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JEE Advanced
Exam
JEE Advanced
Subject
Mathematics
Chapter
Trigonometry

JEE Advanced · Mathematics · Trigonometry

Let y=( x · 2 x+ 3 x · 6 x+ 4 x · 13 x)/( x · 2 x+ 3 x · 6 x+ 4 x ·...

Let y=sin⁡x⋅sin⁡2x+sin⁡3x⋅cos⁡6x+sin⁡4x⋅cos⁡13xsin⁡x⋅cos⁡2x+sin⁡3x⋅cos⁡6x+sin⁡4x⋅cos⁡13xy=\frac{\sin x \cdot \sin 2 x+\sin 3 x \cdot \cos 6 x+\sin 4 x \cdot \cos 13 x}{\sin x \cdot \cos 2 x+\sin 3 x \cdot \cos 6 x+\sin 4 x \cdot \cos 13 x}, then:

Options

  1. Aif x=π72x=\frac{\pi}{72}, then y=2−1y=\sqrt{2}-1
  2. Bif x=π24x=\frac{\pi}{24}, then y=2+1y=\sqrt{2}+1
  3. Cif x=π108x=\frac{\pi}{108}, then y=2+3y=2+\sqrt{3}
  4. Dif x=5π108x=\frac{5 \pi}{108}, then y=2−3y=2-\sqrt{3}

Correct answer

B. if x=π24x=\frac{\pi}{24}, then y=2+1y=\sqrt{2}+1

Step-by-step solution

y=cos⁡x−cos⁡3x+cos⁡3x−cos⁡9x+cos⁡9x−cos⁡17xsin⁡3x−sin⁡x+sin⁡9x−sin⁡3x+sin⁡17x−sin⁡9xy=cos⁡x−cos⁡17xsin⁡17x−sin⁡x=2sin⁡9x⋅sin⁡8x2sin⁡8x⋅cos⁡9x=tan⁡9x∴y=\frac{\cos x-\cos 3 x+\cos 3 x-\cos 9 x+\cos 9 x-\cos 17 x}{\sin 3 x-\sin x+\sin 9 x-\sin 3 x+\sin 17 x-\sin 9 x}y=\frac{\cos x-\cos 17 x}{\sin 17 x-\sin x}=\frac{2 \sin 9 x \cdot \sin 8 x}{2 \sin 8 x \cdot \cos 9 x}=\tan 9 x\therefore \quad (a) and (b) are correct.
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