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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2020 · Mathematics · Trigonometric Ratios & Identities

Paragraph: Read the following information and answer: Let α=β=15^ ^...

Paragraph: Read the following information and answer: Let α=β=15∘\alpha=\beta=15^{{}^{\circ}}.
Question: What is the value of sin⁡7α−cos⁡7β\sin 7 \alpha-\cos 7 \beta ?

Options

  1. A12\frac{1}{\sqrt{2}}
  2. B122\frac{1}{2 \sqrt{2}}
  3. C322\frac{\sqrt{3}}{2 \sqrt{2}}
  4. D32\frac{\sqrt{3}}{\sqrt{2}}

Correct answer

D. 32\frac{\sqrt{3}}{\sqrt{2}}

Step-by-step solution

sin⁡7α−cos⁡7β=sin⁡7(15∘)−cos⁡7(15∘)=sin⁡(105∘)−cos⁡(105∘)=sin⁡(90∘+15∘)−cos⁡(90∘+15∘)\begin{aligned} & \sin 7 \alpha-\cos 7 \beta \\ = & \sin 7\left(15^{{}^{\circ}}\right)-\cos 7\left(15^{{}^{\circ}}\right)=\sin \left(105^{{}^{\circ}}\right)-\cos \left(105^{{}^{\circ}}\right) \\ = & \sin \left(90^{{}^{\circ}}+15^{{}^{\circ}}\right)-\cos \left(90^{{}^{\circ}}+15^{{}^{\circ}}\right)\end{aligned}=\cos 15^{{}^{\circ}}-\left(-\sin \left(15^{{}^{\circ}}\right)\right)=\cos 15^{{}^{\circ}}+\sin 15^{{}^{\circ}}=\frac{\sqrt{3}}{\sqrt{2}}$
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