Question 43 of 50intermediate🔧 ApplyLong Answer4 marks

Prove that:

(cos⁡x+cos⁡y)2+(sin⁡x−sin⁡y)2=4cos⁡2(x+y2)(\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4\cos^2\left(\frac{x + y}{2}\right)
Correct Answer

LHS =(cos⁡x+cos⁡y)2+(sin⁡x−sin⁡y)2= (\cos x + \cos y)^2 + (\sin x - \sin y)^2

Using the transformation formulas:

  • cos⁡x+cos⁡y=2cos⁡(x+y2)cos⁡(x−y2)\cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)
  • sin⁡x−sin⁡y=2cos⁡(x+y2)sin⁡(x−y2)\sin x - \sin y = 2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)

Substituting these formulas:

(cos⁡x+cos⁡y)2=[2cos⁡(x+y2)cos⁡(x−y2)]2=4cos⁡2(x+y2)cos⁡2(x−y2)(\cos x + \cos y)^2 = \left[2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)\right]^2 = 4\cos^2\left(\frac{x+y}{2}\right)\cos^2\left(\frac{x-y}{2}\right) (sin⁡x−sin⁡y)2=[2cos⁡(x+y2)sin⁡(x−y2)]2=4cos⁡2(x+y2)sin⁡2(x−y2)(\sin x - \sin y)^2 = \left[2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right]^2 = 4\cos^2\left(\frac{x+y}{2}\right)\sin^2\left(\frac{x-y}{2}\right)

Adding both terms:

LHS=4cos⁡2(x+y2)cos⁡2(x−y2)+4cos⁡2(x+y2)sin⁡2(x−y2)\text{LHS} = 4\cos^2\left(\frac{x+y}{2}\right)\cos^2\left(\frac{x-y}{2}\right) + 4\cos^2\left(\frac{x+y}{2}\right)\sin^2\left(\frac{x-y}{2}\right) =4cos⁡2(x+y2)[cos⁡2(x−y2)+sin⁡2(x−y2)]= 4\cos^2\left(\frac{x+y}{2}\right)\left[\cos^2\left(\frac{x-y}{2}\right) + \sin^2\left(\frac{x-y}{2}\right)\right] =4cos⁡2(x+y2)×1= 4\cos^2\left(\frac{x+y}{2}\right) \times 1 =4cos⁡2(x+y2)= 4\cos^2\left(\frac{x+y}{2}\right)

∴ Hence, LHS = RHS, which proves the identity.

Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 3 | Q: 3 | (Page 29)
For More Understanding

Explanation

This proof uses the sum-to-product transformation formulas provided in the textbook context (formulas 20(i) and 20(iv)). The key insight is that both (cos⁡x+cos⁡y)(\cos x + \cos y) and (sin⁡x−sin⁡y)(\sin x - \sin y) have a common factor of 2cos⁡(x+y2)2\cos\left(\frac{x+y}{2}\right), which allows for simplification.

The fundamental identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is then applied to eliminate the x−y2\frac{x-y}{2} terms, leaving only the required result.

Solution Steps

  1. Step 1: Apply transformation formula cos⁡x+cos⁡y=2cos⁡(x+y2)cos⁡(x−y2)\cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)

  2. Step 2: Apply transformation formula sin⁡x−sin⁡y=2cos⁡(x+y2)sin⁡(x−y2)\sin x - \sin y = 2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)

  3. Step 3: Square both expressions and add them

  4. Step 4: Factor out 4cos⁡2(x+y2)4\cos^2\left(\frac{x+y}{2}\right)

  5. Step 5: Apply sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 to simplify

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