The shape of the lesson
By the end of the lesson, learners can
- Principal values: In the lab, move the horizontal line \(y = k\) and watch where it cuts the curve.
- Finding every solution: Get the principal value, use the symmetry of the graph to get a second value, then add or subtract the period until you have every solution in the interval.
- Multiple angles and shifts: For \(\sin 2x = k\), or \(\tan(x - 30^\circ) = k\), solve for the whole bracket first.
- Equations that need an identity: If an equation mixes \(\sin\) and \(\cos\), make it a single function.
- One function: use \(\tan x = \dfrac{\sin x}{\cos x}\) or \(\sin^2 x + \cos^2 x = 1\) so only one trig function is left; factorise rather than cancel.
- Bracket interval: for \(\sin 2x\) or \(\cos(x + k)\), rewrite the interval for the whole bracket.
- Principal value: \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\) on the calculator, in the right mode.
- Second value: \(180^\circ - \alpha\) (sin), \(360^\circ - \alpha\) (cos), \(\alpha + 180^\circ\) (tan).
- Period: add or subtract \(360^\circ\) (tan: \(180^\circ\)) until you leave the interval, then convert back to \(x\).
One picture every idea returns to
Find the principal value \(\alpha\) with \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\). Then \(\sin x = k\) also has \(180^\circ - \alpha\); \(\cos x = k\) also has \(360^\circ - \alpha\); \(\tan x = k\) also has \(\alpha + 180^\circ\). Add or subtract whole periods to stay inside the interval, and for \(\sin 2x\) widen the interval for \(2x\) first.
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, One answer from the calculator, the rest from symmetry.. Your calculator gives one solution, the principal value. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 11 min | Teach and check: Principal values | In the lab, move the horizontal line \(y = k\) and watch where it cuts the curve. Project tab 3 of the lab (Trigonometric graphs lab, solving equations) and let a student drive it while the class predicts. Model the worked example Exact principal values on the board, one line at a time. | Whole class, then pairs |
| 11 to 17 min | Teach and check: Finding every solution | Get the principal value, use the symmetry of the graph to get a second value, then add or subtract the period until you have every solution in the interval. Model the worked example Sine, in degrees on the board, one line at a time. | Pairs |
| 17 to 23 min | Teach and check: Multiple angles and shifts | For \(\sin 2x = k\), or \(\tan(x - 30^\circ) = k\), solve for the whole bracket first. Model the worked example A double angle, exact on the board, one line at a time. Your turn: Solve \(\sin(x + 20^\circ) = 0.5\) for \(0^\circ \le x \le 360^\circ\). | Pairs |
| 23 to 30 min | Teach and check: Equations that need an identity | If an equation mixes \(\sin\) and \(\cos\), make it a single function. Model the worked example Sine equals a multiple of cosine on the board, one line at a time. Your turn: Solve \(3\tan x = 2\cos x\) for \(0 \le x \le 2\pi\). | Numbered Heads Together |
| 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, 25 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 Equations that need an identity, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| \(\sin^{-1}x\) is NOT \(\dfrac{1}{\sin x}\). | The \(-1\) means inverse function, just as \(f^{-1}\) does; \((\sin x)^{-1}\) is a different thing. |
| A negative principal value is not automatically wrong, and it is not automatically an answer. | If the interval is \(0^\circ\) to \(360^\circ\) and the calculator gives \(-63.4^\circ\), add \(180^\circ\) (or \(360^\circ\)) to bring it into range. |
| Halving too early loses solutions. | If you solve \(\sin 2x = -\tfrac{1}{2}\) only for \(-\pi < 2x < \pi\) you find two answers instead of four. |
| Never cancel a trig function that could be zero. | In \(\sin x = 2\sin x\cos x\), dividing by \(\sin x\) loses the solutions of \(\sin x = 0\). Instead factorise: \(\sin x(1 - 2\cos x) = 0\). |
Solving trigonometric equations: worksheet
6 questions, 25 marks. Written in the style of the papers; no past paper question is reproduced.
