UPSC NDA Math Model Paper 5

Let us consider B = 18° So, 5B = 90° ⇒ 2B + 3B = 90° ⇒ 2B = 90° - 3B By taking sine on both sides, we get Sin 2B = sin (90° - 3B) = cos 3B ⇒ 2 sin B cos B = 4 cos$^3$ B - 3 cos B ⇒ 2 sin B cos B - 4 cos$^3$ B + 3 cos B = 0 ⇒ cos B (2 sin B - 4 cos B + 3) = 0 Dividing both sides by cos B = cos 18˚ ≠ 0, we get ⇒ 2 sin B - 4 (1 - sin B) + 3 = 0 ⇒ 4 sin B + 2 sin B - 1 = 0 This is a quadratic equation in Sine, So, $\sin B = \frac{-2 \pm \sqrt{4+16}}{2 \times 4} = \frac{-1 \pm \sqrt{5}}{4}$ Now since $18^{\circ}$ is first quadrant so $\sin 18^{\circ}$ is positive Therefore, $\sin B = \frac{-1+\sqrt{5}}{4}$ Now, $\cos 36^{\circ} = \cos (2 \times 18^{\circ})$ $\Rightarrow \cos 36^{\circ} = 1 - 2 \sin^{2} 18^{\circ}$ $\Rightarrow \cos 36^{\circ} = 1 - 2\left(\frac{\sqrt{5}-1}{4}\right)^{2}$ $= \frac{16-2\times(5+1-2\sqrt{5})}{16} = \frac{4+4\sqrt{5}}{16}$ $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ As we know, $\sin^{2} \theta + \cos^{2} \theta = 1$ $\Rightarrow \sin \theta = \sqrt{1 - \cos^{2} \theta}$ Now, $\sin 36^{\circ} = \sqrt{1 - \left(\frac{\sqrt{5}+1}{4}\right)^{2}}$ $= \frac{\sqrt{10-2\sqrt{5}}}{4}$