Addition and double angle formulae and R cos(θ ± α) - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 20 exam-style questions on Addition and double angle formulae and R cos(θ ± α) for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

Express \(\sqrt{7}\sin x + 3\cos x\) in the form \(R\cos(x - \alpha)\), where \(R > 0\) and \(0 < \alpha < 90^\circ\).
Give the exact value of \(R\) and give the value of \(\alpha\), in degrees, to 2 decimal places.

[You may use \(\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B\)]

Model answer

\(R\cos(x - \alpha) = R\cos x\cos\alpha + R\sin x\sin\alpha\)
Comparing with \(3\cos x + \sqrt{7}\sin x\): \(R\cos\alpha = 3\), \(R\sin\alpha = \sqrt{7}\)
\(R = \sqrt{3^2 + (\sqrt{7})^2} = \sqrt{16} = 4\)
\(\tan\alpha = \dfrac{\sqrt{7}}{3} \Rightarrow \alpha = 41.41^\circ\) (2 d.p.)
\(\sqrt{7}\sin x + 3\cos x = 4\cos(x - 41.41^\circ)\)

Explanation

Expand \(R\cos(x - \alpha)\) and compare coefficients to get \(R\cos\alpha = 3\) and \(R\sin\alpha = \sqrt{7}\). Squaring and adding gives R = 4, and dividing gives \(\tan\alpha = \dfrac{\sqrt{7}}{3}\). The common errors are inverting the fraction for \(\tan\alpha\) and working in radians when degrees are asked for.

Mark scheme

1 mark
B1 \(R = 4\); or a correct final answer of \(4\cos(x - 41.41^\circ)\) with no working (scores all 3 marks)
An answer of \(4\cos(x - 41.41^\circ)\) with no working scores 3 marks. Accept \(R = \sqrt{16}\). B0 for \(\pm 4\).
1 mark
M1 \(\tan\alpha = \dfrac{\sqrt{7}}{3}\) (or \(\sin\alpha = \dfrac{\sqrt{7}}{R}\), \(\cos\alpha = \dfrac{3}{R}\) with their R) leading to a value of \(\alpha\)
From \(R\cos\alpha = 3\) and \(R\sin\alpha = \sqrt{7}\). Condone \(\tan\alpha = \pm\dfrac{\sqrt{7}}{3}\) for this mark; \(\tan\alpha = \dfrac{3}{\sqrt{7}}\) (giving 48.59°) is M0. May be implied by awrt 41.4, i.e. in [41.35, 41.45).
1 mark
A1 \(\alpha =\) awrt 41.41, i.e. in [41.405, 41.415)
The degree sign is not required. Answers in radians (0.723) score A0. The final form \(4\cos(x - 41.41^\circ)\) need not be written if R and \(\alpha\) are both stated.

Question 2

5 marks

Figure 1 shows a triangle \(PQR\). The point \(N\) lies on \(PR\) and \(QN\) is perpendicular to \(PR\).
\(QP = a\), \(QR = b\) and \(QN = h\).
Angle \(PQN = A\) and angle \(NQR = B\), where \(A\) and \(B\) are acute angles.

By finding the area of triangle \(PQR\) in two different ways, prove that
\[\sin(A + B) = \sin A\cos B + \cos A\sin B\]

Geometry diagram, marked "Diagram NOT accurately drawn". Triangle PQR. Line segment QN is drawn. N lies on side PR. Angle QNR is marked as a right angle (90°). Angle PQN is marked A. Angle NQR is marked B. Side QP is labelled a. Side QR is labelled b. QN is labelled h.

