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1995

JEE Main 1995 Mathematics question on Trigonometry

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) equals
  1. A.
    11
  2. B.
    12
  3. C.
    13
  4. D.
    14
Show answer and solution

Correct option: C

Solution

Let A = (sinx - cosx)² = 1 - sin2x, B = (sinx + cosx)² = 1 + sin2x.

First term: 3(sinx - cosx)⁴ = 3A² = 3(1 - sin2x)²

Second term: 6(sinx + cosx)² = 6B = 6(1 + sin2x)

Third term: sin⁶x + cos⁶x = 1 - 3sin²x·cos²x = 1 - (3/4)sin²2x, so 4(sin⁶x + cos⁶x) = 4 - 3sin²2x.

Let t = sin2x. Sum = 3(1-t)² + 6(1+t) + 4 - 3t²
= 3 - 6t + 3t² + 6 + 6t + 4 - 3t²
= 13.

The answer is 13.

[1995]

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