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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2024 · Mathematics · Trigonometric Ratios & Identities

What is 12^ ^ 48^ ^ equal to?

What is sin⁡12∘sin⁡48∘\sin 12^{{}^{\circ}} \sin 48^{{}^{\circ}} equal to?

Options

  1. A5−14\frac{\sqrt{5}-1}{4}
  2. B5+14\frac{\sqrt{5}+1}{4}
  3. C5−18\frac{\sqrt{5}-1}{8}
  4. D5+18\frac{\sqrt{5}+1}{8}

Correct answer

C. 5−18\frac{\sqrt{5}-1}{8}

Step-by-step solution

 Let I=sin⁡12∘⋅sin⁡48∘⇒I=2sin⁡12∘sin⁡48∘2⇒I=12[cos⁡(36∘)−cos⁡(60∘)]⇒I=12[5+14−12]=5−18\begin{aligned} & \text{ Let } \mathrm{I}=\sin 12^{{}^{\circ}} \cdot \sin 48^{{}^{\circ}} \Rightarrow \mathrm{I}=\frac{2 \sin 12^{{}^{\circ}} \sin 48^{{}^{\circ}}}{2} \\ & \Rightarrow \mathrm{I}=\frac{1}{2}\left[\cos \left(36^{{}^{\circ}}\right)-\cos \left(60^{{}^{\circ}}\right)\right] \\ & \Rightarrow \mathrm{I}=\frac{1}{2}\left[\frac{\sqrt{5}+1}{4}-\frac{1}{2}\right]=\frac{\sqrt{5}-1}{8}\end{aligned}
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