(a) The gradient of the tangent to the curve (y = 4x^{3}) at points P

Explanation

(a) \(y = 4x^{3}\)

\(\frac{\mathrm d y}{\mathrm d x} = 12x^{2}\)

\(12x^{2} = 108 \implies x^{2} = 9\)

\(\therefore x = \pm 3\)

When x = -3, \(y = 4(-3^{3}) = -108\)

When x = 3, \(y = 4(3^{3}) = 108\)

\(\therefore P = (-3, -108) ; Q = (3, 108)\)

(b)(i) \(\sin (A - B) = \sin A \cos B - \sin B \cos A\)

\(\sin 15 = \sin (45 - 30) = \sin 45 \cos 30 - \sin 30 \cos 45\)

= \((\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) - (\frac{1}{2})(\frac{\sqrt{2}}{2})\)

= \(\frac{\sqrt{6} - \sqrt{2}}{4}\)

\(\cos (A - B) = \cos A \cos B + \sin A \sin B\)

\(\cos 15 = \cos (45 - 30) \)

= \(\cos 45 \cos 30 + \sin 45 \sin 30\)

= \((\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2})\)

= \(\frac{\sqrt{6} + \sqrt{2}}{4}\)

(b)(ii) \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)

\(\tan 15 = \frac{\sqrt{6} - \sqrt{2}}{4} \div \frac{\sqrt{6} + \sqrt{2}}{4}\)

= \(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}\)

Ratinalizing, we have

\((\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}})(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}) = \frac{6 - 2\sqrt{3} - 2\sqrt{3} + 2}{6 - 2}\)

= \(\frac{8 - 4\sqrt{3}}{4}\)

= \(2 - \sqrt{3}\)