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Exercise 3.3 · Q13

Q.Prove that cos⁡22x−cos⁡26x=sin⁡4x sin⁡8x\cos^2 2x - \cos^2 6x = \sin 4x\, \sin 8x.

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Transform the difference of cosine squares using the product-to-sum identity in reverse, then simplify the resulting products to show both sides are equal. The identity holds: cos⁡22x−cos⁡26x=sin⁡4x sin⁡8x\cos^2 2x - \cos^2 6x = \sin 4x\, \sin 8x.

The key insight here is recognizing that a difference of squares of cosines can be rewritten as a product of sines. We'll use the product-to-sum formulas, but in reverse—starting from a difference and working toward a product.

Recall that the product-to-sum identity gives us:

cos⁡Acos⁡B=12[cos⁡(A−B)+cos⁡(A+B)]\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]

This means a difference of two cosine products can be expressed in terms of sines. Specifically, we can factor cos⁡22x−cos⁡26x\cos^2 2x - \cos^2 6x by treating it as a difference of squares.

Proof

  1. Factor the left-hand side as a difference of squares.

    Write cos⁡22x−cos⁡26x=(cos⁡2x−cos⁡6x)(cos⁡2x+cos⁡6x)\cos^2 2x - \cos^2 6x = (\cos 2x - \cos 6x)(\cos 2x + \cos 6x).

  2. Apply sum-to-product formulas to each factor.

    The sum-to-product identities are:

cos⁡C+cos⁡D=2cos⁡(C+D2)cos⁡(C−D2)\cos C + \cos D = 2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)

cos⁡C−cos⁡D=−2sin⁡(C+D2)sin⁡(C−D2)\cos C - \cos D = -2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)

  1. Transform cos⁡2x−cos⁡6x\cos 2x - \cos 6x.

    Using C=2xC = 2x and D=6xD = 6x:

cos⁡2x−cos⁡6x=−2sin⁡(2x+6x2)sin⁡(2x−6x2)\cos 2x - \cos 6x = -2\sin\left(\frac{2x+6x}{2}\right)\sin\left(\frac{2x-6x}{2}\right)

=−2sin⁡(4x)sin⁡(−2x)=−2sin⁡(4x)⋅(−sin⁡2x)=2sin⁡4xsin⁡2x= -2\sin(4x)\sin(-2x) = -2\sin(4x) \cdot (-\sin 2x) = 2\sin 4x \sin 2x

  1. Transform cos⁡2x+cos⁡6x\cos 2x + \cos 6x.

    Using the same values:

cos⁡2x+cos⁡6x=2cos⁡(2x+6x2)cos⁡(2x−6x2)\cos 2x + \cos 6x = 2\cos\left(\frac{2x+6x}{2}\right)\cos\left(\frac{2x-6x}{2}\right)

=2cos⁡(4x)cos⁡(−2x)=2cos⁡4xcos⁡2x= 2\cos(4x)\cos(-2x) = 2\cos 4x \cos 2x

  1. Multiply the two factors. …

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