AS & A Level · Practice questions · Expressing a sin x + b cos x as R sin(x + a)

Find R, find α.

Six original Cambridge-style questions on expressing \(a\sin x + b\cos x\) as a single trigonometric function, at Paper 2 and Paper 3 standard.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Expand, match, square-and-add, divide.

Expand the target form, e.g. \(R\sin(x + \alpha) = R\sin x\cos\alpha + R\cos x\sin\alpha\).
Match coefficients: \(R\cos\alpha = \ldots\), \(R\sin\alpha = \ldots\).
R \(= \sqrt{a^2 + b^2}\); \(\tan\alpha\) \(= \dfrac{R\sin\alpha}{R\cos\alpha}\).
Solve: shift the range by \(\alpha\), find every value of the bracket, then undo the shift.
Max/min: \(-R \le R\sin(x + \alpha) \le R\); for a fraction, the greatest value uses the least denominator.

Recall

[3 marks]

Express \(4\cos x + 3\sin x\) in the form \(R\cos(x - \alpha)\), where \(R > 0\) and \(0^\circ < \alpha < 90^\circ\).

\(R\cos(x - \alpha) = R\cos x\cos\alpha + R\sin x\sin\alpha\), so \(R\cos\alpha = 4\), \(R\sin\alpha = 3\). ✓
\(R = \sqrt{16 + 9} = 5\); \(\tan\alpha = \dfrac{3}{4}\), \(\alpha = 36.87^\circ\). ✓
\(4\cos x + 3\sin x \equiv 5\cos(x - 36.87^\circ)\). ✓

Recall

[3 marks]

Express \(\sin\theta - \sqrt{3}\cos\theta\) in the form \(R\sin(\theta - \alpha)\), where \(R > 0\) and \(0 < \alpha < \dfrac{\pi}{2}\), giving exact values of \(R\) and \(\alpha\).

\(R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha\), so \(R\cos\alpha = 1\), \(R\sin\alpha = \sqrt{3}\). ✓
\(R = \sqrt{1 + 3} = 2\); \(\tan\alpha = \sqrt{3}\), \(\alpha = \dfrac{\pi}{3}\). ✓
\(\sin\theta - \sqrt{3}\cos\theta \equiv 2\sin\left(\theta - \dfrac{\pi}{3}\right)\). ✓

Calculation

[5 marks]

(a) Express \(5\sin x + 12\cos x\) in the form \(R\sin(x + \alpha)\), where \(R > 0\) and \(0^\circ < \alpha < 90^\circ\).

(b) Hence solve the equation \(5\sin x + 12\cos x = 6\) for \(0^\circ \le x \le 360^\circ\).

(a) \(R\cos\alpha = 5\), \(R\sin\alpha = 12\): \(R = 13\), \(\tan\alpha = \dfrac{12}{5}\), \(\alpha = 67.38^\circ\). So \(13\sin(x + 67.38^\circ)\). ✓
(b) \(\sin(x + 67.38^\circ) = \dfrac{6}{13}\), with \(67.38^\circ \le x + 67.38^\circ \le 427.38^\circ\). ✓
\(\sin^{-1}\dfrac{6}{13} = 27.49^\circ\) (too small), so \(x + 67.38^\circ = 152.51^\circ\) or \(387.49^\circ\). ✓
\(x = 85.1^\circ,\ 320.1^\circ\). ✓

Calculation

[5 marks]

(a) Express \(2\cos\theta - 5\sin\theta\) in the form \(R\cos(\theta + \alpha)\), where \(R > 0\) and \(0 < \alpha < \dfrac{\pi}{2}\). Give \(\alpha\) in radians to 3 decimal places.

(b) Hence solve \(2\cos\theta - 5\sin\theta = 3\) for \(0 < \theta < 2\pi\).

(a) \(R\cos\alpha = 2\), \(R\sin\alpha = 5\): \(R = \sqrt{29}\), \(\tan\alpha = \dfrac{5}{2}\), \(\alpha = 1.190\). So \(\sqrt{29}\cos(\theta + 1.190)\). ✓
(b) \(\cos(\theta + 1.190) = \dfrac{3}{\sqrt{29}} = 0.5571\), with \(1.190 < \theta + 1.190 < 7.473\). ✓
\(\cos^{-1} 0.5571 = 0.9799\) (too small), so \(\theta + 1.190 = 2\pi - 0.9799 = 5.3033\) or \(2\pi + 0.9799 = 7.2631\). ✓
\(\theta = 4.11,\ 6.07\) (3 s.f.). ✓

Analysis

[4 marks]

The function f is defined by \(\text{f}(x) = 2 + 7\sin x - 24\cos x\) for \(0^\circ \le x \le 360^\circ\).

State the greatest value of \(\text{f}(x)\) and find the value of \(x\) at which it occurs.

\(7\sin x - 24\cos x = R\sin(x - \alpha)\) with \(R\cos\alpha = 7\), \(R\sin\alpha = 24\): \(R = 25\), \(\tan\alpha = \dfrac{24}{7}\), \(\alpha = 73.74^\circ\). ✓
Greatest value \(= 2 + 25 = 27\). ✓
It occurs when \(x - 73.74^\circ = 90^\circ\), so \(x = 163.7^\circ\). ✓

Analysis

[7 marks]

The depth of water, \(d\) metres, at a harbour entrance \(t\) hours after midnight is modelled by \(d = 6 + 1.5\sin(30t)^\circ + 2\cos(30t)^\circ\), for \(0 \le t \le 12\).

(a) Express \(1.5\sin(30t)^\circ + 2\cos(30t)^\circ\) in the form \(R\sin(30t + \alpha)^\circ\), where \(R > 0\) and \(0 < \alpha < 90\).

(b) Find the greatest depth and the first time it occurs.

(a) \(R\cos\alpha = 1.5\), \(R\sin\alpha = 2\): \(R = \sqrt{2.25 + 4} = 2.5\), \(\tan\alpha = \dfrac{4}{3}\), \(\alpha = 53.13\). So \(2.5\sin(30t + 53.13)^\circ\). ✓
(b) Greatest depth \(= 6 + 2.5 = 8.5\) m, when \(30t + 53.13 = 90\), so \(t = 1.23\) hours (about 01:14). ✓
(c) \(6 + 2.5\sin(30t + 53.13)^\circ = 5\), so \(\sin(30t + 53.13)^\circ = -0.4\), with \(53.13 \le 30t + 53.13 \le 413.13\). ✓
\(30t + 53.13 = 180 + 23.58 = 203.58\) or \(360 - 23.58 = 336.42\). ✓
\(t = 5.01\) hours and \(t = 9.44\) hours. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.