(a) The gradient of the tangent to the curve \(y = 4x^{3}\) at points P and Q is 108. Find the coordinates of P and Q.
(b) Given that \(A = 45°, B = 30°, \sin (A + B) = \sin A \cos B + \sin B \cos A\) and \(\cos (A + B) = \cos A \cos B - \sin A \sin B\)
(i) Show that \(\sin 15° = \frac{\sqrt{6} - \sqrt{2}}{4}\) and \(\cos 15° = \frac{\sqrt{6} + \sqrt{2}}{4}\)
(ii) hence find \(\tan 15°\).
(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
\((\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}})(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}) = \frac{6 - 2\sqrt{3} - 2\sqrt{3} + 2}{6 - 2}\)
= \(\frac{8 - 4\sqrt{3}}{4}\)
= \(2 - \sqrt{3}\)
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