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

Correct Answer is : 3+5+5−54{{ {3 + 5 } + {5 - 5 } }}{4}43+5+5−5

Correct Answer is : 3+5+5−54{{ {3 + 5 } + {5 - 5 } }}{4}43+5+5−5

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UPSC NDA 2 2016 Mathematics Solved Paper

Question : 48 of 120

Marks: +1, -0

\sin 18^\circ = \frac{{\sqrt 5 - 1}}{4}

, then what is the value of sin 81°?

Solution:

Concept:

Use complementary angle and half‑angle formulas:

\sin(90^\circ - \theta) = \cos \theta

and

\cos \theta = \sqrt{\frac{1+\cos 2\theta}{2}}

. Also

\cos^2 \theta = 1 - \sin^2 \theta

Explanation:

Given

\sin 18^\circ = \frac{\sqrt{5}-1}{4}

First, find

\cos 18^\circ

using

\cos^2 18^\circ = 1 - \sin^2 18^\circ

\cos^2 18^\circ = 1 - \left(\frac{\sqrt{5}-1}{4}\right)^2 = 1 - \frac{6-2\sqrt{5}}{16} = \frac{10+2\sqrt{5}}{16}

Thus

\cos 18^\circ = \frac{\sqrt{10+2\sqrt{5}}}{4}

We need

\sin 81^\circ = \sin(90^\circ - 9^\circ) = \cos 9^\circ

Apply half‑angle:

\cos 9^\circ = \sqrt{\frac{1+\cos 18^\circ}{2}} = \sqrt{\frac{1+\frac{\sqrt{10+2\sqrt{5}}}{4}}{2}} = \sqrt{\frac{4+\sqrt{10+2\sqrt{5}}}{8}}

Simplify:

\sqrt{\frac{8+2\sqrt{10+2\sqrt{5}}}{16}}

Notice that

8+2\sqrt{10+2\sqrt{5}} = (\sqrt{3+\sqrt{5}} + \sqrt{5-\sqrt{5}})^2

Therefore,

\sin 81^\circ = \frac{\sqrt{3+\sqrt{5}} + \sqrt{5-\sqrt{5}}}{4}

Answer:

Option A:

\frac{\sqrt{3+\sqrt{5}} + \sqrt{5-\sqrt{5}}}{4}

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