1.

find the value of sin18°

Answer»

Let \(\theta = 18°\)

\(5\theta = 18° \times 5 = 90°\)

⇒ \(2\theta + 3\theta = 90°\)

⇒ \(3\theta = 90° - 2\theta\)

⇒ \(sin3\theta = sin(90° - 2\theta ) = cos2\theta\)

⇒ \(3sin \theta - 4sin^3\theta = 1 - 2sin^2 \theta\)

⇒ \(4sin^3\theta - 2sin^2\theta - 3sin\theta + 1 = 0\)

Let \(sin\theta = x\)

\(\therefore 4x^3 - 2x^2 - 3x + 1 = 0\)

\((x - 1) (4x^2 + 2x - 1) = 0\)

\(\therefore x = 1 , \frac {-2 \pm \sqrt{4 + 16} }{2\times 4}\)

\(= 1, \frac{-2\pm 2\sqrt 5}8\)

\(= 1, \frac{-1\pm \sqrt 5}4\)

\(\because\) sin 18° is a positive value and not equal to 1.

\(\therefore sin 18° = x = \frac {-1 + \sqrt 5}4\).


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