The shape of the lesson
By the end of the lesson, learners can
- The reciprocal functions: Each new function is one over an old one.
- Two Pythagorean identities: Start from \(\sin^2\theta + \cos^2\theta \equiv 1\).
- Proving identities: To prove an identity, start from one side (usually the more complicated one) and work until it becomes the other side.
- Definitions: \(\sec = \dfrac{1}{\cos}\), \(\operatorname{cosec} = \dfrac{1}{\sin}\), \(\cot = \dfrac{\cos}{\sin}\).
- Identities: \(\sec^2\theta \equiv 1 + \tan^2\theta\), \(\operatorname{cosec}^2\theta \equiv 1 + \cot^2\theta\).
- Equations: substitute so that only one function remains, then solve the quadratic.
- Reject impossible roots: \(|\sec\theta| \ge 1\) and \(|\operatorname{cosec}\theta| \ge 1\).
- Proofs: start from one side, convert to \(\sin\) and \(\cos\), combine fractions, finish at the other side.
One picture every idea returns to
\(\sec\theta = \dfrac{1}{\cos\theta}\), \(\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}\), \(\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}\). Dividing \(\sin^2\theta + \cos^2\theta \equiv 1\) by \(\cos^2\theta\) or \(\sin^2\theta\) gives \(1 + \tan^2\theta \equiv \sec^2\theta\) and \(1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta\).
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, Three new functions, three old friends.. 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: The reciprocal functions | Each new function is one over an old one. Model the worked example Exact values on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Two Pythagorean identities | Start from \(\sin^2\theta + \cos^2\theta \equiv 1\). Model the worked example A quadratic in tan on the board, one line at a time. Your turn: Solve \(3\operatorname{cosec}^2\theta = 5 + \cot\theta\) for \(0^\circ \le \theta \le 360^\circ\). | Pairs |
| 21 to 30 min | Teach and check: Proving identities | To prove an identity, start from one side (usually the more complicated one) and work until it becomes the other side. Model the worked example Prove an identity on the board, one line at a time. | Rally Coach |
| 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, 22 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 Proving identities, 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.
Rally Coach
Pairs share one worksheet. One partner solves the first question aloud while the other coaches, checking each line and offering a prompt rather than the answer; they swap roles for the next question.
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 |
|---|---|
| \(\sec x\) is NOT \(\cos^{-1} x\). | The index \(-1\) on a function name means the inverse function; \(\sec x = (\cos x)^{-1}\) is the reciprocal. On the calculator, \(\sec 50^\circ\) is \(1 \div \cos 50^\circ = 1.56\), not \(\cos^{-1} 50\) (which gives an error). |
| The identity is \(\sec^2\theta = 1 + \tan^2\theta\), not \(\sec^2\theta = 1 - \tan^2\theta\) or \(\tan^2\theta = 1 + \sec^2\theta\). | If you forget, re-derive it in one line by dividing \(\sin^2\theta + \cos^2\theta = 1\) by \(\cos^2\theta\). |
| Never work on both sides at once, and never cross-multiply an identity you are asked to prove: that assumes the result. | Write LHS \(= \ldots = \) RHS as one chain. |
Secant, cosecant and cotangent: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Find the exact values of (a) \(\sec 300^\circ\) (b) \(\operatorname{cosec}(-60^\circ)\) (c) \(\cot\dfrac{4\pi}{3}\).
Prove that \(\tan x + \cot x \equiv \sec x \operatorname{cosec} x\).
Solve the equation \(\sec^2\theta = 3\tan\theta - 1\) for \(0^\circ \le \theta \le 360^\circ\).
Solve the equation \(2\cot^2 x + 3\operatorname{cosec} x = 0\) for \(0^\circ < x < 360^\circ\).
(a) Sketch the graph of \(y = \operatorname{cosec} x\) for \(0^\circ < x < 360^\circ\), showing its asymptotes.
