Unit 14 Review Exercises
Trigonometric Functions
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed. Questions marked AHL are HL only.
1 Paper 2 A solid is made from a cylinder of radius \(3\) cm and height \(10\) cm, with a hemisphere of radius \(3\) cm fixed on top. The base of the cylinder is flat.
(a) Find the volume of the solid.
(b) Find the total surface area of the solid.
(a) \(V = \pi(3^2)(10) + \dfrac{2}{3}\pi(3^3) = 90\pi + 18\pi = 108\pi \approx 339\ \text{cm}^3\).
(b) The curved surface of the cylinder is \(2\pi(3)(10) = 60\pi\), the flat base is \(\pi(3^2) = 9\pi\), and the curved surface of the hemisphere is \(2\pi(3^2) = 18\pi\). The top of the cylinder is covered by the hemisphere, so it is not counted.
\[ A = 60\pi + 9\pi + 18\pi = 87\pi \approx 273\ \text{cm}^2 . \]
See 14.1.
2 Paper 1 From a point \(A\) on level ground, the angle of elevation of the top of a tower is \(30^\circ\). After walking \(40\) m straight towards the tower, the angle of elevation is \(60^\circ\). Find the exact height of the tower.
Let the height be \(h\). The horizontal distances from the tower are \(\dfrac{h}{\tan 30^\circ} = \sqrt{3}\,h\) and \(\dfrac{h}{\tan 60^\circ} = \dfrac{h}{\sqrt{3}}\).
\[ \sqrt{3}\,h - \frac{h}{\sqrt{3}} = 40 \;\Longrightarrow\; \frac{2h}{\sqrt{3}} = 40 \;\Longrightarrow\; h = 20\sqrt{3}\ \text{m} . \]
See 14.2.
3 Paper 2 In triangle \(ABC\), \(AB = 9\) cm, \(BC = 7\) cm and \(\hat{A} = 40^\circ\).
(a) Find the two possible values of \(\hat{C}\).
(b) Find the smaller possible area of the triangle.
(a) Sine rule: \(\dfrac{\sin C}{9} = \dfrac{\sin 40^\circ}{7}\), so \(\sin C = 0.8264\ldots\)
\(\hat{C} = 55.7^\circ\) or \(\hat{C} = 180^\circ - 55.7^\circ = 124.3^\circ\). Both are possible, because \(124.3^\circ + 40^\circ < 180^\circ\) (the ambiguous case).
(b) The obtuse \(\hat{C}\) gives the smaller triangle: \(\hat{B} = 180^\circ - 40^\circ - 124.27^\circ = 15.73^\circ\).
\[ \text{Area} = \frac{1}{2}(9)(7)\sin 15.73^\circ \approx 8.54\ \text{cm}^2 . \]
(The other triangle has area \(31.3\ \text{cm}^2\).) See 14.3.
4 Paper 2 A boat sails \(12\) km from a harbour \(H\) on a bearing of \(050^\circ\) to a point \(P\). It then sails \(9\) km on a bearing of \(140^\circ\) to a point \(Q\).
(a) Show that \(\angle HPQ = 90^\circ\).
(b) Find the distance \(HQ\).
(c) Find the bearing of \(Q\) from \(H\).
(a) At \(P\), the bearing back to \(H\) is \(050^\circ + 180^\circ = 230^\circ\). The angle between the bearings \(230^\circ\) and \(140^\circ\) is \(230^\circ - 140^\circ = 90^\circ\).
(b) \(HQ = \sqrt{12^2 + 9^2} = 15\) km.
(c) \(\angle PHQ = \tan^{-1}\left(\dfrac{9}{12}\right) = 36.9^\circ\). The bearing is \(050^\circ + 36.9^\circ = 086.9^\circ\).
See 14.4.
5 Paper 1 A sector of a circle has radius \(6\) cm and perimeter \(20\) cm.
(a) Find the angle of the sector in radians.
(b) Find the area of the sector.
(a) The arc length is \(20 - 2(6) = 8\) cm. From \(l = r\theta\): \(\;\theta = \dfrac{8}{6} = \dfrac{4}{3}\).
(b) \(A = \dfrac{1}{2}(6^2)\left(\dfrac{4}{3}\right) = 24\ \text{cm}^2\).
See 14.5.
6 Paper 1
(a) Find the exact values of \(\sin\dfrac{4\pi}{3}\), \(\;\cos\left(-\dfrac{7\pi}{4}\right)\) and \(\tan\dfrac{11\pi}{6}\).
(b) Given that \(\cos\theta = -\dfrac{2}{5}\) and \(\pi < \theta < \dfrac{3\pi}{2}\), find the exact values of \(\sin\theta\) and \(\tan\theta\).
(a) \(\sin\dfrac{4\pi}{3} = -\dfrac{\sqrt{3}}{2}\) (third quadrant), \(\;\cos\left(-\dfrac{7\pi}{4}\right) = \cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}\), \(\;\tan\dfrac{11\pi}{6} = -\dfrac{1}{\sqrt{3}} = -\dfrac{\sqrt{3}}{3}\) (fourth quadrant).
(b) In the third quadrant \(\sin\theta < 0\): \(\;\sin\theta = -\sqrt{1 - \dfrac{4}{25}} = -\dfrac{\sqrt{21}}{5}\), and \(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\sqrt{21}}{2}\).
See 14.6.
7 Paper 1 Let \(f(x) = 3\sin\left(2\left(x - \dfrac{\pi}{6}\right)\right) + 1\).
(a) Write down the amplitude, the period and the range of \(f\).
(b) Find the coordinates of the maximum point of the graph for \(0 \leq x \leq \pi\).