Model answer

Angle \(PQR = A + B\), so area of \(PQR = \dfrac{1}{2}ab\sin(A + B)\).
In right-angled triangle \(QNR\): \(h = b\cos B\). In right-angled triangle \(QNP\): \(h = a\cos A\).
Area of \(PQN = \dfrac{1}{2} \times QP \times QN \times \sin A = \dfrac{1}{2}a(b\cos B)\sin A\)
Area of \(NQR = \dfrac{1}{2} \times QR \times QN \times \sin B = \dfrac{1}{2}b(a\cos A)\sin B\)
So area of \(PQR = \dfrac{1}{2}ab(\sin A\cos B + \cos A\sin B)\)
Equating: \(\dfrac{1}{2}ab\sin(A + B) = \dfrac{1}{2}ab(\sin A\cos B + \cos A\sin B)\)
Dividing by \(\dfrac{1}{2}ab\): \(\sin(A + B) = \sin A\cos B + \cos A\sin B\), as required.

Explanation

One expression for the area comes from \(\dfrac{1}{2}ab\sin C\) with the whole angle A + B at Q. The other is the sum of the areas of the two right-angled triangles, where the shared height is written as \(b\cos B\) in one and \(a\cos A\) in the other so that every term contains ab. Equating and dividing by \(\dfrac{1}{2}ab\) gives the formula; marks are lost for using the same expression for h in both triangles, which does not lead to the result.

Mark scheme

1 mark
B1 Area of triangle \(PQR = \dfrac{1}{2}ab\sin(A + B)\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: this is a proof. Writing the formula \(\sin(A + B) = \sin A\cos B + \cos A\sin B\) as a quoted result earns nothing.
1 mark
M1 Finds the areas of triangles \(PQN\) and \(NQR\) in terms of the lengths and angles, e.g. \(\dfrac{1}{2}ah\sin A\) and \(\dfrac{1}{2}bh\sin B\), or \(\dfrac{1}{2} \times PN \times h\) and \(\dfrac{1}{2} \times NR \times h\)
Allow \(\dfrac{1}{2}PR \times h\) for the whole triangle with \(PR = PN + NR\).
1 mark
M1 (dM1) Uses the right-angled triangles to write the lengths in terms of a, b, A and B: \(h = b\cos B\) and \(h = a\cos A\) (or \(PN = a\sin A\), \(NR = b\sin B\))
Depends on the previous M mark. The substitutions must be chosen so that every term contains ab.
1 mark
A1 Total area \(= \dfrac{1}{2}ab\sin A\cos B + \dfrac{1}{2}ab\cos A\sin B\)
E.g. \(\dfrac{1}{2}a(b\cos B)\sin A + \dfrac{1}{2}b(a\cos A)\sin B\), or \(\dfrac{1}{2}(a\sin A)(b\cos B) + \dfrac{1}{2}(b\sin B)(a\cos A)\).
1 mark
A1* Equates the two expressions for the area and divides by \(\dfrac{1}{2}ab\) to reach \(\sin(A + B) = \sin A\cos B + \cos A\sin B\), with a conclusion, cso
All previous marks must have been scored. A minimal conclusion such as 'as required' or 'proved' is needed.

Question 3

7 marks

In this question you must show all stages of your working.

\[g(\theta) = \dfrac{30}{21 + 8\cos\theta + 15\sin\theta}, \quad 0 \leqslant \theta < 360^\circ\]

By writing \(8\cos\theta + 15\sin\theta\) in the form \(R\cos(\theta - \alpha)\), where \(R > 0\) and \(0 < \alpha < 90^\circ\), find
• the greatest value of \(g(\theta)\)
• the least value of \(g(\theta)\)
• the value of \(\theta\) at which the greatest value occurs, to one decimal place.