Find the exact values of (a) \(\sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)\) (b) \(\cos^{-1}\left(-\dfrac{\sqrt{2}}{2}\right)\) (c) \(\tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)\).
Solve the equation \(5\cos x + 2 = 0\) for \(0^\circ \le x \le 360^\circ\).
Solve the equation \(5\sin 2\theta = 2\) for \(0 \le \theta \le \pi\), giving your answers in radians correct to 3 significant figures.
Solve the equation \(2\tan^2 x - \tan x - 6 = 0\) for \(0^\circ \le x \le 180^\circ\).
(a) Show that the equation \(4\tan\theta\sin\theta = 15\) can be written as \(4\cos^2\theta + 15\cos\theta - 4 = 0\).
(b) Hence solve \(4\tan\theta\sin\theta = 15\) for \(0^\circ \le \theta \le 360^\circ\).
The temperature in a greenhouse \(t\) hours after midnight is modelled by \(T = 14 - 6\cos\left(\dfrac{\pi t}{12}\right)\) degrees Celsius, for \(0 \le t \le 24\).
(a) Find the exact temperature at 3 a.m.
(b) Find the times at which the temperature is \(17^\circ\text{C}\).
(c) For how many hours of the day is the temperature at least \(17^\circ\text{C}\)?
Answer key and marking notes
- (a) \(-\dfrac{\pi}{3}\) (the range of \(\sin^{-1}\) is \(-\dfrac{\pi}{2}\) to \(\dfrac{\pi}{2}\))
- (b) \(\pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}\) (the range of \(\cos^{-1}\) is 0 to \(\pi\))
- (c) \(\dfrac{\pi}{6}\)
- \(\cos x = -0.4\)
- \(x = \cos^{-1}(-0.4) = 113.6^\circ\)
- \(x = 360^\circ - 113.6^\circ = 246.4^\circ\)
- \(\sin 2\theta = 0.4\), with \(0 \le 2\theta \le 2\pi\).
- \(2\theta = \sin^{-1}0.4 = 0.41152\) or \(2\theta = \pi - 0.41152 = 2.73008\).
- (Adding \(2\pi\) to either exceeds \(2\pi\).)
- \(\theta = 0.206\) or \(\theta = 1.37\)
- Factorise: \((2\tan x + 3)(\tan x - 2) = 0\), so \(\tan x = 2\) or \(\tan x = -1.5\).
- \(\tan x = 2\): \(x = 63.4^\circ\) (\(243.4^\circ\) is outside the interval).
- \(\tan x = -1.5\): principal value \(-56.3^\circ\), add \(180^\circ\): \(x = 123.7^\circ\).
- \(x = 63.4^\circ\) or \(123.7^\circ\)
- (a) \(4 \times \dfrac{\sin\theta}{\cos\theta} \times \sin\theta = 15\), so \(4\sin^2\theta = 15\cos\theta\).
- \(4(1 - \cos^2\theta) = 15\cos\theta\), so \(4\cos^2\theta + 15\cos\theta - 4 = 0\).
- (b) \((4\cos\theta - 1)(\cos\theta + 4) = 0\); \(\cos\theta = -4\) is impossible, so \(\cos\theta = \tfrac{1}{4}\).
- \(\theta = 75.5^\circ\) or \(360^\circ - 75.5^\circ = 284.5^\circ\)
- (a) \(t = 3\): \(T = 14 - 6\cos\dfrac{\pi}{4} = 14 - 6 \times \dfrac{\sqrt{2}}{2} = 14 - 3\sqrt{2}\ ^\circ\text{C}\) (about 9.76).
- (b) \(14 - 6\cos\dfrac{\pi t}{12} = 17\) gives \(\cos\dfrac{\pi t}{12} = -\tfrac{1}{2}\), with \(0 \le \dfrac{\pi t}{12} \le 2\pi\).
- \(\dfrac{\pi t}{12} = \dfrac{2\pi}{3}\) or \(\dfrac{4\pi}{3}\), so \(t = 8\) or \(t = 16\): 8 a.m. and 4 p.m.
