Solution :
The value of sin 18 degrees is \(\sqrt{5} – 1\over 4\).
Proof :
Let \(\theta\) = 18 degrees. Then,
\(5\theta\) = 90 degrees
\(\implies\) \(2\theta\) + \(3\theta\) = 90
\(\implies\) \(2\theta\) = 90 – \(3\theta\)
\(\implies\) \(sin 2\theta\) = \(sin (90 – 3\theta)\)
\(\implies\) \(sin 2\theta\) = \(cos 3\theta\)
By using the formula of \(sin 2\theta\) and \(cos 3\theta\)
\(\implies\) \(2 sin \theta cos \theta\) = \(4 cos^3 \theta – 3 cos \theta\)
\(\implies\) \(cos \theta\) (\(2 sin \theta – 4 cos^2 \theta + 3\) = 0
\(\implies\) \(2 sin \theta\) – \(4 cos^2 \theta\) + 3 = 0
[ \(\because\) \(cos \theta\) = cos 18 \(\ne\) 0 ]
\(\implies\) \(2 sin \theta\) – \(4(1 – sin^2 \theta)\) + 3 = 0
\(\implies\) \(4 sin^2 \theta\) + \(2 sin \theta\) – 1 = 0
Solving this quadratic equation by using quadratic formula,
\(\implies\) \(sin \theta\) = \(-2 \pm \sqrt{4 + 16}\over 8\)
\(\implies\) \(sin \theta\) = \(-1 \pm \sqrt{5}\over 4\)
Since \(\theta\) lies in 1st quadrant \(\therefore\) \(sin \theta\) > 0
\(\implies\) \(sin \theta\) = \(-1 + \sqrt{5}\over 4\) = \(\sqrt{5} – 1\over 4\)
Hence, sin 18 degrees = \(\sqrt{5} – 1\over 4\)
