Skip to content
Exercise 3.3 · Q25

Q.Prove that cos⁡6x=32cos⁡6x−48cos⁡4x+18cos⁡2x−1\cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1.

Chhattisgarh CgbseTextbookSubjective· 3mImportance★★★★★est
34% · 51/150 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

We'll express cos⁡6x\cos 6x in terms of cos⁡x\cos x by repeatedly applying the double-angle formula, then expand and simplify to show that cos⁡6x=32cos⁡6x−48cos⁡4x+18cos⁡2x−1\cos 6x = 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1.

The heart of this proof lies in recognizing that 6x=2(3x)6x = 2(3x), which lets us use the double-angle formula for cosine. Once we have cos⁡3x\cos 3x in terms of cos⁡x\cos x, we can build up to cos⁡6x\cos 6x systematically. The key is patience with algebraic expansion and a willingness to collect like terms carefully.

We'll need two fundamental identities:

  • Double-angle formula: cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1
  • Triple-angle formula: cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta

Let me first derive the triple-angle formula since we'll need it.

›Proof

Deriving cos⁡3x\cos 3x in terms of cos⁡x\cos x:

Write 3x=2x+x3x = 2x + x and use the addition formula:

cos⁡3x=cos⁡(2x+x)=cos⁡2xcos⁡x−sin⁡2xsin⁡x\cos 3x = \cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x

Substitute cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1 and sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x \cos x:

cos⁡3x=(2cos⁡2x−1)cos⁡x−(2sin⁡xcos⁡x)sin⁡x\cos 3x = (2\cos^2 x - 1)\cos x - (2\sin x \cos x)\sin x

=2cos⁡3x−cos⁡x−2sin⁡2xcos⁡x= 2\cos^3 x - \cos x - 2\sin^2 x \cos x

Replace sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x:

cos⁡3x=2cos⁡3x−cos⁡x−2(1−cos⁡2x)cos⁡x\cos 3x = 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x

=2cos⁡3x−cos⁡x−2cos⁡x+2cos⁡3x= 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x

=4cos⁡3x−3cos⁡x= 4\cos^3 x - 3\cos x

Now for the main proof:

  1. Express cos⁡6x\cos 6x using the double-angle formula.

    Since 6x=2(3x)6x = 2(3x), we have:

cos⁡6x=2cos⁡23x−1\cos 6x = 2\cos^2 3x - 1

  1. Substitute the triple-angle formula.

    From our derivation above, cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x = 4\cos^3 x - 3\cos x. Therefore:

cos⁡6x=2(4cos⁡3x−3cos⁡x)2−1\cos 6x = 2(4\cos^3 x - 3\cos x)^2 - 1

  1. Expand the square.

    Let's compute (4cos⁡3x−3cos⁡x)2(4\cos^3 x - 3\cos x)^2 carefully: …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.