- (c) \(T\) rises from its minimum \(8^\circ\text{C}\) at midnight to \(20^\circ\text{C}\) at noon, so \(T \ge 17\) for \(8 \le t \le 16\): 8 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.
Solve \(\cos x = 0.7\) for \(0^\circ \le x \le 360^\circ\).
- A\(45.6^\circ,\ 314.4^\circ\)
- B\(45.6^\circ,\ 134.4^\circ\)
- C\(45.6^\circ,\ 225.6^\circ\)
- D\(45.6^\circ\) only
Solve \(\sin x = -0.35\) for \(0 \le x \le 2\pi\), giving your answers in radians to 3 significant figures.
- A\(3.50,\ 5.93\)
- B\(0.358,\ 2.78\)
- C\(-0.358,\ 3.50\)
- D\(2.78,\ 5.93\)
Solve \(\tan 2x = 1\) for \(0 \le x \le \pi\), giving exact answers.
- A\(\dfrac{\pi}{8},\ \dfrac{5\pi}{8}\)
- B\(\dfrac{\pi}{8}\) only
- C\(\dfrac{\pi}{8},\ \dfrac{9\pi}{8}\)
- D\(\dfrac{\pi}{4},\ \dfrac{5\pi}{4}\)
Find the exact principal value of \(\tan^{-1}\left(-\dfrac{1}{\sqrt{3}}\right)\).
- A\(-\dfrac{\pi}{6}\)
- B\(\dfrac{5\pi}{6}\)
- C\(\dfrac{11\pi}{6}\)
- D\(-\dfrac{\pi}{3}\)
Solve \(4\sin x\cos x = 3\cos x\) for \(0^\circ \le x \le 360^\circ\).
- A\(48.6^\circ,\ 90^\circ,\ 131.4^\circ,\ 270^\circ\)
- B\(48.6^\circ,\ 131.4^\circ\)
- C\(48.6^\circ,\ 90^\circ,\ 131.4^\circ\)
- D\(48.6^\circ,\ 131.4^\circ,\ 228.6^\circ,\ 311.4^\circ\)
Solve \(2\cos^2\theta = 1 + \sin\theta\) for \(0^\circ \le \theta \le 360^\circ\).
- A\(30^\circ,\ 150^\circ,\ 270^\circ\)
- B\(30^\circ,\ 150^\circ\)
- C\(90^\circ,\ 210^\circ,\ 330^\circ\)
- D\(30^\circ,\ 150^\circ,\ 210^\circ,\ 330^\circ\)
Solve \(\cos(x + 40^\circ) = -0.5\) for \(0^\circ \le x \le 360^\circ\).
- A\(80^\circ,\ 200^\circ\)
- B\(160^\circ,\ 280^\circ\)
- C\(20^\circ,\ 260^\circ\)
- D\(80^\circ\) only
Solve \(2\sin\left(2x - \dfrac{\pi}{3}\right) = 1\) for \(0 \le x \le \pi\), giving exact answers.
- A\(\dfrac{\pi}{4},\ \dfrac{7\pi}{12}\)
- B\(\dfrac{\pi}{12},\ \dfrac{5\pi}{12}\)
- C\(\dfrac{\pi}{2},\ \dfrac{7\pi}{6}\)
- D\(\dfrac{\pi}{4}\) only
Quiz key
\(\cos^{-1} 0.7 = 45.6^\circ\). Cosine is positive in the first and fourth quadrants, so the other solution is \(360^\circ - 45.6^\circ = 314.4^\circ\). The tempting \(134.4^\circ\) uses the sine rule \(180^\circ - x\); \(\cos 134.4^\circ = -0.7\). \(225.6^\circ\) uses the tan pattern \(180^\circ + x\), and stopping at \(45.6^\circ\) misses the second solution.
\(\sin^{-1}(-0.35) = -0.3576\), which is outside the interval. Sine is negative in the third and fourth quadrants: \(x = \pi + 0.3576 = 3.50\) and \(x = 2\pi - 0.3576 = 5.93\). The tempting \(-0.358,\ 3.50\) keeps the principal value, which is less than 0. \(0.358,\ 2.78\) solve \(\sin x = +0.35\), and \(2.78,\ 5.93\) mixes the two.
