A-level · Trigonometry

Modelling with trigonometryExam-style questions

Using a sine or cosine model for something that repeats — a tide, a day's temperature, the hours of daylight, a turning wheel. What the midline, amplitude, period and shift mean in context; reading greatest and least values; building a complete equation from a description; finding every time a value is reached, to the nearest minute; using the R-form when the model is a sum of two waves; and saying where a model breaks down.

10 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAllLevel 4Level 5
  1. 1
    A ship is moored beside a floating pontoon in a harbour. Waves make the deck of the ship and the surface of the pontoon rise and fall.

    The heights are modelled for a period of 60 seconds. At time tt seconds after the start of the period, the deck of the ship is ss metres above mean sea level and the surface of the pontoon is pp metres above mean sea level, where

    s=4.2−0.35sin⁡ts = 4.2 - 0.35\sin t

    and

    p=0.8−0.6cos⁡tp = 0.8 - 0.6\cos t
    (a) Show that, at the start of the period, the deck is 4 metres above the surface of the pontoon.

    (1 mark)
    (b) At time tt seconds the deck is dd metres above the surface of the pontoon.

    Show that

    d=3.4+Rcos⁡(t+α)d = 3.4 + R\cos(t + \alpha)

    stating the values of the positive constants RR and α\alpha

    (5 marks)
    (c) Hence find the least height of the deck above the surface of the pontoon during the 60 seconds, giving your answer to the nearest centimetre.

    (2 marks)
    Level 4 · 8 marksTry it
  2. 2
    A child is playing on a swing in a park.

    The seat of the swing moves backwards and forwards. In a model of this motion
    • the seat is at its greatest height, 1.71.7 m above the ground, at each end of a swing
    • the seat is at its least height, 0.50.5 m above the ground, as it passes through the lowest point of a swing
    The height of the seat above the ground, HH m, tt seconds after the seat is released from one end of a swing, is modelled by the equation

    H=a+b(t−1)2H = a + b(t - 1)^2

    where aa and bb are positive constants.
    (a) Find a complete equation for the model.

    (3 marks)
    (b) Find, according to the model, the height of the seat above the ground when t=0.5t = 0.5

    (1 mark)

    A second model for the height of the seat above the ground, HH m, tt seconds after the seat is released, is

    H=0.6sin⁡(180t+α)∘+βH = 0.6\sin(180t + \alpha)^\circ + \beta

    where α\alpha and β\beta are constants and 0⩽α<3600 \leqslant \alpha < 360
    (c) Find a complete equation for the second model.

    (2 marks)

    The child keeps swinging in the same way for 33 minutes.
    (d) Explain why the second model is more appropriate than the first model over this time.

    (1 mark)
    Level 4 · 7 marksTry it
  3. 3
    A cyclist rides at a constant speed along a straight horizontal road. A small reflector is fixed to one of the spokes of the front wheel of the bicycle.

    The height of the reflector above the road, HH cm, tt seconds after the cyclist passes a lamp post, is modelled by the equation

    H=∣Asin⁡(bt+α)∘∣H = \left|A\sin(bt + \alpha)^\circ\right|

    where AA, bb and α\alpha are constants.

    In the model
    • the reflector reaches its greatest height above the road, 5858 cm, once in each revolution of the wheel
    • the wheel makes one complete revolution every 0.60.6 seconds
    • when t=0t = 0 the reflector is 2020 cm above the road and is rising
    (a) Find a complete equation for the model. Give AA and bb as exact values and give α\alpha correct to 3 significant figures.

    (4 marks)

    A second model for the height of the reflector above the road is

    H=∣Asin⁡(bt+α)∘∣+dH = \left|A\sin(bt + \alpha)^\circ\right| + d

    where dd is a positive constant.
    (b) Give a reason why the second model is more realistic than the model in part (a).

    (1 mark)
    Level 4 · 5 marksTry it
  4. 4
    (a) Express 8cos⁡θ+15sin⁡θ8\cos\theta + 15\sin\theta in the form Rcos⁡(θ−α)R\cos(\theta - \alpha), where R>0R > 0 and 0<α<90∘0 < \alpha < 90^\circ

    State the value of RR and give the value of α\alpha, in degrees, to 2 decimal places.

    (3 marks)

    The air temperature, T ∘T\ ^\circC, at a weather station in a desert, tt hours after midnight on a particular day, is modelled by the equation

    T=15−8cos⁡(15t)∘−15sin⁡(15t)∘0⩽t<24T = 15 - 8\cos(15t)^\circ - 15\sin(15t)^\circ \qquad 0 \leqslant t < 24
    (b) Using your answer to part (a), find

    (i) the maximum temperature at the weather station on that day according to the model,

    (ii) the time of day at which this maximum temperature occurs according to the model, giving your answer to the nearest minute.

    (3 marks)

    On that day, the lowest temperature recorded at the weather station was 4 ∘4\ ^\circC, at 06:30
    (c) Use this information to comment on the model.

    (2 marks)
    Level 4 · 8 marksTry it
  5. 5
    The number of hours of daylight, DD, in a town on the day that is tt days after 21 March is modelled by the equation

    D=A+Bsin⁡(kt)∘D = A + B\sin(kt)^\circ

    where AA, BB and kk are positive constants.

    Given that, according to the model,
    • DD is periodic with period 365365 days
    • the greatest number of hours of daylight in the town is 16.416.4
    • the least number of hours of daylight in the town is 7.67.6
    (a) find a complete equation for the model.

    (3 marks)
    (b) Write down the value of tt on which, according to the model, the town has the greatest number of hours of daylight.

    (1 mark)
    (c) Use the model to find the number of hours of daylight in the town when t=41t = 41, giving your answer in hours and minutes to the nearest minute.

    (2 marks)

    On 21 December, when t=275t = 275, the town actually had 7.97.9 hours of daylight.
    (d) Use this information to comment on the model.

    (2 marks)
    Level 4 · 8 marksTry it

2 more questions on modelling with trigonometry

Free with an account — just an email, no card. The whole bank, every mark scheme and worked solution.

Start free