(b) State the set of values of \(k\) for which the equation \(\operatorname{cosec} x = k\) has no solution.
(a) Show that the equation \(2\operatorname{cosec}^2\theta + 5\cot\theta = 5\) can be written as \(2\cot^2\theta + 5\cot\theta - 3 = 0\).
(b) Hence solve the equation for \(0^\circ < \theta < 180^\circ\).
Answer key and marking notes
- (a) \(\cos 300^\circ = \dfrac{1}{2}\), so \(\sec 300^\circ = 2\).
- (b) \(\sin(-60^\circ) = -\dfrac{\sqrt{3}}{2}\), so \(\operatorname{cosec}(-60^\circ) = -\dfrac{2}{\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}\).
- (c) \(\tan\dfrac{4\pi}{3} = \tan\dfrac{\pi}{3} = \sqrt{3}\), so \(\cot\dfrac{4\pi}{3} = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\).
- LHS \(= \dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} = \dfrac{\sin^2 x + \cos^2 x}{\sin x \cos x}\).
- \(= \dfrac{1}{\sin x\cos x} = \dfrac{1}{\cos x} \times \dfrac{1}{\sin x} = \sec x \operatorname{cosec} x\) = RHS.
- \(1 + \tan^2\theta = 3\tan\theta - 1\), so \(\tan^2\theta - 3\tan\theta + 2 = 0\).
- \((\tan\theta - 1)(\tan\theta - 2) = 0\), so \(\tan\theta = 1\) or \(\tan\theta = 2\).
- \(\tan\theta = 1\): \(\theta = 45^\circ, 225^\circ\). \(\tan\theta = 2\): \(\theta = 63.4^\circ, 243.4^\circ\).
- \(\theta = 45^\circ,\ 63.4^\circ,\ 225^\circ,\ 243.4^\circ\).
- \(2(\operatorname{cosec}^2 x - 1) + 3\operatorname{cosec} x = 0\), so \(2\operatorname{cosec}^2 x + 3\operatorname{cosec} x - 2 = 0\).
- \((2\operatorname{cosec} x - 1)(\operatorname{cosec} x + 2) = 0\).
- \(\operatorname{cosec} x = \dfrac{1}{2}\) would need \(\sin x = 2\): no solutions.
- \(\operatorname{cosec} x = -2\): \(\sin x = -\dfrac{1}{2}\), so \(x = 210^\circ, 330^\circ\).
- (a) Vertical asymptotes at \(x = 0^\circ, 180^\circ, 360^\circ\). For \(0^\circ < x < 180^\circ\) the curve is U-shaped above the axis with minimum point \((90^\circ, 1)\); for \(180^\circ < x < 360^\circ\) it is an upside-down U below the axis with maximum point \((270^\circ, -1)\).
- (b) \(\operatorname{cosec} x\) never lies strictly between \(-1\) and \(1\), so there is no solution for \(-1 < k < 1\).
- (a) \(2(1 + \cot^2\theta) + 5\cot\theta = 5\), so \(2\cot^2\theta + 5\cot\theta - 3 = 0\).
- (b) \((2\cot\theta - 1)(\cot\theta + 3) = 0\), so \(\cot\theta = \dfrac{1}{2}\) or \(\cot\theta = -3\).
- \(\cot\theta = \dfrac{1}{2}\): \(\tan\theta = 2\), \(\theta = 63.4^\circ\).
- \(\cot\theta = -3\): \(\tan\theta = -\dfrac{1}{3}\); the calculator gives \(-18.4^\circ\), so in range \(\theta = 180^\circ - 18.4^\circ = 161.6^\circ\).
- \(\theta = 63.4^\circ,\ 161.6^\circ\).
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.
Find the exact value of \(\operatorname{cosec} 240^\circ\).
- A\(-\dfrac{2\sqrt{3}}{3}\)
- B\(\dfrac{2\sqrt{3}}{3}\)
- C\(-2\)
- D\(-\dfrac{\sqrt{3}}{2}\)
For which values of the constant \(k\) does the equation \(\sec x = k\) have no solutions?
