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Exercise 3.3 · Q10

Q.Prove that sin⁡(n+1)x sin⁡(n+2)x+cos⁡(n+1)x cos⁡(n+2)x=cos⁡x\sin(n+1)x\, \sin(n+2)x + \cos(n+1)x\, \cos(n+2)x = \cos x.

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Recognize the left-hand side as the cosine of a difference: cos⁡[(n+2)x−(n+1)x]=cos⁡x\cos[(n+2)x - (n+1)x] = \cos x. The product-to-sum structure immediately collapses to the desired result.

The key insight here is pattern recognition. When you see a product of sines plus a product of cosines, your mind should jump to the cosine addition formula—but in reverse. The standard identity

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B

is exactly what we have, just with the terms written in the opposite order.

Notice that our left-hand side has sin⁡(n+1)x⋅sin⁡(n+2)x+cos⁡(n+1)x⋅cos⁡(n+2)x\sin(n+1)x \cdot \sin(n+2)x + \cos(n+1)x \cdot \cos(n+2)x. If we rearrange to match the cosine difference formula, we get cos⁡(n+1)x⋅cos⁡(n+2)x+sin⁡(n+1)x⋅sin⁡(n+2)x\cos(n+1)x \cdot \cos(n+2)x + \sin(n+1)x \cdot \sin(n+2)x, which is precisely cos⁡[(n+2)x−(n+1)x]\cos[(n+2)x - (n+1)x].

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B

Let me show you the proof step by step:

  1. Identify the angles. Set A=(n+2)xA = (n+2)x and B=(n+1)xB = (n+1)x. We want to apply the cosine difference formula to these angles.

  2. Write out the cosine difference. Using the formula above:

cos⁡[(n+2)x−(n+1)x]=cos⁡(n+2)xcos⁡(n+1)x+sin⁡(n+2)xsin⁡(n+1)x\cos[(n+2)x - (n+1)x] = \cos(n+2)x \cos(n+1)x + \sin(n+2)x \sin(n+1)x

  1. Recognize commutativity. Since multiplication is commutative, cos⁡(n+2)xcos⁡(n+1)x=cos⁡(n+1)xcos⁡(n+2)x\cos(n+2)x \cos(n+1)x = \cos(n+1)x \cos(n+2)x and sin⁡(n+2)xsin⁡(n+1)x=sin⁡(n+1)xsin⁡(n+2)x\sin(n+2)x \sin(n+1)x = \sin(n+1)x \sin(n+2)x. So:

cos⁡[(n+2)x−(n+1)x]=cos⁡(n+1)xcos⁡(n+2)x+sin⁡(n+1)xsin⁡(n+2)x\cos[(n+2)x - (n+1)x] = \cos(n+1)x \cos(n+2)x + \sin(n+1)x \sin(n+2)x

This is exactly our left-hand side. …

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