Correct Answer - Option 1 :
\(\frac{\sqrt{5} - 1}{4}\)
Concept:
\(\sin 18^\circ = \frac{{√ 5 - 1}}{4}\)
\({\rm{cos\;}}36^\circ = \frac{{√ 5\; + \;1}}{4}\)
Derivation:
Let θ = 18°
5θ = 5 × 18° = 90°
2θ + 3θ = 90°
2θ = 90° - 3θ
sin 2θ = sin(90° - 3θ)
sin 2θ = cos 3θ
2 sinθ cosθ = 4 cos3θ - 3 cosθ
2 sinθ cosθ - 4 cos3θ + 3 cosθ = 0
\( (As \ \cos(90 - x) = \sin x) \\\ \sin 2x = 2 \sin x \cos x \\\ \cos 3x = 4 \cos^3 x - 3 \cos x\)
cosθ(2 sinθ - 4 cos2θ + 3) = 0
2 sinθ - 4 cos2θ + 3 = 0
2 sinθ - 4(1 - sin2θ) + 3 = 0
2 sinθ - 4 + 4 sin2θ + 3 = 0
4 sin2θ + 2 sinθ - 1 = 0
Let sinθ = x
4x2 + 2x - 1 = 0
Roots can be calculated as:
\( x = \dfrac{-b \pm √{b^2 - 4ac}}{2a}\)
\(x = \dfrac{-2 \pm \sqrt{2^2 - 4(4)(-1)}}{2(4)} \)
\(= \dfrac{-2 \pm \sqrt{4 + 16}}{8} = \dfrac{-2 \pm\sqrt{20}}{8} \)
\(x= \dfrac{-2 \pm \sqrt{4 \times 5}}{8}\)
\(x = \dfrac{-2 \pm 2 √ 5}{8} = \dfrac{-1 \pm √5}{4}\)
∴ sin θ = sin 18° = \(\dfrac{-1 \pm √5}{4}\)
since, sin is positive
\(\sin 18^\circ = \dfrac{-1 + √ 5}{4}\) (since √5 = 2.236 > 1)
