Question 30 of 50intermediate🔧 ApplyLong Answer4 marks

Prove that:

cos⁡22x−cos⁡26x=sin⁡4xsin⁡8x\cos^2 2x - \cos^2 6x = \sin 4x \sin 8x
Correct Answer

L.H.S. =cos⁡22x−cos⁡26x= \cos^2 2x - \cos^2 6x

Using the identity cos⁡2A=1+cos⁡2A2\cos^2 A = \frac{1 + \cos 2A}{2}, we get:

cos⁡22x=1+cos⁡4x2\cos^2 2x = \frac{1 + \cos 4x}{2} cos⁡26x=1+cos⁡12x2\cos^2 6x = \frac{1 + \cos 12x}{2}

Therefore,

L.H.S.=1+cos⁡4x2−1+cos⁡12x2\textbf{L.H.S.} = \frac{1 + \cos 4x}{2} - \frac{1 + \cos 12x}{2} =cos⁡4x−cos⁡12x2= \frac{\cos 4x - \cos 12x}{2}

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

cos⁡4x−cos⁡12x=−2sin⁡(4x+12x2)sin⁡(4x−12x2)\cos 4x - \cos 12x = -2 \sin\left(\frac{4x+12x}{2}\right) \sin\left(\frac{4x-12x}{2}\right) =−2sin⁡(8x)sin⁡(−4x)= -2 \sin(8x) \sin(-4x) =2sin⁡(8x)sin⁡(4x)[since sin⁡(−θ)=−sin⁡θ]= 2 \sin(8x) \sin(4x) \quad [\text{since } \sin(-\theta) = -\sin \theta]

Therefore,

L.H.S.=2sin⁡8xsin⁡4x2=sin⁡4xsin⁡8x=R.H.S.\textbf{L.H.S.} = \frac{2 \sin 8x \sin 4x}{2} = \sin 4x \sin 8x = \textbf{R.H.S.}

Hence proved.

Exercise: EXERCISE 3.3 | Q: 13 | (Page 25)
For More Understanding

Explanation

This question uses the half-angle identity for cos⁡2A=1+cos⁡2A2\cos^2 A = \frac{1 + \cos 2A}{2} shown in Example 22, and the sum-to-product identity for cos⁡A−cos⁡B\cos A - \cos B demonstrated in Example 16. The key insight is converting squared terms to linear terms using the half-angle formula, then applying the transformation formula to factor the expression into the required product form.

Solution Steps

  1. Step 1: Apply the identity cos⁡2A=1+cos⁡2A2\cos^2 A = \frac{1 + \cos 2A}{2} to both terms

  2. Step 2: Simplify to get cos⁡4x−cos⁡12x2\frac{\cos 4x - \cos 12x}{2}

  3. Step 3: Apply the identity cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)

  4. Step 4: Simplify using sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin \theta to get sin⁡4xsin⁡8x\sin 4x \sin 8x

  5. Step 5: Conclude L.H.S. = R.H.S.

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