Edexcel A level Maths (9MA0) · Pure mathematics › Trigonometry
Practise Trigonometry in context. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy7 marks
The depth of water, \(D\) metres, at the entrance to a river estuary is modelled by the equation \[D = 5 + 3.5\sin(30t)^\circ,\] where \(t\) is the time in hours after midnight, \(0 \le t \lt 24\).
(a) Find the maximum depth of water predicted by the model and the value of \(t\) when this maximum first occurs.[2]
(b) Find the depth of water at 01:30, giving your answer to 2 decimal places.[2]
(c) Find the first time after midnight at which the depth of water is 6.1 m. Give your answer to the nearest minute.[3]
Show the answer and mark scheme
(a)Answer: Maximum \(8.5\) m at \(t = 3\)
B1 for \(8.5\) m
B1 for \(t = 3\) (03:00)
Worked solution: The maximum of \(\sin\) is 1, so the maximum depth is \(5 + 3.5 = 8.5\) m, first when \(30t = 90\), i.e. \(t = 3\).
(b)Answer: \(7.47\) m
M1 for substituting \(t = 1.5\)
A1 for \(7.47\)
Worked solution: \(D = 5 + 3.5\sin(30 \times 1.5)^\circ = 7.47\) m.
(c)Answer: 00:37 (\(t = 0.61\))
M1 for \(\sin(30t)^\circ = \frac{11}{35}\) (oe)
M1 for a correct method for the smallest positive value of \(30t\), e.g. \(30t = 18.31\ldots^\circ\)
A1 for 00:37 (accept \(t = 0.61\))
Worked solution: \(\sin(30t)^\circ = 0.3143\). The principal value gives \(30t = 18.31\ldots^\circ\), so \(t = 0.610\ldots\) hours, i.e. 00:37.
Question 2Medium8 marks
The temperature, \(T\) °C, inside a greenhouse \(t\) hours after midnight is modelled by \[T = 18 + 3\sin\left(\frac{\pi t}{12}\right) - 4\cos\left(\frac{\pi t}{12}\right), \quad 0 \le t \lt 24\]
(a) Express \(3\sin x - 4\cos x\) in the form \(R\sin(x - \alpha)\), where \(R \gt 0\) and \(0 \lt \alpha \lt \frac{\pi}{2}\). Give the value of \(\alpha\) to 4 decimal places.[3]
(b) Find the maximum temperature predicted by the model and the time of day, to the nearest minute, at which it occurs.[3]
(c) Give two reasons why this model may not be suitable for predicting the temperature in the greenhouse on every day of a month.[2]
Show the answer and mark scheme
(a)Answer: \(5\sin(x - 0.9273)\)
M1 for \(R = \sqrt{3^2 + 4^2} = 5\)
M1 for \(\tan\alpha = \frac{4}{3}\)
A1 for \(\alpha = 0.9273\)
Worked solution: \(R\sin(x - \alpha) = R\sin x\cos\alpha - R\cos x\sin\alpha\), so \(R\cos\alpha = 3\), \(R\sin\alpha = 4\). \(R = 5\), \(\tan\alpha = \frac{4}{3}\), \(\alpha = 0.9273\). So \(3\sin x - 4\cos x = 5\sin(x - 0.9273)\).
(b)Answer: 23 °C at 09:33
B1 for 23 °C
M1 for \(\frac{\pi t}{12} - 0.9273 = \frac{\pi}{2}\)
A1 for \(t = 9.54\ldots\), i.e. 09:33 (or 9:33 am)
Worked solution: The maximum of \(\sin\) is 1, so the maximum temperature is \(18 + 5 = 23\) °C. It occurs when \(\frac{\pi t}{12} - 0.9273 = \frac{\pi}{2}\), so \(t = \frac{12}{\pi}(1.5708 + 0.9273) = 9.542\) hours, which is 9 hours 33 minutes after midnight: 09:33.
(c)Answer: It predicts exactly the same temperatures every day, but the weather varies (cloudy or sunny days); temperatures change with the season over a month; the maximum is predicted at 09:33, whereas a greenhouse is usually hottest in the early afternoon.
B1 for one sensible reason in context
B1 for a second, different reason in context
Worked solution: For example: the model gives exactly the same temperature pattern every day, but cloudy and sunny days give very different temperatures; and over a month the season changes, so the average temperature and the hours of daylight change.
Question 3Hard12 marks
An observation wheel has diameter 120 m and its centre is 62 m above the ground. The wheel rotates at a constant speed, completing one revolution every 30 minutes. Sofia gets on at the lowest point of the wheel, at time \(t = 0\).
(a) Show that Sofia's height, \(h\) metres, above the ground \(t\) minutes later can be modelled by \[h = 62 - 60\cos(12t)^\circ.\][3]
(b) Find the length of time, during each complete revolution, for which Sofia is more than 74 m above the ground. Give your answer in minutes to 1 decimal place.[4]
(c) Erin gets on at the lowest point a quarter of a revolution after Sofia. Find the first time after Erin gets on at which Sofia and Erin are at the same height, and find this height. Give your answers to 3 significant figures.[5]
Show the answer and mark scheme
(a)Answer: \(h = 62 - 60\cos(12t)^\circ\) (shown)
B1 for the radius 60 m and the centre height 62 m used as \(62 - 60\cos(\ldots)\)
M1 for a negative cosine term because Sofia starts at the lowest point
A1* for \(12t\) from \(360 \div 30\)
Worked solution: The wheel turns through \(360^\circ\) every 30 minutes, i.e. \(12^\circ\) per minute. After \(t\) minutes Sofia has turned through \(12t^\circ\) from the lowest point, so is \(60\cos(12t)^\circ\) below the centre. Hence \(h = 62 - 60\cos(12t)^\circ\).
(b)Answer: \(13.1\) minutes
M1 for \(62 - 60\cos(12t)^\circ = 74\) leading to \(\cos(12t)^\circ = -0.2\)
A1 for \(t = 8.46\) (or \(12t = 101.54^\circ\))
dM1 for finding the second value \(t = 21.54\) and subtracting
A1 for \(13.1\) minutes
Worked solution: \(\cos(12t)^\circ = \frac{62 - 74}{60} = -0.2\), so \(12t = 101.53\ldots^\circ\) or \(258.46\ldots^\circ\), i.e. \(t = 8.461\ldots\) or \(21.538\ldots\). Sofia is above 74 m between these times, for \(21.538\ldots - 8.461\ldots = 13.1\) minutes.
(c)Answer: \(t = 18.8\) minutes after Sofia gets on (\(11.3\) minutes after Erin), height \(104\) m
B1 for Erin's height \(62 - 60\cos(12t - 90)^\circ\) for \(t \ge 7.5\)
M1 for equating the heights: \(\cos(12t)^\circ = \cos(12t - 90)^\circ = \sin(12t)^\circ\)
M1 for \(\tan(12t)^\circ = 1\) and choosing the solution with \(12t \ge 90\)
A1 for \(t = 18.8\)
A1 for height \(104\) m (exact \(62 + 30\sqrt{2}\))