[You may use \(\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B\)]

Model answer

\(8\cos\theta + 15\sin\theta = R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha\)
\(R\cos\alpha = 8\), \(R\sin\alpha = 15\)
\(R = \sqrt{64 + 225} = 17\), \(\tan\alpha = \dfrac{15}{8} \Rightarrow \alpha = 61.93^\circ\)
\(g(\theta) = \dfrac{30}{21 + 17\cos(\theta - 61.93^\circ)}\)
Since \(-1 \leqslant \cos(\theta - 61.93^\circ) \leqslant 1\), the denominator lies between 4 and 38.
Greatest value of \(g(\theta) = \dfrac{30}{4} = 7.5\)
Least value of \(g(\theta) = \dfrac{30}{38} = \dfrac{15}{19}\)
Greatest value when \(\cos(\theta - 61.93^\circ) = -1\): \(\theta - 61.93^\circ = 180^\circ\), \(\theta = 241.9^\circ\) (1 d.p.)

Explanation

Writing \(8\cos\theta + 15\sin\theta\) as \(17\cos(\theta - 61.93^\circ)\) shows that the denominator varies between 21 - 17 = 4 and 21 + 17 = 38. Because the numerator is fixed and positive, g is greatest when the denominator is smallest, which happens when the cosine equals -1. A common error is to use the largest denominator for the greatest value, or to set \(\theta - \alpha = 0\), which gives the angle of the least value.

Mark scheme

1 mark
M1 Attempts the R form: \(R = \sqrt{8^2 + 15^2} = 17\) or \(\tan\alpha = \dfrac{15}{8}\). An answer with no working scores 0 marks.
An answer of greatest value 7.5, least value \(\dfrac{15}{19}\) and \(\theta = 241.9^\circ\) with no working scores 0 marks: the question demands working.
1 mark
A1 \(17\cos(\theta - 61.93^\circ)\) (R = 17 and \(\alpha\) awrt 61.9, i.e. in [61.85, 61.95))
\(\tan\alpha = \dfrac{8}{15}\) (giving 28.07°) is A0, but the later method marks are available.
1 mark
M1 Uses \(-1 \leqslant \cos(\theta - \alpha) \leqslant 1\): the denominator lies between \(21 - 17 = 4\) and \(21 + 17 = 38\) (their R)
Not dependent on the first two marks.
1 mark
A1 Greatest value of \(g(\theta) = \dfrac{30}{4} = 7.5\)
Accept \(\dfrac{15}{2}\). A common error is to divide by the largest denominator (38) for the greatest value: A0.
1 mark
A1 Least value of \(g(\theta) = \dfrac{30}{38} = \dfrac{15}{19}\)
Depends on the second M mark (not the point just before). Accept awrt 0.789, i.e. in [0.7885, 0.7895).
1 mark
M1 The greatest value occurs when \(\cos(\theta - \alpha) = -1\): \(\theta - \alpha = 180^\circ\), so \(\theta = 180^\circ + \alpha\)
Not dependent on the earlier marks. \(\theta - \alpha = 0\) (which gives the least value, at 61.9°) is M0.
1 mark
A1 \(\theta =\) awrt 241.9°, i.e. in [241.85, 241.95)
The degree sign is not required. Answers in radians score A0.

Question 4

9 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

The blood pressure, \(P\) mmHg, in an artery of a healthy adult at rest is modelled by the equation
\[P = 100 + 15\sin(7.5t) - 8\cos(7.5t), \quad t \geqslant 0\]
where \(t\) is the time in seconds and the angles are in radians. Each cycle of the model represents one heartbeat.

Express \(15\sin(7.5t) - 8\cos(7.5t)\) in the form \(R\sin(7.5t - \alpha)\), where \(R > 0\) and \(0 < \alpha < \dfrac{\pi}{2}\), giving \(\alpha\) to 3 significant figures.

Hence, according to the model, find
• the maximum and the minimum blood pressure
• the number of heartbeats per minute, to the nearest whole number
• the length of time, in seconds to 3 significant figures, in each heartbeat for which the blood pressure is above 110 mmHg.

