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Edexcel Pure Paper 1 2024 Question 12

Edexcel Pure Paper 1 2024 Question 12

Edexcel Pure Paper 1 2024 Question 12 – Trigonometric Modelling and Interpretation

❓ The Question

🧠 Before you start

This one looks long, but it’s really just the same idea repeated.

Everything comes from rewriting:

140\cos\theta – 480\sin\theta

into a single cosine form.

Once that’s done, the rest is just using it carefully.

Where people struggled wasn’t the maths — it was:

  • mixing degrees and radians

  • not using earlier parts properly

or losing track of what the model actually means

✏️ Working

Part (a)

We want:

140\cos\theta – 480\sin\theta = K\cos(\theta + \alpha)

Compare coefficients:

K\cos\alpha = 140
 K\sin\alpha = 480

Find K:

K = \sqrt{140^2 + 480^2}

= \sqrt{19600 + 230400}

= \sqrt{250000} = 500

Now find \alpha:

\tan\alpha = \frac{480}{140}

\alpha \approx 73.74^\circ

So:

140\cos\theta – 480\sin\theta = 500\cos(\theta + 73.74^\circ)

Part (b)(i)

Model is:

R = A + 140\cos(30t) – 480\sin(30t)

Use part (a):

R = A + 500\cos(30t + 73.74^\circ)

We are told the maximum value is 1500.

Max of cosine = 1, so:

A + 500 = 1500

A = 1000

So the model becomes:

R = 1000 + 500\cos(30t + 73.74^\circ)

Part (b)(ii)

Minimum occurs when cosine = -1:

R_{\min} = 1000 – 500 = 500

Part (c)

Minimum is said to happen in middle of April.

That corresponds to:

t = 3.5

Substitute:

R = 1000 + 500\cos(30(3.5) + 73.74)

= 1000 + 500\cos(178.74^\circ)

This is approximately:

R \approx 500

So the model predicts the minimum at about the right time.

Conclusion:
 The model is reasonable / reliable.

Part (d)

We are told foxes are minimum.

So:

\sin(30t + 70) = -1

That happens at:

30t + 70 = 270

30t = 200

t = \frac{20}{3}

Now substitute into rabbit model:

R = 1000 + 500\cos(30 \cdot \frac{20}{3} + 73.74)

= 1000 + 500\cos(200 + 73.74)

= 1000 + 500\cos(273.74^\circ)

\approx 1032

✅ Final Answer

1032 \text{ (or } 1033\text{)}

🎯 Where the Marks Are

Part (a):

  • B1 correct K

  • M1 correct trig setup

  • A1 correct angle

Part (b):

  • B1 correct model

  • B1 correct minimum

Part (c):

  • M1 correct substitution

  • A1 valid interpretation

Part (d):

  • M1 solving trig equation

  • A1 correct t

  • M1 substitution

  • A1 final answer

⚠️ What Went Wrong

A lot of issues here were quite small, but they added up.

In part (a), some used radians instead of degrees. That immediately gives the wrong angle. Others reversed the tangent fraction and used \frac{140}{480} instead.

Part (b) caused more problems than expected.

Many didn’t use the result from part (a), even though it made things much easier. Instead, they tried to work directly with sine and cosine, which made it harder to identify the maximum.

In part (c), interpretation was the issue.

Some used values like t = 4 or t = 4.5, rather than recognising “middle of April” as about 3.5 months.

That small detail made a difference.

Part (d) was generally done well, but mistakes still appeared.

Some solved:

\sin(…) = 1

instead of -1.

Others used angles like 90° or 180°, which don’t give the minimum.

A few ended up with negative time values, which should have been a clear sign something had gone wrong.

💡 The takeaway

This question is about consistency.

Once you get the harmonic form, everything else follows from it.

But you need to:

  • stay in degrees

  • use earlier results

and interpret carefully

🚀 If This Felt Difficult

If this felt a bit scattered, that’s normal.

This type of modelling question mixes algebra, trig and interpretation.

Working through more of these in a maths revision support course helps bring those ideas together.

And if you want more targeted help with A level maths, especially on trig modelling, regular practice makes a big difference.

🔗 Next Steps

👨‍🏫Author Bio

S. Mahandru is an experienced A Level Maths teacher and founder of Exam.Tips, specialising in exam-focused revision techniques and helping students achieve top grades.

❓ Frequently Asked Questions

📌 Why use harmonic form?

It makes maximum and minimum values much easier to identify.

The question is set in degrees — using radians gives wrong answers.

Use cosine = -1 once in harmonic form.

Not using part (a) properly in later parts.

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