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1986

JEE Main 1986 Mathematics question on Trigonometry

The expression
3[sin⁡4(3π2−α)+sin⁡4(3π+α)]−2[sin⁡6(π2+α)+sin⁡6(5π−α)]3\left[\sin^4\left(\frac{3\pi}{2}-\alpha\right)+\sin^4(3\pi+\alpha)\right]-2\left[\sin^6\left(\frac{\pi}{2}+\alpha\right)+\sin^6(5\pi-\alpha)\right]
is equal to
  1. A.
    0
  2. B.
    1
  3. C.
    3
  4. D.
    sin⁡4α+cos⁡6α\sin 4\alpha + \cos 6\alpha
Show answer and solution

Correct option: B

Solution

Simplify each trig term using standard identities:

- sin(3π/2 - α) = -cos α, so sin⁴(3π/2 - α) = cos⁴α
- sin(3π + α) = -sin α, so sin⁴(3π + α) = sin⁴α
- sin(π/2 + α) = cos α, so sin⁶(π/2 + α) = cos⁶α
- sin(5π - α) = sin α, so sin⁶(5π - α) = sin⁶α

The expression becomes:
3[cos⁴α + sin⁴α] - 2[cos⁶α + sin⁶α]

Use the identities:
- sin⁴α + cos⁴α = 1 - 2sin²α cos²α
- sin⁶α + cos⁶α = 1 - 3sin²α cos²α

So the expression = 3(1 - 2sin²α cos²α) - 2(1 - 3sin²α cos²α)
= 3 - 6sin²α cos²α - 2 + 6sin²α cos²α = 1

[1986]

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