\(0 \le 2x \le 2\pi\). \(\tan 2x = 1\) gives \(2x = \dfrac{\pi}{4}\) or \(\dfrac{\pi}{4} + \pi = \dfrac{5\pi}{4}\), so \(x = \dfrac{\pi}{8}\) or \(\dfrac{5\pi}{8}\). The tempting \(\dfrac{\pi}{8}\) only forgets that doubling the angle doubles the range. \(\dfrac{9\pi}{8}\) adds \(\pi\) after halving (the period of \(\tan 2x\) is \(\dfrac{\pi}{2}\)) and is outside the interval, and \(\dfrac{\pi}{4},\ \dfrac{5\pi}{4}\) are the values of \(2x\), not \(x\).
\(\tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}\), and the principal value must lie in \(\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\), so it is \(-\dfrac{\pi}{6}\). The tempting \(\dfrac{5\pi}{6}\) and \(\dfrac{11\pi}{6}\) are also solutions of \(\tan x = -\dfrac{1}{\sqrt{3}}\), but neither is in the principal range. \(-\dfrac{\pi}{3}\) has \(\tan = -\sqrt{3}\).
\(4\sin x\cos x - 3\cos x = 0\), so \(\cos x(4\sin x - 3) = 0\). \(\cos x = 0\) gives \(90^\circ, 270^\circ\); \(\sin x = 0.75\) gives \(48.6^\circ\) and \(180^\circ - 48.6^\circ = 131.4^\circ\). The tempting \(48.6^\circ, 131.4^\circ\) divides both sides by \(\cos x\), which loses the solutions where \(\cos x = 0\). Stopping at \(90^\circ\) misses \(270^\circ\), and \(228.6^\circ, 311.4^\circ\) solve \(\sin x = -0.75\).
\(2(1 - \sin^2\theta) = 1 + \sin\theta\) gives \(2\sin^2\theta + \sin\theta - 1 = 0\), so \((2\sin\theta - 1)(\sin\theta + 1) = 0\). \(\sin\theta = \tfrac{1}{2}\) gives \(30^\circ, 150^\circ\), and \(\sin\theta = -1\) gives \(270^\circ\). The tempting \(30^\circ, 150^\circ\) drops the root \(\sin\theta = -1\), which is a valid value of sine. \(90^\circ, 210^\circ, 330^\circ\) come from the wrong factorisation \((2\sin\theta + 1)(\sin\theta - 1)\), and \(210^\circ, 330^\circ\) solve \(\sin\theta = -\tfrac{1}{2}\).
\(40^\circ \le u \le 400^\circ\). \(\cos u = -0.5\) gives \(u = 120^\circ\) or \(240^\circ\) (the next, \(480^\circ\), is too big). Then \(x = u - 40^\circ = 80^\circ\) or \(200^\circ\). The tempting \(160^\circ, 280^\circ\) adds \(40^\circ\) instead of subtracting. \(20^\circ, 260^\circ\) solve \(\cos u = +0.5\), and \(80^\circ\) only misses the second value of \(u\).
\(-\dfrac{\pi}{3} \le u \le \dfrac{5\pi}{3}\). \(\sin u = \tfrac{1}{2}\) gives \(u = \dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\). Then \(2x = u + \dfrac{\pi}{3} = \dfrac{\pi}{2}\) or \(\dfrac{7\pi}{6}\), so \(x = \dfrac{\pi}{4}\) or \(\dfrac{7\pi}{12}\). The tempting \(\dfrac{\pi}{12}, \dfrac{5\pi}{12}\) halves \(u\) without adding \(\dfrac{\pi}{3}\) first. \(\dfrac{\pi}{2}, \dfrac{7\pi}{6}\) are the values of \(2x\), not \(x\), and \(\dfrac{\pi}{4}\) only misses \(u = \dfrac{5\pi}{6}\).
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
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.