- A\(-1 < k < 1\)
- B\(k < -1\) or \(k > 1\)
- C\(-1 \le k \le 1\)
- D\(k < 0\)
Solve \(\operatorname{cosec} x = 4\) for \(0^\circ \le x \le 360^\circ\).
- A\(x = 14.5^\circ,\ 165.5^\circ\)
- B\(x = 14.5^\circ,\ 345.5^\circ\)
- C\(x = 75.5^\circ,\ 284.5^\circ\)
- D\(x = 14.5^\circ\) only
Given that \(\tan\theta = \dfrac{12}{5}\) and \(180^\circ < \theta < 270^\circ\), find the exact value of \(\sec\theta\).
- A\(-\dfrac{13}{5}\)
- B\(\dfrac{13}{5}\)
- C\(-\dfrac{13}{12}\)
- D\(-\dfrac{5}{13}\)
Solve \(2\sec^2\theta - 5\tan\theta = 0\) for \(0^\circ \le \theta \le 360^\circ\).
- A\(\theta = 26.6^\circ,\ 63.4^\circ,\ 206.6^\circ,\ 243.4^\circ\)
- B\(\theta = 26.6^\circ,\ 63.4^\circ\)
- C\(\theta = 116.6^\circ,\ 153.4^\circ,\ 296.6^\circ,\ 333.4^\circ\)
- D\(\theta = 26.6^\circ,\ 63.4^\circ,\ 153.4^\circ,\ 116.6^\circ\)
Solve \(2\cot^2 x = 7\operatorname{cosec} x - 8\) for \(0^\circ < x < 360^\circ\).
- A\(x = 30^\circ,\ 41.8^\circ,\ 138.2^\circ,\ 150^\circ\)
- B\(x = 30^\circ,\ 41.8^\circ\)
- Cno solutions
- D\(x = 48.2^\circ,\ 60^\circ,\ 300^\circ,\ 311.8^\circ\)
Which expression is identical to \(\dfrac{\operatorname{cosec} x}{\operatorname{cosec} x - \sin x}\)?
- A\(\sec^2 x\)
- B\(\operatorname{cosec}^2 x\)
- C\(\cos^2 x\)
- D\(\sec x\)
Solve \(\tan\theta + 3\cot\theta = 4\) for \(0^\circ < \theta < 360^\circ\).
- A\(\theta = 45^\circ,\ 71.6^\circ,\ 225^\circ,\ 251.6^\circ\)
- B\(\theta = 45^\circ,\ 71.6^\circ\)
- C\(\theta = 45^\circ,\ 225^\circ\)
- D\(\theta = 18.4^\circ,\ 45^\circ,\ 198.4^\circ,\ 225^\circ\)
Quiz key
\(\sin 240^\circ = -\sin 60^\circ = -\dfrac{\sqrt{3}}{2}\), so \(\operatorname{cosec} 240^\circ = -\dfrac{2}{\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}\). The tempting positive value forgets that sine is negative in the third quadrant. \(-2\) is \(\sec 240^\circ\) (it uses cosine), and \(-\dfrac{\sqrt{3}}{2}\) is \(\sin 240^\circ\) itself, not its reciprocal.
\(\cos x\) lies between \(-1\) and \(1\), so \(\sec x = \dfrac{1}{\cos x}\) is always \(\le -1\) or \(\ge 1\). There is no solution exactly when \(-1 < k < 1\). The tempting \(k < -1\) or \(k > 1\) is the range of \(\sec x\), where solutions DO exist. \(-1 \le k \le 1\) wrongly excludes \(k = \pm 1\), which give \(x = 0^\circ\) and \(180^\circ\); \(\sec x\) can be negative, so \(k < 0\) is wrong too.
