Question 40 of 50advanced🔧 ApplyLong Answer4 marks

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
Correct Answer

To Prove:

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

Proof:

Step 1: We know that cos⁡6x\cos 6x can be written as cos⁡2(3x)\cos 2(3x).

Using the formula cos⁡2A=2cos⁡2A−1\cos 2A = 2\cos^2 A - 1 (from the context), we get:

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

Step 2: Now, using the formula cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x = 4\cos^3 x - 3\cos x (from the context), we substitute:

cos⁡6x=2[4cos⁡3x−3cos⁡x]2−1\cos 6x = 2[4\cos^3 x - 3\cos x]^2 - 1

Step 3: Expanding [4cos⁡3x−3cos⁡x]2[4\cos^3 x - 3\cos x]^2:

=(4cos⁡3x)2−2(4cos⁡3x)(3cos⁡x)+(3cos⁡x)2= (4\cos^3 x)^2 - 2(4\cos^3 x)(3\cos x) + (3\cos x)^2 =16cos⁡6x−24cos⁡4x+9cos⁡2x= 16\cos^6 x - 24\cos^4 x + 9\cos^2 x

Step 4: Substituting back:

cos⁡6x=2[16cos⁡6x−24cos⁡4x+9cos⁡2x]−1\cos 6x = 2[16\cos^6 x - 24\cos^4 x + 9\cos^2 x] - 1 =32cos⁡6x−48cos⁡4x+18cos⁡2x−1= 32\cos^6 x - 48\cos^4 x + 18\cos^2 x - 1

Hence Proved.

Exercise: EXERCISE 3.3 | Q: 25 | (Page 26)
For More Understanding

Explanation

This proof requires applying two standard identities from the chapter: cos⁡2A=2cos⁡2A−1\cos 2A = 2\cos^2 A - 1 and cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x = 4\cos^3 x - 3\cos x. The key insight is recognizing that 6x=2(3x)6x = 2(3x), allowing us to use the double angle formula on 3x3x.

After substitution, expanding the perfect square and simplifying yields the required expression. This is a 4-mark question testing multiple identity applications.

Solution Steps

  1. Step 1: Express cos⁡6x\cos 6x as cos⁡2(3x)\cos 2(3x) and apply cos⁡2A\cos 2A formula

  2. Step 2: Substitute cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x = 4\cos^3 x - 3\cos x

  3. Step 3: Expand the square (4cos⁡3x−3cos⁡x)2(4\cos^3 x - 3\cos x)^2

  4. Step 4: Multiply by 2 and subtract 1 to get the final expression

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