The shape of the lesson
By the end of the lesson, learners can
- Matching coefficients: Fig. 1 shows that \(3\sin x + 4\cos x\) is just a sine wave of amplitude 5, shifted left.
- Solving equations: Once the left side is \(R\sin(x + \alpha)\), the equation has only one trig term.
- Greatest and least values: Because \(-1 \le \sin(x + \alpha) \le 1\), the expression \(R\sin(x + \alpha)\) lies between \(-R\) and \(R\).
- 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.
One picture every idea returns to
\(a\sin x + b\cos x \equiv R\sin(x + \alpha)\) where \(R = \sqrt{a^2 + b^2}\), \(R\cos\alpha = a\), \(R\sin\alpha = b\), so \(\tan\alpha = \dfrac{b}{a}\). Expand the target form, match coefficients of \(\sin x\) and \(\cos x\), then square-and-add for \(R\) and divide for \(\alpha\).
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Two waves make one.. Papers 2 and 3. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Matching coefficients | Fig. 1 shows that \(3\sin x + 4\cos x\) is just a sine wave of amplitude 5, shifted left. Model the worked example R sin(x + α) on the board, one line at a time. Your turn: Express \(5\sin x - 12\cos x\) in the form \(R\sin(x - \alpha)\), where \(R > 0\) and \(0^\circ < \alpha < 90^\circ\). | Whole class, then pairs |
| 13 to 21 min | Teach and check: Solving equations | Once the left side is \(R\sin(x + \alpha)\), the equation has only one trig term. Model the worked example Degrees on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Greatest and least values | Because \(-1 \le \sin(x + \alpha) \le 1\), the expression \(R\sin(x + \alpha)\) lies between \(-R\) and \(R\). Model the worked example Greatest and least of a fraction on the board, one line at a time. | Jigsaw |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 27 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Greatest and least values, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| \(\tan\alpha = \dfrac{R\sin\alpha}{R\cos\alpha}\), so it is the coefficient matched with \(\sin\alpha\) on top. | For \(3\sin x + 4\cos x = R\sin(x + \alpha)\), \(\tan\alpha = \dfrac{4}{3}\), not \(\dfrac{3}{4}\). Always write the two matching equations first. |
| Solve for the whole bracket \(x + \alpha\) in its shifted range FIRST, and only then subtract \(\alpha\). | Taking the calculator value \(23.58^\circ\) and subtracting \(53.13^\circ\) gives a negative angle, and students often then miss \(330.4^\circ\). |
| For a fraction with the trig expression underneath, the greatest value of the fraction comes from the LEAST value of the denominator. | Also check the denominator can never be zero. |
Expressing a sin x + b cos x as R sin(x + a): worksheet
6 questions, 27 marks. Written in the style of the papers; no past paper question is reproduced.
Express \(4\cos x + 3\sin x\) in the form \(R\cos(x - \alpha)\), where \(R > 0\) and \(0^\circ < \alpha < 90^\circ\).
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 \(\alpha\).
(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) 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 \(0 < \theta < 2\pi\).
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.
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.
(c) Find the times when the depth is 5 metres.
Answer key and marking notes
- \(R\cos(x - \alpha) = R\cos x\cos\alpha + R\sin x\sin\alpha\), so \(R\cos\alpha = 4\), \(R\sin\alpha = 3\).
- \(R = \sqrt{16 + 9} = 5\); \(\tan\alpha = \dfrac{3}{4}\), \(\alpha = 36.87^\circ\).
- \(4\cos x + 3\sin x \equiv 5\cos(x - 36.87^\circ)\).
- \(R\sin(\theta - \alpha) = R\sin\theta\cos\alpha - R\cos\theta\sin\alpha\), so \(R\cos\alpha = 1\), \(R\sin\alpha = \sqrt{3}\).
- \(R = \sqrt{1 + 3} = 2\); \(\tan\alpha = \sqrt{3}\), \(\alpha = \dfrac{\pi}{3}\).
