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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2023 · Mathematics · Trigonometric Ratios & Identities

Paragraph: Let β be the GM of α and α ; γ be the AM of α and α...

Paragraph: Let sin⁡β\sin \beta be the GM of sin⁡α\sin \alpha and cos⁡α;tan⁡γ\cos \alpha ; \tan \gamma be the AM of sin⁡α\sin\alpha and cos⁡α\cos \alpha
Question: What is cos⁡2β\cos 2 \beta equal to?

Options

  1. A(cos⁡α−sin⁡α)2(\cos \alpha-\sin \alpha)^2
  2. B(cos⁡α+sin⁡α)2(\cos \alpha+\sin \alpha)^2
  3. C(cos⁡α−sin⁡α)3(\cos \alpha-\sin \alpha)^3
  4. D(cos⁡α−sin⁡α)22\frac{(\cos \alpha-\sin \alpha)^2}{2}

Correct answer

A. (cos⁡α−sin⁡α)2(\cos \alpha-\sin \alpha)^2

Step-by-step solution

(sin⁡β)2=sin⁡α⋅cos⁡α(\sin \beta)^2=\sin \alpha \cdot \cos \alpha....(i) tan⁡γ=sin⁡α+cos⁡α2cos⁡2β=1−2sin⁡2β=1−2sin⁡αcos⁡α[∵\tan \gamma=\frac{\sin \alpha+\cos \alpha}{2}\cos 2 \beta=1-2 \sin ^2 \beta=1-2 \sin \alpha \cos \alpha \quad[\because By (i) ]=sin⁡2α+cos⁡2α−2sin⁡αcos⁡α=(cos⁡α−sin⁡α)2]=\sin ^2 \alpha+\cos ^2 \alpha-2 \sin \alpha \cos \alpha=(\cos \alpha-\sin \alpha)^2
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