AQA A-Level Mathematics Paper 3, June 2025: Question 10
10 marks · Medium difficulty · Modelling
Express $2\sin x + 5\cos x$ in the form $R\sin(x + \alpha)$ and use this model to find the minimum/maximum water temperature and the number of weeks the temperature exceeds 15 °C.
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Mark scheme
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How to answer it
Harmonic Form and Trigonometric Temperature Modelling
📋 What this question tests
- Expressing linear combinations of sine and cosine in the harmonic form R sin(x + α) using compound angle expansions.
- Applying trigonometric transformations to a real-world periodic context (water temperature over a year).
- Deducing minimum and maximum values of a negated sine model: T = A - R sin(θ) .
- Solving trigonometric inequalities and converting calendar units (days to weeks).
Part (a): Expressing 2 sin x + 5 cos x in Harmonic Form
Target: R sin(x + α) with R > 0 and 0° ≤ α ≤ 90° [3 Marks]
📐 Step-by-Step Calculation
- Expand the target identity:
R sin(x + α) = R sin x cos α + R cos x sin α - Equate coefficients with 2 sin x + 5 cos x:
• R cos α = 2
• R sin α = 5 - Calculate R:
R² = 2² + 5² = 4 + 25 = 29
R = √29 (or awfw 5.38 to 5.4) - Calculate α:
tan α = (R sin α) / (R cos α) = 5 / 2 = 2.5
α = arctan(2.5) = 68.198...° ≈ 68.2°
✅ Final Answer & Mark Breakdown
√29 sin(x + 68.2°)
[B1] (AO 1.1b): Correct value of R = √29 .
[M1] (AO 1.1a): Correct method for α, showing tan α = 5/2 or using compound angles to find α ≈ 68.2° .
[A1/R1] (AO 2.1): Complete expression written correctly in the required form.
[M1] (AO 1.1a): Correct method for α, showing tan α = 5/2 or using compound angles to find α ≈ 68.2° .
[A1/R1] (AO 2.1): Complete expression written correctly in the required form.
🧠 Exam Technique
Always write out the expanded form R sin x cos α + R cos x sin α directly below the given expression. Lining up terms prevents mixing up whether tan α is 5/2 or 2/5.
❌ Common Errors
- Inverting the ratio: Writing tan α = 2/5 , which gives α = 21.8° (this corresponds to R cos(x - α) , not R sin(x + α) ).
- Radian confusion: Giving α in radians (1.19 rad) without reading the question specification: 0° ≤ α ≤ 90° .
Part (b)(i) & (b)(ii): Minimum and Maximum Values
Model: T = 14.2 - √29 sin(d + 68.2)° [1 Mark + 1 Mark]
💡 Key Knowledge: The Subtraction Flip
Because the expression is subtracted, the behaviour is inverted:
- Lowest Temperature: Occurs when the subtracted sine term is at its maximum positive value ( +1 ).
- Highest Temperature: Occurs when the subtracted sine term is at its minimum negative value ( -1 ).
📐 Calculations
Part (b)(i): Lowest T occurs when sin(d + 68.2°) = +1
- d + 68.2 = 90
- d = 90 - 68.2 = 21.8
Part (b)(ii): Maximum temperature occurs when sin(d + 68.2°) = -1
- T_max = 14.2 - √29(-1) = 14.2 + √29
- T_max = 14.2 + 5.385... = 19.585... ≈ 19.6 °C
✅ Answers & Marks
(b)(i) Value of d: d = 21.8 (or follow-through 90 - their α )
[B1F] (AO 2.2a): 1 mark for deducing d = 21.8
(b)(ii) Maximum Temperature: 19.6 °C (to 1 d.p.)
[B1F] (AO 2.2a): 1 mark for 19.6 °C (deduces 14.2 + √29). Units not penalised.
❌ Common Errors
- Finding max when min was requested: Setting d + 68.2 = 270° for the minimum temperature instead of recognizing the negative sign in front of the bracket.
- Rounding too early: Calculating 14.2 + 5.4 = 19.6 happens to give the same answer here, but always maintain exact values ( √29 ) until the final round-off.
Part (b)(iii): Solving the Inequality for Time in Weeks
Target: Find the number of weeks where T > 15 °C (nearest whole number) [5 Marks]
📐 Step-by-Step Solution
- Set up the inequality/equation:
14.2 - √29 sin(d + 68.2)° > 15 - Rearrange to isolate the sine function:
-√29 sin(d + 68.2)° > 15 - 14.2
-√29 sin(d + 68.2)° > 0.8
Dividing by -√29 reverses the inequality symbol:
sin(d + 68.2)° < -0.8 / √29 ≈ -0.14855 - Find the critical boundary angles:
Principal value: arcsin(-0.14855) = -8.544°
In the cycle from 0° to 360°:
• Angle 1: 180° - (-8.544°) = 188.544°
• Angle 2: 360° + (-8.544°) = 351.456° - Solve for d (number of days):
• d₁ + 68.2 = 188.544° ⇒ d₁ = 188.544 - 68.2 = 120.34 days
• d₂ + 68.2 = 351.456° ⇒ d₂ = 351.456 - 68.2 = 283.26 days - Calculate duration and convert to weeks:
Total days above 15 °C: 283.26 - 120.34 = 162.92 days
(Alternatively: difference in angles = 351.456 - 188.544 = 162.91°)
Number of weeks: 162.92 / 7 = 23.27 weeks
To nearest whole number: 23 weeks
✅ Final Answer & Mark Breakdown
23 weeks
[M1] (AO 3.4): Equating complete temperature model to 15 (or using inequality).
[M1] (AO 1.1a): Correct rearrangement to make sin(d + 68.2) = -0.8/√29 .
[A1] (AO 1.1b): Correct critical values: angles in [188.2, 189] and [351, 351.5] or days in [120, 120.8] and [282.8, 283.3].
[M1] (AO 3.1b): Subtracting positive d values (or angles) and dividing by 7.
[A1] (AO 1.1b): Final answer of 23 (accept 23.1 to 23.4 before rounding).
[M1] (AO 1.1a): Correct rearrangement to make sin(d + 68.2) = -0.8/√29 .
[A1] (AO 1.1b): Correct critical values: angles in [188.2, 189] and [351, 351.5] or days in [120, 120.8] and [282.8, 283.3].
[M1] (AO 3.1b): Subtracting positive d values (or angles) and dividing by 7.
[A1] (AO 1.1b): Final answer of 23 (accept 23.1 to 23.4 before rounding).
❌ Calculation Traps & Examiner Insights
- Forgetting to divide by 7: Leaving the answer as ~163 days loses the final 2 marks. The question explicitly asks for weeks.
- Sign slip when rearranging: When solving 14.2 - ... = 15 , many students mistakenly get +0.8 instead of -0.8 . Always track negative coefficients carefully.
- Finding negative d values: Using -8.54° - 68.2° = -76.7° gives a negative number of days, which lies in the previous year (2009), not in 2010. You must find the two positive solutions within [0, 365].
Topics
Pure Mathematics · E: Trigonometry
Question and mark scheme from the AQA A-Level Mathematics examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.