The shape of the lesson
By the end of the lesson, learners can
- The six formulae: Fig. 1 shows where \(\sin(A + B)\) comes from: the height of P is built from two pieces.
- Using given ratios: A common question gives one ratio of each angle and asks for a compound ratio as an exact fraction.
- Simplifying and solving: When an equation contains \(\sin(x + 30^\circ)\) or \(\cos(x - 60^\circ)\), expand the bracket with exact values for the known angle, collect the \(\sin x\) and \(\cos x\) terms, and divide to get a single \(\tan x\).
- The tangent formula: With \(\tan(A \pm B)\), an equation in \(\tan x\) usually turns into a quadratic once you clear the fraction.
- Exact values: split the angle into two from \(30^\circ, 45^\circ, 60^\circ\) (or \(\dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}\)).
- Signs: sine keeps the sign, cosine flips it; tangent keeps it on top and flips it underneath.
- Given ratios: find the missing ratios from a right-angled triangle, then fix signs by quadrant.
- Equations: expand, collect \(\sin x\) and \(\cos x\), divide to get \(\tan x = k\).
- Tan equations: clear the fraction and solve the quadratic in \(\tan x\).
One picture every idea returns to
\(\sin(A \pm B) \equiv \sin A\cos B \pm \cos A\sin B\), \(\cos(A \pm B) \equiv \cos A\cos B \mp \sin A\sin B\), \(\tan(A \pm B) \equiv \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}\). In the cosine formula the sign flips.
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 angles, one formula.. Papers 2 and 3. \(\sin(A + B)\) is not \(\sin A + \sin B\). 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: The six formulae | Fig. 1 shows where \(\sin(A + B)\) comes from: the height of P is built from two pieces. Model the worked example Exact value of sin 75° on the board, one line at a time. | Whole class, then pairs |
| 11 to 17 min | Teach and check: Using given ratios | A common question gives one ratio of each angle and asks for a compound ratio as an exact fraction. Model the worked example Two acute angles on the board, one line at a time. Your turn: Given that \(\tan A = 3\) and \(\tan B = \dfrac{1}{2}\), with both angles acute, find \(\tan(A - B)\) and hence the angle \(A - B\). | Pairs |
| 17 to 23 min | Teach and check: Simplifying and solving | When an equation contains \(\sin(x + 30^\circ)\) or \(\cos(x - 60^\circ)\), expand the bracket with exact values for the known angle, collect the \(\sin x\) and \(\cos x\) terms, and divide to get a single \(\tan x\). Model the worked example Simplify on the board, one line at a time. | Pairs |
| 23 to 30 min | Teach and check: The tangent formula | With \(\tan(A \pm B)\), an equation in \(\tan x\) usually turns into a quadratic once you clear the fraction. Model the worked example A quadratic in tan x on the board, one line at a time. | 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, 21 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 The tangent formula, 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 |
|---|---|
| \(\sin(A + B) \ne \sin A + \sin B\). | Test it: \(\sin(30^\circ + 60^\circ) = 1\), but \(\sin 30^\circ + \sin 60^\circ = 1.37\). Brackets after sin, cos or tan never multiply out. |
| Signs come from the quadrant, not from the triangle. | If \(A\) is obtuse and \(\sin A = \dfrac{5}{13}\), then \(\cos A = -\dfrac{12}{13}\); forgetting the minus sign wrecks every later line. |
| Do not write \(\sin(x + 30^\circ) = \sin x + \dfrac{1}{2}\). | Expand properly: the known angle's sine AND cosine both appear. |
| The tangent formula has a 1 in the denominator: \(\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}\), with a MINUS sign for the plus case. | Return to The tangent formula: ask a pair to state the correct version and show it on the lab before moving on. |
Compound angle formulae: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
Find the exact value of \(\cos 15^\circ\).
Find the exact value of \(\tan 75^\circ\), giving your answer in the form \(a + \sqrt{b}\).
Prove that \(\cos(x + 60^\circ) + \cos(x - 60^\circ) \equiv \cos x\).
Solve the equation \(\cos(x - 60^\circ) = 3\sin x\) for \(0^\circ \le x \le 360^\circ\).
Given that \(\tan(x + y) = 3\) and \(\tan x = 2\), find the exact value of \(\tan y\).
(a) Show that \(\sin(\theta + 60^\circ) + \cos(\theta + 30^\circ) \equiv \sqrt{3}\cos\theta\).
(b) Hence solve the equation \(\sin(\theta + 60^\circ) + \cos(\theta + 30^\circ) = 1\) for \(0^\circ < \theta < 360^\circ\).
Answer key and marking notes
- \(\cos(45^\circ - 30^\circ) = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ\).
- \(= \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}\).
- \(\tan(45^\circ + 30^\circ) = \dfrac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}\).
- Rationalise: \(\dfrac{(\sqrt{3} + 1)^2}{3 - 1} = \dfrac{4 + 2\sqrt{3}}{2}\).
