Question 39 of 50intermediate🔧 ApplyLong Answer4 marks

Prove that:

cos⁡4x=1−8sin⁡2xcos⁡2x\cos 4x = 1 - 8 \sin^2 x \cos^2 x
Correct Answer

L.H.S. =cos⁡4x= \cos 4x

We can write cos⁡4x\cos 4x as cos⁡2(2x)\cos 2(2x).

Using the identity cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x, we get:

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

From the identity sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x, we have:

sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2 \sin x \cos x

Therefore,

sin⁡2(2x)=(2sin⁡xcos⁡x)2=4sin⁡2xcos⁡2x\sin^2(2x) = (2 \sin x \cos x)^2 = 4 \sin^2 x \cos^2 x

Substituting this value:

cos⁡4x=1−2×4sin⁡2xcos⁡2x\cos 4x = 1 - 2 \times 4 \sin^2 x \cos^2 x cos⁡4x=1−8sin⁡2xcos⁡2x=R.H.S.\cos 4x = 1 - 8 \sin^2 x \cos^2 x = \text{R.H.S.}

Hence proved.

Exercise: EXERCISE 3.3 | Q: 24 | (Page 26)
For More Understanding

Explanation

This proof uses the double angle identities provided in the textbook context. The key insight is to recognize that cos⁡4x\cos 4x can be treated as cos⁡2(2x)\cos 2(2x), allowing application of the double angle formula.

The identity cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x is applied with 2x2x as the argument, and then sin⁡(2x)\sin(2x) is substituted using sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x.

Squaring this gives 4sin⁡2xcos⁡2x4 \sin^2 x \cos^2 x, which when multiplied by 22 yields the required 8sin⁡2xcos⁡2x8 \sin^2 x \cos^2 x term.

Solution Steps

  1. Step 1: Write cos⁡4x\cos 4x as cos⁡2(2x)\cos 2(2x)

  2. Step 2: Apply the identity cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x with argument 2x2x

  3. Step 3: Use sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x to find sin⁡2(2x)=4sin⁡2xcos⁡2x\sin^2(2x) = 4 \sin^2 x \cos^2 x

  4. Step 4: Substitute and simplify to get 1−8sin⁡2xcos⁡2x1 - 8 \sin^2 x \cos^2 x

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