Skip to content
Exercise 3.3 · Q24

Q.Prove that cos⁡4x=1−8sin⁡2x cos⁡2x\cos 4x = 1 - 8\sin^2 x\, \cos^2 x.

Yanam BieapTextbookSubjective· 3mImportance★★★★★est
33% · 50/150 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 prove cos⁡4x=1−8sin⁡2xcos⁡2x\cos 4x = 1 - 8\sin^2 x \cos^2 x by applying the double-angle identity for cosine twice — first to express cos⁡4x\cos 4x in terms of cos⁡2x\cos 2x, then to rewrite cos⁡2x\cos 2x in terms of sin⁡x\sin x and cos⁡x\cos x, followed by a simple algebraic simplification.


The core idea here is that compound-angle identities let you break a larger angle into smaller, manageable pieces. Since 4x=2(2x)4x = 2(2x), we can apply the double-angle formula for cosine twice. The first application gives cos⁡4x\cos 4x in terms of cos⁡2x\cos 2x; the second expresses cos⁡2x\cos 2x in terms of sin⁡x\sin x and cos⁡x\cos x. Then it's just algebra.

Let's walk through it.

  1. Start with the double-angle identity for cosine. The standard formula is cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1. Here, let θ=2x\theta = 2x. Then:

cos⁡4x=cos⁡(2⋅2x)=2cos⁡2(2x)−1.\cos 4x = \cos(2 \cdot 2x) = 2\cos^2(2x) - 1.

  1. Now we need cos⁡2x\cos 2x in terms of sin⁡x\sin x and cos⁡x\cos x.

    Another form of the double-angle identity is cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x.

    But there's a more useful version for this problem: cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x (or 2cos⁡2x−12\cos^2 x - 1).

    However, notice the target has sin⁡2xcos⁡2x\sin^2 x \cos^2 x — a product. So we should use the form that keeps both sin⁡\sin and cos⁡\cos visible.

    The identity cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x is perfect.

  2. Substitute cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x into step 1.

cos⁡4x=2(cos⁡2x−sin⁡2x)2−1.\cos 4x = 2(\cos^2 x - \sin^2 x)^2 - 1.

  1. Expand the square.

(cos⁡2x−sin⁡2x)2=cos⁡4x−2sin⁡2xcos⁡2x+sin⁡4x.(\cos^2 x - \sin^2 x)^2 = \cos^4 x - 2\sin^2 x \cos^2 x + \sin^4 x.

So:

cos⁡4x=2(cos⁡4x+sin⁡4x−2sin⁡2xcos⁡2x)−1.\cos 4x = 2(\cos^4 x + \sin^4 x - 2\sin^2 x \cos^2 x) - 1.

  1. Simplify the cos⁡4x+sin⁡4x\cos^4 x + \sin^4 x term. A neat trick: cos⁡4x+sin⁡4x=(cos⁡2x+sin⁡2x)2−2sin⁡2xcos⁡2x\cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\sin^2 x \cos^2 x. Since cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1, this becomes: cos⁡4x+sin⁡4x=1−2sin⁡2xcos⁡2x.\cos^4 x + \sin^4 x = 1 - 2\sin^2 x \cos^2 x. …

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.