The shape of the lesson
By the end of the lesson, learners can
- From compound to double: Put \(B = A\) in each compound angle formula.
- Solving equations: Replace the double angle so that only one angle, \(x\), remains.
- Identities and triple angles: In proofs, choose the version of \(\cos 2x\) that cancels the 1 that is already there: \(1 - \cos 2x = 2\sin^2 x\) and \(1 + \cos 2x = 2\cos^2 x\).
- sin: \(\sin 2A = 2\sin A\cos A\).
- cos: \(\cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A\).
- tan: \(\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}\).
- Equations: one angle, one function; factorise rather than divide.
- Proofs: \(1 - \cos 2x = 2\sin^2 x\) and \(1 + \cos 2x = 2\cos^2 x\) remove the 1.
One picture every idea returns to
\(\sin 2A \equiv 2\sin A\cos A\), \(\cos 2A \equiv \cos^2 A - \sin^2 A \equiv 2\cos^2 A - 1 \equiv 1 - 2\sin^2 A\), \(\tan 2A \equiv \dfrac{2\tan A}{1 - \tan^2 A}\). Choose the form of \(\cos 2A\) that matches the other terms in the equation.
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, Put B = A and see what falls out.. 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: From compound to double | Put \(B = A\) in each compound angle formula. Model the worked example All three from one ratio on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Solving equations | Replace the double angle so that only one angle, \(x\), remains. Model the worked example Factorise, do not divide on the board, one line at a time. Your turn: Solve \(\tan 2x = 3\tan x\) for \(0^\circ \le x \le 360^\circ\). | Pairs |
| 21 to 30 min | Teach and check: Identities and triple angles | In proofs, choose the version of \(\cos 2x\) that cancels the 1 that is already there: \(1 - \cos 2x = 2\sin^2 x\) and \(1 + \cos 2x = 2\cos^2 x\). Model the worked example Prove an identity on the board, one line at a time. | Round Table |
| 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 Identities and triple angles, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| \(\sin 2A \ne 2\sin A\). | With \(A = 90^\circ\): \(\sin 180^\circ = 0\), but \(2\sin 90^\circ = 2\). The 2 belongs inside the function, so you need the formula. |
| Dividing both sides of \(2\sin x\cos x = \cos x\) by \(\cos x\) throws away the solutions \(\cos x = 0\) (here \(90^\circ\) and \(270^\circ\)). | Bring everything to one side and factorise instead. |
| In a 'show that' question every step must be visible. | Writing down \(3\sin x - 4\sin^3 x\) straight after \(\sin(2x + x)\) earns nothing. |
Double angle formulae: worksheet
6 questions, 22 marks. Written in the style of the papers; no past paper question is reproduced.
Given that \(\sin A = \dfrac{1}{4}\), find the exact value of \(\cos 2A\).
Without a calculator, find the exact values of (a) \(\cos^2 75^\circ - \sin^2 75^\circ\) (b) \(\sin 75^\circ\cos 75^\circ\).
Solve the equation \(3\cos 2x + 7\cos x = 0\) for \(0^\circ \le x \le 360^\circ\).
Prove that \(\dfrac{\sin 2x}{1 + \cos 2x} \equiv \tan x\).
Show that \(\cos 3x \equiv 4\cos^3 x - 3\cos x\).
(a) Prove that \(\tan x + \cot x \equiv 2\operatorname{cosec} 2x\).
(b) Hence solve the equation \(\tan x + \cot x = 8\) for \(0^\circ < x < 180^\circ\).
Answer key and marking notes
- \(\cos 2A = 1 - 2\sin^2 A = 1 - 2 \times \dfrac{1}{16}\).
- \(= \dfrac{7}{8}\).
- (a) \(\cos^2 75^\circ - \sin^2 75^\circ = \cos 150^\circ = -\dfrac{\sqrt{3}}{2}\).
- (b) \(\sin 75^\circ\cos 75^\circ = \dfrac{1}{2}\sin 150^\circ = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\).
- \(3(2\cos^2 x - 1) + 7\cos x = 0\), so \(6\cos^2 x + 7\cos x - 3 = 0\).
- \((3\cos x - 1)(2\cos x + 3) = 0\).
- \(\cos x = -\dfrac{3}{2}\) is impossible.