[You may use \(\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B\)]

Model answer

\(R\sin(7.5t - \alpha) = R\sin(7.5t)\cos\alpha - R\cos(7.5t)\sin\alpha\)
\(R\cos\alpha = 15\), \(R\sin\alpha = 8\)
\(R = \sqrt{225 + 64} = 17\), \(\tan\alpha = \dfrac{8}{15} \Rightarrow \alpha = 0.490\) (3 s.f.)
\(P = 100 + 17\sin(7.5t - 0.490)\)
Maximum \(= 100 + 17 = 117\) mmHg; minimum \(= 100 - 17 = 83\) mmHg
Period \(= \dfrac{2\pi}{7.5} = 0.8377...\) s, so heartbeats per minute \(= \dfrac{60}{0.8377...} = 71.6... \approx 72\)
\(P > 110 \Rightarrow \sin(7.5t - \alpha) > \dfrac{10}{17}\)
\(\sin u = \dfrac{10}{17}\) at \(u = 0.6288...\) and \(u = \pi - 0.6288... = 2.5127...\)
Time above 110 in each beat \(= \dfrac{2.5127... - 0.6288...}{7.5} = \dfrac{1.8838...}{7.5} = 0.251\) s (3 s.f.)

Explanation

The R form \(P = 100 + 17\sin(7.5t - 0.490)\) shows that the pressure varies by 17 mmHg either side of 100, so the maximum and minimum are 117 and 83, and one heartbeat takes one period, \(\dfrac{2\pi}{7.5}\) seconds. The pressure is above 110 while the sine is above \(\dfrac{10}{17}\), which happens between the principal value and \(\pi\) minus it, and dividing that angle interval by 7.5 gives the time. Common errors are adding 15 and 8 for the maximum and finding a single time instead of the length of the interval.

Mark scheme

1 mark
B1 \(R = 17\). An answer with no working scores 0 marks.
An answer of maximum 117, minimum 83, 72 beats per minute and 0.251 s with no working scores 0 marks: the question demands all stages of working.
1 mark
M1 \(\tan\alpha = \dfrac{8}{15}\) (from \(R\cos\alpha = 15\), \(R\sin\alpha = 8\)), leading to a value of \(\alpha\)
\(\tan\alpha = \dfrac{15}{8}\) is M0.
1 mark
A1 \(\alpha =\) awrt 0.490, i.e. in [0.4895, 0.4905), so \(P = 100 + 17\sin(7.5t - 0.490)\)
Degrees (28.07°) score A0.
1 mark
B1 Maximum blood pressure = 117 mmHg (ft 100 + their R)
Units not required, but wrong units lose this mark.
1 mark
B1 Minimum blood pressure = 83 mmHg (ft 100 - their R)
Adding 15 and 8 to 100 (giving 123) is B0 for the maximum; both B marks need the R form.
1 mark
B1 One heartbeat lasts \(\dfrac{2\pi}{7.5} = 0.8377...\) s, so the model gives \(\dfrac{60}{0.8377...} = 71.6...\), i.e. 72 heartbeats per minute
Accept 72 only as the final answer (71.6 seen and rounded). Using a period of \(\dfrac{360}{7.5}\) is B0.
1 mark
M1 Sets \(100 + 17\sin(7.5t - \alpha) = 110\) (or > 110) and reaches \(\sin(7.5t - \alpha) = \dfrac{10}{17}\), with principal value \(7.5t - \alpha = 0.6288...\)
Follow through their R. Accept awrt 0.629, i.e. in [0.6285, 0.6295).
1 mark
M1 (dM1) Uses the second value \(\pi - 0.6288... = 2.5127...\) and finds the difference of the two values of \(7.5t - \alpha\) (or of the two times) and divides by 7.5
Depends on the previous M mark. The two times are awrt \(t = 0.149\) and awrt \(t = 0.400\) s (in [0.1485, 0.1495) and [0.3995, 0.4005)). Equivalent: \(\dfrac{\pi - 2\arcsin\frac{10}{17}}{7.5}\).
1 mark
A1 awrt 0.251 seconds, i.e. in [0.2505, 0.2515)
Accept 251 milliseconds.

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