Question 29 of 50intermediate🔧 ApplyLong Answer4 marks

Prove that:

sin⁡26x−sin⁡24x=sin⁡2xsin⁡10x\sin^2 6x - \sin^2 4x = \sin 2x \sin 10x
Correct Answer

L.H.S. =sin⁡26x−sin⁡24x= \sin^2 6x - \sin^2 4x

Using the identity: sin⁡2A−sin⁡2B=sin⁡(A+B)⋅sin⁡(A−B)\sin^2 A - \sin^2 B = \sin(A+B) \cdot \sin(A-B)

Let A=6xA = 6x and B=4xB = 4x

Therefore,

sin⁡26x−sin⁡24x=sin⁡(6x+4x)⋅sin⁡(6x−4x)\sin^2 6x - \sin^2 4x = \sin(6x + 4x) \cdot \sin(6x - 4x) =sin⁡10x⋅sin⁡2x= \sin 10x \cdot \sin 2x =sin⁡2xsin⁡10x= \sin 2x \sin 10x =R.H.S.= \textbf{R.H.S.}

Hence proved.

Exercise: EXERCISE 3.3 | Q: 12 | (Page 25)
For More Understanding

Explanation

This question uses the standard trigonometric identity sin⁡2A−sin⁡2B=sin⁡(A+B)sin⁡(A−B)\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B), which is derived from the factorization formula. The proof follows the same methodology shown in Example 19 of the textbook context, where products of trigonometric functions are manipulated. By substituting A=6xA = 6x and B=4xB = 4x, we directly obtain sin⁡(10x)sin⁡(2x)\sin(10x) \sin(2x), which equals the R.H.S.

Solution Steps

  1. Step 1: Identify the L.H.S. as sin⁡26x−sin⁡24x\sin^2 6x - \sin^2 4x

  2. Step 2: Apply the identity sin⁡2A−sin⁡2B=sin⁡(A+B)sin⁡(A−B)\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)

  3. Step 3: Substitute A=6xA = 6x and B=4xB = 4x to get sin⁡(10x)sin⁡(2x)\sin(10x) \sin(2x)

  4. Step 4: Rearrange to match R.H.S. format: sin⁡2xsin⁡10x\sin 2x \sin 10x

  5. Step 5: Conclude L.H.S. = R.H.S., hence proved

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