

sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)

cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)

Let me work through the proof step by step.

  1. Apply sum-to-product to the numerator sin⁡x+sin⁡3x\sin x + \sin 3x.

    Here A=3xA = 3x and B=xB = x, so:

A+B2=3x+x2=2x,A−B2=3x−x2=x\frac{A+B}{2} = \frac{3x+x}{2} = 2x, \quad \frac{A-B}{2} = \frac{3x-x}{2} = x

Therefore:

sin⁡x+sin⁡3x=2sin⁡(2x)cos⁡(x)\sin x + \sin 3x = 2\sin(2x)\cos(x)

  1. Apply sum-to-product to the denominator cos⁡x+cos⁡3x\cos x + \cos 3x.

    Using the same values A=3xA = 3x and B=xB = x:

cos⁡x+cos⁡3x=2cos⁡(2x)cos⁡(x)\cos x + \cos 3x = 2\cos(2x)\cos(x)

  1. Form the ratio and simplify.

    Substituting both results: …

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