(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°).
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 – 2sqrt{3} – 2sqrt{3} + 2}{6 – 2})
= (frac{8 – 4sqrt{3}}{4})
= (2 – sqrt{3})