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JEE Main 20199 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2019 · 9 January · Morning Shift · Mathematics · Trigonometric Ratios & Identities

For any θ ∈ ((π)/(4), (π)/(2)), the expression 3 (sinθ - cosθ) ^ 4...

For any θ∈(π4,π2),\theta \in \left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right), the expression 3(sinθ−cosθ)4+6(sinθ+cosθ)2+4sin⁡6θ3{\left(sin\theta - cos\theta\right)}^{4} + 6{\left(sin\theta + cos\theta\right)}^{2} + 4\sin^{6}\theta equals:

Options

  1. A13−4cos⁡2θ+6cos⁡4θ13 - 4\cos^{2}\theta + 6\cos^{4}\theta
  2. B13−4cos⁡2θ+6sin⁡2θcos⁡2θ13 - 4\cos^{2}\theta + 6\sin^{2}\theta\cos^{2}\theta
  3. C13−4cos⁡6θ13 - 4\cos^{6}\theta
  4. D13−4cos⁡4θ+2sin⁡2θcos⁡2θ13 - 4\cos^{4}\theta + 2\sin^{2}\theta\cos^{2}\theta

Correct answer

C. 13−4cos⁡6θ13 - 4\cos^{6}\theta

Step-by-step solution

3(sinθ−cosθ)4+6(cosθ+sinθ)2+4sin⁡6θ3{\left(sin\theta - cos\theta\right)}^{4} + 6{\left(cos\theta + sin\theta\right)}^{2} + 4\sin^{6}\theta

=3((sinθ−cosθ)2)2+6(cosθ+sinθ)2+4sin⁡6θ= 3{\left({\left(sin\theta - cos\theta\right)}^{2}\right)}^{2} + 6{\left(cos\theta + sin\theta\right)}^{2} + 4\sin^{6}\theta

=3(cos⁡2θ+sin⁡2θ−2sinθcosθ)2+6(cos⁡2θ+sin⁡2θ+2sinθcosθ)+4sin⁡6θ= 3{\left(\cos^{2}\theta + \sin^{2}\theta - 2sin\theta cos\theta\right)}^{2} + 6\left(\cos^{2}\theta + \sin^{2}\theta + 2sin\theta cos\theta\right) + 4\sin^{6}\theta

=3(1−2sinθcosθ)2+6(1+2sinθcosθ)+4sin⁡6θ= 3{\left(1 - 2sin\theta cos\theta\right)}^{2} + 6\left(1 + 2sin\theta cos\theta\right) + 4\sin^{6}\theta

=3(1−4sinθcosθ+4sin⁡2θcos⁡2θ)+6(1+2sinθcosθ)+4sin⁡6θ= 3\left(1 - 4sin\theta cos\theta + 4\sin^{2}\theta\cos^{2}\theta\right) + 6\left(1 + 2sin\theta cos\theta\right) + 4\sin^{6}\theta

=9+12sin⁡2θcos⁡2θ+4sin⁡6θ= 9 + 12\sin^{2}\theta\cos^{2}\theta + 4\sin^{6}\theta

=9+12(1−cos⁡2θ)cos⁡2θ+4(1−cos⁡2θ)3= 9 + 12\left(1 - \cos^{2}\theta\right)\cos^{2}\theta + 4{\left(1 - \cos^{2}\theta\right)}^{3}

=9+12cos⁡2θ−12cos⁡4θ+4(1−3cos⁡2θ+3cos⁡4θ−cos⁡6θ)= 9 + 12\cos^{2}\theta - 12\cos^{4}\theta + 4\left(1 - 3\cos^{2}\theta + 3\cos^{4}\theta - \cos^{6}\theta\right)

=13−4cos⁡6θ= 13 - 4\cos^{6}\theta

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