- \(= 2 + \sqrt{3}\).
- LHS \(= (\cos x\cos 60^\circ - \sin x\sin 60^\circ) + (\cos x\cos 60^\circ + \sin x\sin 60^\circ)\).
- The \(\sin x\) terms cancel: \(= 2\cos x\cos 60^\circ = 2\cos x \times \dfrac{1}{2} = \cos x\) = RHS.
- \(\dfrac{1}{2}\cos x + \dfrac{\sqrt{3}}{2}\sin x = 3\sin x\).
- \(\dfrac{1}{2}\cos x = \left(3 - \dfrac{\sqrt{3}}{2}\right)\sin x\), so \(\tan x = \dfrac{1/2}{3 - \sqrt{3}/2} = \dfrac{1}{6 - \sqrt{3}} = 0.2343\).
- \(x = 13.2^\circ\) or \(13.2^\circ + 180^\circ = 193.2^\circ\).
- Let \(t = \tan y\): \(\dfrac{2 + t}{1 - 2t} = 3\).
- \(2 + t = 3 - 6t\), so \(7t = 1\).
- \(\tan y = \dfrac{1}{7}\).
- (a) \(\sin(\theta + 60^\circ) = \dfrac{1}{2}\sin\theta + \dfrac{\sqrt{3}}{2}\cos\theta\).
- \(\cos(\theta + 30^\circ) = \dfrac{\sqrt{3}}{2}\cos\theta - \dfrac{1}{2}\sin\theta\).
- Add: the \(\sin\theta\) terms cancel, leaving \(\sqrt{3}\cos\theta\), as required.
- (b) \(\sqrt{3}\cos\theta = 1\), so \(\cos\theta = \dfrac{1}{\sqrt{3}} = 0.5774\).
- \(\theta = 54.7^\circ\) or \(360^\circ - 54.7^\circ = 305.3^\circ\).
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 is the correct expansion of \(\cos(A - B)\)?
- A\(\cos A\cos B + \sin A\sin B\)
- B\(\cos A\cos B - \sin A\sin B\)
- C\(\sin A\cos B - \cos A\sin B\)
- D\(\cos A - \cos B\)
Find the exact value of \(\sin 15^\circ\).
- A\(\dfrac{\sqrt{6} - \sqrt{2}}{4}\)
- B\(\dfrac{\sqrt{6} + \sqrt{2}}{4}\)
- C\(\dfrac{\sqrt{2} - 1}{2}\)
- D\(\dfrac{\sqrt{2} - \sqrt{6}}{4}\)
Find the exact value of \(\cos 80^\circ\cos 20^\circ + \sin 80^\circ\sin 20^\circ\).
- A\(\dfrac{1}{2}\)
- B\(\dfrac{\sqrt{3}}{2}\)
- C\(-0.174\)
- D\(1\)
Angles \(A\) and \(B\) are acute, with \(\tan A = \dfrac{3}{4}\) and \(\tan B = \dfrac{1}{7}\). Find the value of \(\tan(A + B)\).
- A\(1\)
- B\(\dfrac{25}{31}\)
- C\(\dfrac{25}{28}\)
- D\(\dfrac{17}{31}\)
Angle \(A\) is obtuse with \(\sin A = \dfrac{3}{5}\), and angle \(B\) is acute with \(\cos B = \dfrac{12}{13}\). Find the exact value of \(\sin(A + B)\).
- A\(\dfrac{16}{65}\)
- B\(\dfrac{56}{65}\)
- C\(-\dfrac{63}{65}\)
- D\(-\dfrac{16}{65}\)
Solve \(\sin(x + 60^\circ) = 2\cos x\) for \(0^\circ \le x \le 360^\circ\).
- A\(x = 66.2^\circ,\ 246.2^\circ\)
- B\(x = 80.1^\circ,\ 260.1^\circ\)
- C\(x = 48.6^\circ,\ 228.6^\circ\)
- D\(x = 66.2^\circ\) only
Solve \(\tan(x + 45^\circ) = 6\tan x\) for \(0^\circ < x < 180^\circ\).
- A\(x = 18.4^\circ,\ 26.6^\circ\)
- B\(x = 11.3^\circ\)
- C\(x = 9.5^\circ\)
- D\(x = 18.4^\circ,\ 26.6^\circ,\ 198.4^\circ,\ 206.6^\circ\)
Given that \(\sin(x + 30^\circ) = 2\cos(x + 60^\circ)\), find the exact value of \(\tan x\).
- A\(\dfrac{\sqrt{3}}{9}\)
- B\(-\dfrac{\sqrt{3}}{3}\)
- C\(3\sqrt{3}\)
- D\(0\)
Quiz key
\(\cos(A - B) \equiv \cos A\cos B + \sin A\sin B\): the sign flips. The tempting \(\cos A\cos B - \sin A\sin B\) is \(\cos(A + B)\). \(\sin A\cos B - \cos A\sin B\) is \(\sin(A - B)\), and \(\cos A - \cos B\) multiplies out a bracket after a function, which is never allowed (try \(A = B\): the left side is 1, the right side is 0).