- \(\cos x = \dfrac{1}{3}\): \(x = 70.5^\circ\) or \(360^\circ - 70.5^\circ = 289.5^\circ\).
- \(1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x\).
- LHS \(= \dfrac{2\sin x\cos x}{2\cos^2 x} = \dfrac{\sin x}{\cos x} = \tan x\) = RHS.
- \(\cos 3x = \cos(2x + x) = \cos 2x\cos x - \sin 2x\sin x\).
- \(= (2\cos^2 x - 1)\cos x - 2\sin^2 x\cos x\).
- \(= 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x\).
- \(= 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x = 4\cos^3 x - 3\cos x\).
- (a) LHS \(= \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \dfrac{1}{\sin x\cos x}\).
- \(\sin x\cos x = \dfrac{1}{2}\sin 2x\), so LHS \(= \dfrac{2}{\sin 2x} = 2\operatorname{cosec} 2x\) = RHS.
- (b) \(2\operatorname{cosec} 2x = 8\), so \(\sin 2x = \dfrac{1}{4}\).
- For \(0^\circ < 2x < 360^\circ\): \(2x = 14.48^\circ\) or \(180^\circ - 14.48^\circ = 165.52^\circ\).
- \(x = 7.2^\circ,\ 82.8^\circ\).
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.
Which expression is identical to \(\cos 2A\), written in terms of \(\cos A\) only?
- A\(2\cos^2 A - 1\)
- B\(1 - 2\cos^2 A\)
- C\(\cos^2 A - 1\)
- D\(2\cos A - 1\)
Given that \(\sin x = \dfrac{3}{5}\) and \(x\) is acute, find the exact value of \(\sin 2x\).
- A\(\dfrac{24}{25}\)
- B\(\dfrac{6}{5}\)
- C\(\dfrac{7}{25}\)
- D\(\dfrac{12}{25}\)
Which expression is identical to \(\dfrac{1 - \cos 2x}{\sin 2x}\)?
- A\(\tan x\)
- B\(\cot x\)
- C\(\tan 2x\)
- D\(\dfrac{1}{2}\tan x\)
Given that \(\tan\theta = \dfrac{1}{3}\), find the exact value of \(\tan 2\theta\).
- A\(\dfrac{3}{4}\)
- B\(\dfrac{2}{3}\)
- C\(\dfrac{3}{5}\)
- D\(1\)
Solve \(\sin 2x = \sin x\) for \(0^\circ \le x \le 360^\circ\).
- A\(x = 0^\circ,\ 60^\circ,\ 180^\circ,\ 300^\circ,\ 360^\circ\)
- B\(x = 60^\circ,\ 300^\circ\)
- C\(x = 0^\circ,\ 180^\circ,\ 360^\circ\)
- D\(x = 0^\circ,\ 30^\circ,\ 150^\circ,\ 180^\circ,\ 360^\circ\)
Solve \(\cos 2x + 3\cos x + 2 = 0\) for \(0^\circ \le x \le 360^\circ\).
- A\(x = 120^\circ,\ 180^\circ,\ 240^\circ\)
- B\(x = 120^\circ,\ 240^\circ\)
- C\(x = 0^\circ,\ 60^\circ,\ 300^\circ,\ 360^\circ\)
- Dno solutions
Given that \(\cos x = -\dfrac{1}{4}\) and \(180^\circ < x < 270^\circ\), find the exact value of \(\sin 2x\).
- A\(\dfrac{\sqrt{15}}{8}\)
- B\(-\dfrac{\sqrt{15}}{8}\)
- C\(-\dfrac{7}{8}\)
- D\(-\dfrac{1}{2}\)
Which expression is identical to \(\cos 3x\)?
- A\(4\cos^3 x - 3\cos x\)
- B\(3\cos x - 4\cos^3 x\)
- C\(4\cos^3 x - \cos x\)
- D\(3\cos x\)
Quiz key
\(\cos 2A = \cos^2 A - (1 - \cos^2 A) = 2\cos^2 A - 1\). The tempting \(1 - 2\cos^2 A\) has the signs reversed; the matching form is \(1 - 2\sin^2 A\), with SINE. \(\cos^2 A - 1\) forgets to double, and \(2\cos A - 1\) loses the square. Check with \(A = 0\): \(\cos 0 = 1\), and only \(2\cos^2 A - 1\) gives 1.
