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Exercise 3.3 · Q12

Q.Prove that sin⁡26x−sin⁡24x=sin⁡2x sin⁡10x\sin^2 6x - \sin^2 4x = \sin 2x\, \sin 10x.

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The identity sin⁡26x−sin⁡24x=sin⁡2xsin⁡10x\sin^2 6x - \sin^2 4x = \sin 2x \sin 10x is proved by applying the difference-of-squares factorization and the sum-to-product formulas, reducing the left side to a product of sines that matches the right side.

This problem is a classic trigonometric identity proof. The key insight is that the left side is a difference of two squares, which can be factored into a product of sums and differences. Then, using sum-to-product formulas, we can rewrite those sums and differences as products of sines and cosines, which simplifies directly to the right side.

Let’s work through it step by step.

  1. Factor the difference of squares Recall that a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b). Here, a=sin⁡6xa = \sin 6x and b=sin⁡4xb = \sin 4x, so:

sin⁡26x−sin⁡24x=(sin⁡6x−sin⁡4x)(sin⁡6x+sin⁡4x)\sin^2 6x - \sin^2 4x = (\sin 6x - \sin 4x)(\sin 6x + \sin 4x)

  1. Apply sum-to-product formulas The sum-to-product identities are:

sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2 \cos\frac{A+B}{2} \sin\frac{A-B}{2}

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2 \sin\frac{A+B}{2} \cos\frac{A-B}{2}

For A=6xA = 6x and B=4xB = 4x:

  • sin⁡6x−sin⁡4x=2cos⁡6x+4x2sin⁡6x−4x2=2cos⁡5xsin⁡x\sin 6x - \sin 4x = 2 \cos\frac{6x+4x}{2} \sin\frac{6x-4x}{2} = 2 \cos 5x \sin x
  • sin⁡6x+sin⁡4x=2sin⁡6x+4x2cos⁡6x−4x2=2sin⁡5xcos⁡x\sin 6x + \sin 4x = 2 \sin\frac{6x+4x}{2} \cos\frac{6x-4x}{2} = 2 \sin 5x \cos x
  1. Multiply the two factors Substituting back:

(sin⁡6x−sin⁡4x)(sin⁡6x+sin⁡4x)=(2cos⁡5xsin⁡x)(2sin⁡5xcos⁡x)(\sin 6x - \sin 4x)(\sin 6x + \sin 4x) = (2 \cos 5x \sin x)(2 \sin 5x \cos x)

=4sin⁡xcos⁡xsin⁡5xcos⁡5x= 4 \sin x \cos x \sin 5x \cos 5x

  1. Use the double-angle identity Recall sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2 \sin \theta \cos \theta. Here, we have two such pairs:
    • 2sin⁡xcos⁡x=sin⁡2x2 \sin x \cos x = \sin 2x
    • 2sin⁡5xcos⁡5x=sin⁡10x2 \sin 5x \cos 5x = \sin 10x …

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