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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2016 · Mathematics · Trigonometric Ratios & Identities

If 18^ ^ = 5 -1 4 , then what is the value of 81^ ^ ?

If sin⁡18∘=5−14\sin 18^{{}^{\circ}}=\frac{\sqrt{5}-1}{4}, then what is the value of sin⁡81∘\sin 81^{{}^{\circ}} ?

Options

  1. A3+5+5−54\frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}
  2. B3+5+5+54\frac{\sqrt{3+\sqrt{5}}+\sqrt{5+\sqrt{5}}}{4}
  3. C3−5+5−54\frac{\sqrt{3-\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}
  4. D3+5−5−54\frac{\sqrt{3+\sqrt{5}}-\sqrt{5-\sqrt{5}}}{4}

Correct answer

A. 3+5+5−54\frac{\sqrt{3+\sqrt{5}}+\sqrt{5-\sqrt{5}}}{4}

Step-by-step solution

<br/><br/>∵sin⁡18∘=5−14x2=42−(5−1)2⇒x=10+25⇒cos⁡18∘=10+254⇒2cos⁡29−1=10+254⇒cos⁡29=10+25+48⇒sin⁡281=4+10+258\begin{aligned} \because \quad & \sin 18^{{}^{\circ}}=\frac{\sqrt{5}-1}{4} \\ & \mathrm{x}^2=4^2-(\sqrt{5}-1)^2 \\ \Rightarrow & \mathrm{x}=\sqrt{10+2 \sqrt{5}} \\ \Rightarrow & \cos 18^{{}^{\circ}}=\frac{\sqrt{10+2 \sqrt{5}}}{4} \\ \Rightarrow & 2 \cos ^2 9-1=\frac{\sqrt{10+2 \sqrt{5}}}{4} \\ \Rightarrow & \cos ^2 9=\frac{\sqrt{10+2 \sqrt{5}}+4}{8} \\ \Rightarrow & \sin ^2 81=\frac{4+\sqrt{10+2 \sqrt{5}}}{8} \end{aligned}$$ After squaring all the options available, we come to a conclusion that option (1) is correct.
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