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NCERT Exemplar · Q65

Q.3(sin⁡x−cos⁡x)4+6(sin⁡x+cos⁡x)2+4(sin⁡6x+cos⁡6x)=3(\sin x - \cos x)^4 + 6(\sin x + \cos x)^2 + 4(\sin^6 x + \cos^6 x) = ______.

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The expression simplifies to a constant independent of xx by expanding powers, using identities like sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, and combining terms. The final value is 13.


We start with the expression:

3(sin⁡x−cos⁡x)4+6(sin⁡x+cos⁡x)2+4(sin⁡6x+cos⁡6x)3(\sin x - \cos x)^4 + 6(\sin x + \cos x)^2 + 4(\sin^6 x + \cos^6 x)

The key insight: this looks like a polynomial in sin⁡x\sin x and cos⁡x\cos x that might collapse to a constant. The Binomial Theorem and basic trigonometric identities are our tools.


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

Using (a−b)4=a4−4a3b+6a2b2−4ab3+b4(a-b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4:

(sin⁡x−cos⁡x)4=sin⁡4x−4sin⁡3xcos⁡x+6sin⁡2xcos⁡2x−4sin⁡xcos⁡3x+cos⁡4x(\sin x - \cos x)^4 = \sin^4 x - 4\sin^3 x \cos x + 6\sin^2 x \cos^2 x - 4\sin x \cos^3 x + \cos^4 x

So the first term becomes:

3(sin⁡x−cos⁡x)4=3sin⁡4x−12sin⁡3xcos⁡x+18sin⁡2xcos⁡2x−12sin⁡xcos⁡3x+3cos⁡4x3(\sin x - \cos x)^4 = 3\sin^4 x - 12\sin^3 x \cos x + 18\sin^2 x \cos^2 x - 12\sin x \cos^3 x + 3\cos^4 x

2. Expand (sin⁡x+cos⁡x)2(\sin x + \cos x)^2

(sin⁡x+cos⁡x)2=sin⁡2x+2sin⁡xcos⁡x+cos⁡2x=1+2sin⁡xcos⁡x(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + 2\sin x \cos x

Thus:

6(sin⁡x+cos⁡x)2=6+12sin⁡xcos⁡x6(\sin x + \cos x)^2 = 6 + 12\sin x \cos x

3. Simplify sin⁡6x+cos⁡6x\sin^6 x + \cos^6 x

Use the identity a3+b3=(a+b)3−3ab(a+b)a^3 + b^3 = (a+b)^3 - 3ab(a+b). Let a=sin⁡2xa = \sin^2 x, b=cos⁡2xb = \cos^2 x:

sin⁡6x+cos⁡6x=(sin⁡2x)3+(cos⁡2x)3=(sin⁡2x+cos⁡2x)3−3sin⁡2xcos⁡2x(sin⁡2x+cos⁡2x)\sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)^3 - 3\sin^2 x \cos^2 x (\sin^2 x + \cos^2 x)

Since sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1:

sin⁡6x+cos⁡6x=1−3sin⁡2xcos⁡2x\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x

Therefore:

4(sin⁡6x+cos⁡6x)=4−12sin⁡2xcos⁡2x4(\sin^6 x + \cos^6 x) = 4 - 12\sin^2 x \cos^2 x

Tip

The identity sin⁡6x+cos⁡6x=1−3sin⁡2xcos⁡2x\sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x is a neat shortcut — it avoids expanding the sixth powers directly.

4. Add all three parts

Collect the terms from steps 1, 2, and 3:

  • From step 1: 3sin⁡4x+3cos⁡4x+18sin⁡2xcos⁡2x−12sin⁡3xcos⁡x−12sin⁡xcos⁡3x3\sin^4 x + 3\cos^4 x + 18\sin^2 x \cos^2 x - 12\sin^3 x \cos x - 12\sin x \cos^3 x
  • From step 2: 6+12sin⁡xcos⁡x6 + 12\sin x \cos x
  • From step 3: 4−12sin⁡2xcos⁡2x4 - 12\sin^2 x \cos^2 x

Now combine like terms:

  • Constant terms: 6+4=106 + 4 = 10
  • sin⁡4x+cos⁡4x\sin^4 x + \cos^4 x terms: 3sin⁡4x+3cos⁡4x3\sin^4 x + 3\cos^4 x
  • sin⁡2xcos⁡2x\sin^2 x \cos^2 x terms: 18sin⁡2xcos⁡2x−12sin⁡2xcos⁡2x=6sin⁡2xcos⁡2x18\sin^2 x \cos^2 x - 12\sin^2 x \cos^2 x = 6\sin^2 x \cos^2 x
  • sin⁡3xcos⁡x\sin^3 x \cos x and sin⁡xcos⁡3x\sin x \cos^3 x terms: −12sin⁡3xcos⁡x−12sin⁡xcos⁡3x-12\sin^3 x \cos x - 12\sin x \cos^3 x
  • sin⁡xcos⁡x\sin x \cos x term: +12sin⁡xcos⁡x+12\sin x \cos x

So far we have: …

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