- \(\sin\theta - \sqrt{3}\cos\theta \equiv 2\sin\left(\theta - \dfrac{\pi}{3}\right)\).
- (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 \(387.49^\circ\).
- \(x = 85.1^\circ,\ 320.1^\circ\).
- (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.).
- \(7\sin x - 24\cos x = R\sin(x - \alpha)\) 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 \(x = 163.7^\circ\).
- (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.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Given that \(4\sin x + 3\cos x \equiv 5\sin(x + \alpha)\), where \(0^\circ < \alpha < 90^\circ\), find \(\alpha\).
- A\(36.9^\circ\)
- B\(53.1^\circ\)
- C\(48.6^\circ\)
- D\(41.4^\circ\)
The expression \(\sqrt{5}\sin x - 2\cos x\) is written as \(R\sin(x - \alpha)\), where \(R > 0\). Find \(R\).
- A\(3\)
- B\(1\)
- C\(9\)
- D\(\sqrt{5} + 2\)
Find the least value of \(5\sin x + 12\cos x + 4\).
- A\(-9\)
- B\(-13\)
- C\(17\)
- D\(-1\)
Given that \(3\cos x - 4\sin x \equiv 5\cos(x + \alpha)\), where \(0^\circ < \alpha < 90^\circ\), find \(\alpha\).
- A\(53.1^\circ\)
- B\(36.9^\circ\)
- C\(-53.1^\circ\)
- D\(126.9^\circ\)
Solve \(4\sin x + 3\cos x = 2\) for \(0^\circ \le x \le 360^\circ\).
- A\(x = 119.6^\circ,\ 346.7^\circ\)
- B\(x = 119.6^\circ\) only
- C\(x = 103.3^\circ,\ 330.4^\circ\)
- D\(x = 60.4^\circ,\ 193.3^\circ\)
Find the greatest value of \(\dfrac{26}{17 - 5\sin x - 12\cos x}\).
- A\(\dfrac{13}{2}\)
- B\(\dfrac{13}{15}\)
- C\(2\)
- D\(\dfrac{26}{17}\)
Find the greatest value of \(8\cos x + 6\sin x\) and the value of \(x\), \(0^\circ \le x \le 360^\circ\), at which it occurs.
- A10, at \(x = 36.9^\circ\)
- B10, at \(x = 53.1^\circ\)
- C14, at \(x = 0^\circ\)
- D10, at \(x = 216.9^\circ\)
Given that \(\sqrt{3}\sin x - \cos x \equiv 2\sin\left(x - \dfrac{\pi}{6}\right)\), solve \(\sqrt{3}\sin x - \cos x = \sqrt{2}\) for \(0 \le x \le 2\pi\).
- A\(x = \dfrac{5\pi}{12},\ \dfrac{11\pi}{12}\)
- B\(x = \dfrac{\pi}{12},\ \dfrac{7\pi}{12}\)
- C\(x = \dfrac{5\pi}{12}\) only
- D\(x = \dfrac{5\pi}{12},\ \dfrac{17\pi}{12}\)
Quiz key
\(5\sin(x + \alpha) = 5\cos\alpha\sin x + 5\sin\alpha\cos x\). Matching: \(5\cos\alpha = 4\) and \(5\sin\alpha = 3\), so \(\tan\alpha = \dfrac{3}{4}\) and \(\alpha = 36.9^\circ\). The tempting \(53.1^\circ\) uses \(\tan\alpha = \dfrac{4}{3}\), the fraction upside down. \(48.6^\circ\) is \(\sin^{-1}\dfrac{3}{4}\) and \(41.4^\circ\) is \(\cos^{-1}\dfrac{3}{4}\); both forget to divide by \(R = 5\).
\(R = \sqrt{(\sqrt{5})^2 + 2^2} = \sqrt{5 + 4} = 3\). The tempting 1 subtracts the squares (\(\sqrt{5 - 4}\)) because of the minus sign, but \(R\cos\alpha\) and \(R\sin\alpha\) are squared and ADDED, so the sign never matters. 9 is \(R^2\), and \(\sqrt{5} + 2\) adds the coefficients without squaring.
