What is the Value of Sin 18 Degrees ?

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 = \(\sqrt{5} – 1\over 4\)

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