(a) The gradient of the tangent to the curve \(y =... | WAEC SSCE - Further Mathematics - 2006 (Essay)

Question 1 Report

(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 \(\tan 15°\).

Answer

(a) For \(y=4x^3\), the gradient is \(\dfrac{dy}{dx}=12x^2\). Set it equal to 108:

\[12x^2=108\Rightarrow x^2=9\Rightarrow x=\pm 3.\]

At \(x=3:\ y=4(3)^3=108\); at \(x=-3:\ y=4(-3)^3=-108\).

So \(P(3,\ 108)\) and \(Q(-3,\ -108)\).

(b)(i) Use \(A=45^\circ,\ B=30^\circ\) so that \(A+B=75^\circ\). Since \(\sin 15^\circ=\cos 75^\circ\) and \(\cos 15^\circ=\sin 75^\circ\):

\[\sin 75^\circ=\sin A\cos B+\sin B\cos A=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\] \[\cos 75^\circ=\cos A\cos B-\sin A\sin B=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\cdot\frac12=\frac{\sqrt6-\sqrt2}{4}.\]

Hence \(\sin 15^\circ=\cos 75^\circ=\dfrac{\sqrt6-\sqrt2}{4}\) and \(\cos 15^\circ=\sin 75^\circ=\dfrac{\sqrt6+\sqrt2}{4}\), as required.

(ii) \(\tan 15^\circ=\dfrac{\sin 15^\circ}{\cos 15^\circ}=\dfrac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}\). Rationalise:

\[\tan 15^\circ=\frac{(\sqrt6-\sqrt2)^2}{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)}=\frac{6-2\sqrt{12}+2}{6-2}=\frac{8-4\sqrt3}{4}=2-\sqrt3.\]

So \(\tan 15^\circ=2-\sqrt3\).

Read lesson note on Trigonometry (WAEC)

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