\(R = \sqrt{25 + 144} = 13\), so the expression is \(13\sin(x + \alpha) + 4\). Its least value is \(-13 + 4 = -9\). The tempting \(-13\) forgets the \(+4\). 17 is the greatest value, and \(-1\) treats the least of \(5\sin x + 12\cos x\) as \(-5\), but \(\sin x\) and \(\cos x\) do not reach their extremes separately.
\(5\cos(x + \alpha) = 5\cos\alpha\cos x - 5\sin\alpha\sin x\). Matching: \(5\cos\alpha = 3\) and \(5\sin\alpha = 4\) (the minus signs match), so \(\tan\alpha = \dfrac{4}{3}\) and \(\alpha = 53.1^\circ\). The tempting \(36.9^\circ\) uses \(\tan\alpha = \dfrac{3}{4}\), the fraction upside down. \(-53.1^\circ\) carries the minus sign into \(\alpha\), but it is already in the expansion, and \(126.9^\circ\) is outside the given range.
\(4\sin x + 3\cos x = 5\sin(x + 36.87^\circ)\), so \(\sin(x + 36.87^\circ) = 0.4\). The calculator gives \(23.58^\circ\), which is below \(36.87^\circ\); in the range \(36.87^\circ\) to \(396.87^\circ\) the values are \(180^\circ - 23.58^\circ = 156.42^\circ\) and \(360^\circ + 23.58^\circ = 383.58^\circ\). Subtracting: \(x = 119.6^\circ, 346.7^\circ\). The tempting "\(119.6^\circ\) only" rejects \(23.58^\circ - 36.87^\circ < 0\) and never looks for \(383.58^\circ\). \(103.3^\circ, 330.4^\circ\) uses \(\alpha = 53.13^\circ\) (tan the wrong way up), and \(60.4^\circ, 193.3^\circ\) adds \(\alpha\) instead of subtracting it.
The denominator is \(17 - 13\sin(x + \alpha)\), which lies between \(17 - 13 = 4\) and \(17 + 13 = 30\). It is always positive, and the fraction is greatest when the denominator is least: \(\dfrac{26}{4} = \dfrac{13}{2}\). The tempting \(\dfrac{13}{15} = \dfrac{26}{30}\) uses the GREATEST denominator, which gives the least value of the fraction. 2 is \(\dfrac{26}{13}\), forgetting the 17, and \(\dfrac{26}{17}\) ignores the trigonometric terms.
\(R\cos(x - \alpha) = R\cos\alpha\cos x + R\sin\alpha\sin x\), so \(R\cos\alpha = 8\), \(R\sin\alpha = 6\): \(R = 10\) and \(\tan\alpha = \dfrac{6}{8}\), \(\alpha = 36.9^\circ\). The greatest value 10 occurs when \(x - 36.9^\circ = 0^\circ\), so \(x = 36.9^\circ\). The tempting \(53.1^\circ\) uses \(\tan\alpha = \dfrac{8}{6}\). 14 adds the coefficients, but \(\cos x\) and \(\sin x\) are never both 1 together, and \(216.9^\circ\) is where the LEAST value \(-10\) occurs.
\(\sin\left(x - \dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{2}}\), so \(x - \dfrac{\pi}{6} = \dfrac{\pi}{4}\) or \(\pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}\). Adding \(\dfrac{\pi}{6}\): \(x = \dfrac{5\pi}{12}\) or \(\dfrac{11\pi}{12}\). The tempting \(\dfrac{\pi}{12}, \dfrac{7\pi}{12}\) subtracts \(\dfrac{\pi}{6}\) instead of adding it. \(\dfrac{5\pi}{12}\) only misses the second-quadrant angle, and \(\dfrac{17\pi}{12}\) adds \(\pi\) as if it were a tangent equation.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.