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JEE Main 20209 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
9 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2020 · 9 January · Morning Shift · Mathematics · Trigonometric Ratios & Identities

The value of ^ 3 ((π)/(8)) . cos ((3π)/(8)) + ^ 3 ((π)/(8)) ....

The value of cos⁡3\cos^{3}\left(\dfrac{\pi}{8}\right) . cos(3π8)+sin⁡3(π8).sin⁡(3π8)\left(\dfrac{3\pi}{8}\right) + \sin^{3}\left(\dfrac{\pi}{8}\right) . \sin \left(\dfrac{3\pi}{8}\right) is:

Options

  1. A12\dfrac{1}{\sqrt{2}}
  2. B122\dfrac{1}{2\sqrt{2}}
  3. C12\dfrac{1}{2}
  4. D14\dfrac{1}{4}

Correct answer

B. 122\dfrac{1}{2\sqrt{2}}

Step-by-step solution

⇒cos⁡3(π8)cos⁡(π2−π8)+sin⁡3(π8)sin⁡(π2−π8)⇒ \cos^{3}\left(\dfrac{\pi}{8}\right)\cos \left(\dfrac{\pi}{2} - \dfrac{\pi}{8}\right) + \sin^{3}\left(\dfrac{\pi}{8}\right)\sin \left(\dfrac{\pi}{2} - \dfrac{\pi}{8}\right)

⇒cos⁡3π8sinπ8+sin⁡3π8cosπ8⇒ \cos^{3}\dfrac{\pi}{8}sin\dfrac{\pi}{8} + \sin^{3}\dfrac{\pi}{8}cos\dfrac{\pi}{8}

⇒sinπ8cosπ8(sin⁡2π8+cos⁡2π8)⇒ sin\dfrac{\pi}{8}cos\dfrac{\pi}{8}\left(\sin^{2}\dfrac{\pi}{8} + \cos^{2}\dfrac{\pi}{8}\right)

⇒12sinπ4⇒ \dfrac{1}{2}sin\dfrac{\pi}{4}

⇒122⇒ \dfrac{1}{2\sqrt{2}}

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