\(\sin x = \dfrac{1}{4}\), so \(x = \sin^{-1} 0.25 = 14.5^\circ\) or \(180^\circ - 14.5^\circ = 165.5^\circ\). The tempting \(14.5^\circ, 345.5^\circ\) uses the cosine pattern (\(360^\circ - x\)) for the second solution. \(75.5^\circ, 284.5^\circ\) solves \(\cos x = 0.25\), treating cosec as the reciprocal of cosine, and \(14.5^\circ\) only misses the second-quadrant solution.
\(\sec^2\theta = 1 + \dfrac{144}{25} = \dfrac{169}{25}\), so \(\sec\theta = \pm\dfrac{13}{5}\). In the third quadrant \(\cos\theta\) is negative, so \(\sec\theta = -\dfrac{13}{5}\). The tempting \(\dfrac{13}{5}\) takes the positive square root without checking the quadrant. \(-\dfrac{13}{12}\) is \(\operatorname{cosec}\theta\), and \(-\dfrac{5}{13}\) is \(\cos\theta\), not its reciprocal.
\(2(1 + \tan^2\theta) - 5\tan\theta = 0\) gives \(2\tan^2\theta - 5\tan\theta + 2 = 0\), so \((2\tan\theta - 1)(\tan\theta - 2) = 0\). \(\tan\theta = \dfrac{1}{2}\): \(26.6^\circ, 206.6^\circ\). \(\tan\theta = 2\): \(63.4^\circ, 243.4^\circ\). The tempting \(26.6^\circ, 63.4^\circ\) stops at the calculator values and forgets the \(+180^\circ\) solutions. \(116.6^\circ, 153.4^\circ, \ldots\) come from a sign slip giving \(\tan\theta = -2\) and \(-\dfrac{1}{2}\); the last option uses the sine pattern \(180^\circ - x\), which is wrong for tangent.
\(2(\operatorname{cosec}^2 x - 1) - 7\operatorname{cosec} x + 8 = 0\) gives \(2\operatorname{cosec}^2 x - 7\operatorname{cosec} x + 6 = 0\), so \((2\operatorname{cosec} x - 3)(\operatorname{cosec} x - 2) = 0\). Both roots are allowed (\(\ge 1\)). \(\operatorname{cosec} x = 2\): \(\sin x = \dfrac{1}{2}\), \(x = 30^\circ, 150^\circ\). \(\operatorname{cosec} x = \dfrac{3}{2}\): \(\sin x = \dfrac{2}{3}\), \(x = 41.8^\circ, 138.2^\circ\). The tempting "no solutions" sets \(\sin x = 2\) and \(\sin x = \dfrac{3}{2}\), forgetting to take reciprocals. \(30^\circ, 41.8^\circ\) misses the second-quadrant angles, and the last option solves \(\cos x = \dfrac{1}{2}\) and \(\dfrac{2}{3}\).
Multiply top and bottom by \(\sin x\): \(\dfrac{1}{1 - \sin^2 x} = \dfrac{1}{\cos^2 x} = \sec^2 x\). The tempting \(\cos^2 x\) simplifies the denominator correctly but forgets that it is still underneath, so the answer is its reciprocal. \(\operatorname{cosec}^2 x\) comes from using \(1 - \sin^2 x = \sin^2 x\), and \(\sec x\) forgets that \(\cos^2 x\) is a square.
\(t + \dfrac{3}{t} = 4\) gives \(t^2 - 4t + 3 = 0\), so \((t - 1)(t - 3) = 0\). \(\tan\theta = 1\): \(45^\circ, 225^\circ\). \(\tan\theta = 3\): \(71.6^\circ, 251.6^\circ\). The tempting \(45^\circ, 71.6^\circ\) forgets that tangent repeats every \(180^\circ\). \(45^\circ, 225^\circ\) loses the root \(t = 3\), and the last option solves \(\tan\theta = \dfrac{1}{3}\) (that is, \(\cot\theta = 3\)) in place of \(\tan\theta = 3\).
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.