\(\cos x = \dfrac{4}{5}\) (acute), so \(\sin 2x = 2 \times \dfrac{3}{5} \times \dfrac{4}{5} = \dfrac{24}{25}\). The tempting \(\dfrac{6}{5}\) is \(2\sin x\), which is not \(\sin 2x\) (and is bigger than 1, so impossible for a sine). \(\dfrac{7}{25}\) is \(\cos 2x\), and \(\dfrac{12}{25}\) forgets the factor 2.
Use \(\cos 2x = 1 - 2\sin^2 x\): the top is \(2\sin^2 x\). The bottom is \(2\sin x\cos x\). So the fraction is \(\dfrac{2\sin^2 x}{2\sin x\cos x} = \dfrac{\sin x}{\cos x} = \tan x\). The tempting \(\cot x\) comes from using \(1 - \cos 2x = 2\cos^2 x\), the wrong way round. \(\tan 2x\) treats the 1 as if it vanished, and \(\dfrac{1}{2}\tan x\) loses the 2 in \(2\sin^2 x\).
\(\tan 2\theta = \dfrac{2/3}{1 - 1/9} = \dfrac{2/3}{8/9} = \dfrac{3}{4}\). The tempting \(\dfrac{2}{3}\) is just \(2\tan\theta\). \(\dfrac{3}{5}\) uses \(1 + \tan^2\theta\) in the denominator, and 1 forgets to square \(\tan\theta\) (\(\dfrac{2/3}{1 - 1/3}\)).
\(2\sin x\cos x - \sin x = 0\), so \(\sin x(2\cos x - 1) = 0\). \(\sin x = 0\): \(0^\circ, 180^\circ, 360^\circ\). \(\cos x = \dfrac{1}{2}\): \(60^\circ, 300^\circ\). The tempting \(60^\circ, 300^\circ\) divides both sides by \(\sin x\) and throws away the solutions where \(\sin x = 0\). \(0^\circ, 180^\circ, 360^\circ\) loses the other factor, and the last option solves \(\sin x = \dfrac{1}{2}\) instead of \(\cos x = \dfrac{1}{2}\).
\(2\cos^2 x - 1 + 3\cos x + 2 = 0\) gives \(2\cos^2 x + 3\cos x + 1 = 0\), so \((2\cos x + 1)(\cos x + 1) = 0\). \(\cos x = -\dfrac{1}{2}\): \(120^\circ, 240^\circ\). \(\cos x = -1\): \(180^\circ\). The tempting \(120^\circ, 240^\circ\) forgets the root \(\cos x = -1\). \(0^\circ, 60^\circ, 300^\circ, 360^\circ\) comes from a sign slip giving \(\cos x = \dfrac{1}{2}\) and \(1\), and "no solutions" uses \(\cos 2x = 2\cos^2 x\) (forgetting the \(-1\)), which gives \(2\cos^2 x + 3\cos x + 2 = 0\) with a negative discriminant.
\(\sin^2 x = 1 - \dfrac{1}{16} = \dfrac{15}{16}\). In the third quadrant sine is negative, so \(\sin x = -\dfrac{\sqrt{15}}{4}\). Then \(\sin 2x = 2\left(-\dfrac{\sqrt{15}}{4}\right)\left(-\dfrac{1}{4}\right) = \dfrac{\sqrt{15}}{8}\). The tempting \(-\dfrac{\sqrt{15}}{8}\) takes \(\sin x\) as positive; two negatives multiply to a positive. \(-\dfrac{7}{8}\) is \(\cos 2x = 2\cos^2 x - 1\), and \(-\dfrac{1}{2}\) is \(2\cos x\).
\(\cos(2x + x) = \cos 2x\cos x - \sin 2x\sin x = (2\cos^2 x - 1)\cos x - 2\sin^2 x\cos x\). Replace \(\sin^2 x\): \(= 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x = 4\cos^3 x - 3\cos x\). The tempting \(3\cos x - 4\cos^3 x\) copies the pattern of \(\sin 3x = 3\sin x - 4\sin^3 x\) with the signs the wrong way round for cosine (check \(x = 0\): \(\cos 0 = 1\), and only \(4 - 3 = 1\) works). \(4\cos^3 x - \cos x\) misses the \(-2\cos x\) from \(\sin^2 x\), and \(3\cos x\) is the "multiply out the 3" error.
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.