(c) Write down the period of \(g(x) = 2\tan 3x\).
(a) Amplitude \(3\), period \(\dfrac{2\pi}{2} = \pi\), range \(-2 \leq f(x) \leq 4\).
(b) The maximum is when \(2\left(x - \dfrac{\pi}{6}\right) = \dfrac{\pi}{2}\), so \(x = \dfrac{\pi}{6} + \dfrac{\pi}{4} = \dfrac{5\pi}{12}\). The point is \(\left(\dfrac{5\pi}{12}, 4\right)\).
(c) The period of \(\tan x\) is \(\pi\), so the period of \(\tan 3x\) is \(\dfrac{\pi}{3}\).
8 Paper 1 AHL Find the exact value of
(a) \(\sec\dfrac{2\pi}{3}\)
(b) \(\arcsin\left(-\dfrac{1}{2}\right)\)
(c) \(\cos\left(\arcsin\dfrac{3}{5}\right)\)
(a) \(\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}\), so \(\sec\dfrac{2\pi}{3} = -2\).
(b) The range of \(\arcsin\) is \(\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\), so \(\arcsin\left(-\dfrac{1}{2}\right) = -\dfrac{\pi}{6}\).
(c) Let \(\alpha = \arcsin\dfrac{3}{5}\), an angle between \(0\) and \(\dfrac{\pi}{2}\). Then \(\cos\alpha = \sqrt{1 - \dfrac{9}{25}} = \dfrac{4}{5}\).
See 14.9.
9 Paper 2 The depth of water in a harbour is \(d\) metres, \(t\) hours after high tide. It is modelled by \(d(t) = a\cos(bt) + c\). At high tide (\(t = 0\)) the depth is \(9\) m, and the next low tide is \(6.2\) hours later, with depth \(3\) m.
(a) Find \(a\), \(b\) and \(c\).
(b) Find the first time after high tide when the depth is \(5\) m.
(a) \(c = \dfrac{9 + 3}{2} = 6\) and \(a = \dfrac{9 - 3}{2} = 3\) (positive, because \(d\) is largest at \(t = 0\)). The period is \(2 \times 6.2 = 12.4\) hours, so \(b = \dfrac{2\pi}{12.4} = \dfrac{\pi}{6.2} \approx 0.507\).
(b) \(3\cos(bt) + 6 = 5\), so \(\cos(bt) = -\dfrac{1}{3}\) and \(bt = \cos^{-1}\left(-\dfrac{1}{3}\right) = 1.9106\ldots\)
\(t = \dfrac{1.9106\ldots \times 6.2}{\pi} \approx 3.77\) hours.
See 14.10.
10 Paper 1 Show that \(\tan x + \dfrac{1}{\tan x} = \dfrac{1}{\sin x\cos x}\).
\[ \tan x + \frac{1}{\tan x} = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \frac{1}{\sin x\cos x} . \;\blacksquare \]
See 14.11.
11 Paper 1 Solve \(2\sin^2 x + 3\cos x = 3\) for \(0 \leq x \leq 2\pi\).
Replace \(\sin^2 x\) by \(1 - \cos^2 x\): \(\;2 - 2\cos^2 x + 3\cos x - 3 = 0\), so
\[ 2\cos^2 x - 3\cos x + 1 = 0 \;\Longrightarrow\; (2\cos x - 1)(\cos x - 1) = 0 . \]
\(\cos x = 1\) gives \(x = 0, 2\pi\); \(\;\cos x = \dfrac{1}{2}\) gives \(x = \dfrac{\pi}{3}, \dfrac{5\pi}{3}\).
See 14.12.
12 Paper 1 Given that \(\cos x = \dfrac{1}{3}\) and \(0 < x < \dfrac{\pi}{2}\), find the exact value of
(a) \(\sin 2x\) \(\qquad\) (b) \(\cos 2x\) \(\qquad\) (c) \(\tan 2x\)
\(\sin x = \sqrt{1 - \dfrac{1}{9}} = \dfrac{2\sqrt{2}}{3}\).
(a) \(\sin 2x = 2 \cdot \dfrac{2\sqrt{2}}{3} \cdot \dfrac{1}{3} = \dfrac{4\sqrt{2}}{9}\).
(b) \(\cos 2x = 2\left(\dfrac{1}{9}\right) - 1 = -\dfrac{7}{9}\).
(c) \(\tan 2x = \dfrac{4\sqrt{2}}{9} \div \left(-\dfrac{7}{9}\right) = -\dfrac{4\sqrt{2}}{7}\).
See 14.13.
13 Paper 1 AHL
(a) By writing \(3x = 2x + x\), show that \(\sin 3x = 3\sin x - 4\sin^3 x\).
(b) Hence solve \(\sin 3x = 2\sin x\) for \(0 \leq x \leq \pi\).
(a)
\[ \begin{aligned} \sin 3x &= \sin 2x\cos x + \cos 2x\sin x \\ &= 2\sin x\cos^2 x + (1 - 2\sin^2 x)\sin x \\ &= 2\sin x(1 - \sin^2 x) + \sin x - 2\sin^3 x \\ &= 3\sin x - 4\sin^3 x . \;\blacksquare \end{aligned} \]
(b) \(3\sin x - 4\sin^3 x = 2\sin x\), so \(\sin x\,(1 - 4\sin^2 x) = 0\).
\(\sin x = 0\) gives \(x = 0, \pi\). \(\;\sin x = \pm\dfrac{1}{2}\): in \([0, \pi]\) only \(\sin x = \dfrac{1}{2}\) is possible, giving \(x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}\).
So \(x = 0, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \pi\).
See 14.13.