\(\sin(45^\circ - 30^\circ) = \sin 45^\circ\cos 30^\circ - \cos 45^\circ\sin 30^\circ = \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6} - \sqrt{2}}{4}\) (about 0.259). The tempting \(\dfrac{\sqrt{6} + \sqrt{2}}{4}\) uses a plus sign; it is \(\cos 15^\circ\). \(\dfrac{\sqrt{2} - 1}{2}\) is \(\sin 45^\circ - \sin 30^\circ\), the forbidden "multiply out the bracket", and \(\dfrac{\sqrt{2} - \sqrt{6}}{4}\) is negative, which is impossible for an acute angle.
\(\cos A\cos B + \sin A\sin B = \cos(A - B)\), so the expression is \(\cos(80^\circ - 20^\circ) = \cos 60^\circ = \dfrac{1}{2}\). The tempting \(-0.174\) is \(\cos 100^\circ\): it uses \(\cos(A + B)\), but the plus sign in the expansion belongs to \(\cos(A - B)\). \(\dfrac{\sqrt{3}}{2}\) is \(\sin 60^\circ\), and 1 confuses the pattern with \(\cos^2\theta + \sin^2\theta\).
Top: \(\dfrac{3}{4} + \dfrac{1}{7} = \dfrac{25}{28}\). Bottom: \(1 - \dfrac{3}{28} = \dfrac{25}{28}\). So \(\tan(A + B) = 1\) and \(A + B = 45^\circ\). The tempting \(\dfrac{25}{31}\) uses \(1 + \tan A\tan B\) in the denominator, the sign belonging to \(\tan(A - B)\). \(\dfrac{25}{28}\) is just \(\tan A + \tan B\), and \(\dfrac{17}{31}\) is \(\tan(A - B)\).
\(A\) is obtuse, so \(\cos A = -\dfrac{4}{5}\); \(B\) is acute, so \(\sin B = \dfrac{5}{13}\). Then \(\sin(A + B) = \sin A\cos B + \cos A\sin B = \dfrac{3}{5} \cdot \dfrac{12}{13} - \dfrac{4}{5} \cdot \dfrac{5}{13} = \dfrac{36 - 20}{65} = \dfrac{16}{65}\). The tempting \(\dfrac{56}{65}\) takes \(\cos A = +\dfrac{4}{5}\), ignoring the quadrant. \(-\dfrac{63}{65}\) is \(\cos(A + B)\), and \(-\dfrac{16}{65}\) is a sign slip in the final line.
\(\dfrac{1}{2}\sin x + \dfrac{\sqrt{3}}{2}\cos x = 2\cos x\). Multiply by 2: \(\sin x = (4 - \sqrt{3})\cos x\), so \(\tan x = 4 - \sqrt{3} = 2.268\). Then \(x = 66.2^\circ\) or \(66.2^\circ + 180^\circ = 246.2^\circ\). The tempting \(80.1^\circ, 260.1^\circ\) comes from \(\tan x = 4 + \sqrt{3}\), a sign slip when moving \(\sqrt{3}\cos x\) across. \(48.6^\circ, 228.6^\circ\) drops the \(\dfrac{1}{2}\) in front of \(\sin x\), giving \(\tan x = 2 - \dfrac{\sqrt{3}}{2}\), and \(66.2^\circ\) only misses the second solution of the tangent equation.
\(\dfrac{1 + t}{1 - t} = 6t\) gives \(1 + t = 6t - 6t^2\), so \(6t^2 - 5t + 1 = 0\) and \((2t - 1)(3t - 1) = 0\). \(\tan x = \dfrac{1}{2}\): \(x = 26.6^\circ\); \(\tan x = \dfrac{1}{3}\): \(x = 18.4^\circ\). Adding \(180^\circ\) leaves the range, so these are the only solutions. The tempting \(11.3^\circ\) writes \(\tan(x + 45^\circ) = \tan x + 1\), so \(\tan x = 0.2\). \(9.5^\circ\) uses \(1 + t\) in the denominator, which cancels to \(6t = 1\), and the last option includes angles outside \(0^\circ < x < 180^\circ\).
LHS \(= \dfrac{\sqrt{3}}{2}\sin x + \dfrac{1}{2}\cos x\). RHS \(= 2\left(\dfrac{1}{2}\cos x - \dfrac{\sqrt{3}}{2}\sin x\right) = \cos x - \sqrt{3}\sin x\). Collect: \(\dfrac{3\sqrt{3}}{2}\sin x = \dfrac{1}{2}\cos x\), so \(\tan x = \dfrac{1}{3\sqrt{3}} = \dfrac{\sqrt{3}}{9}\). The tempting \(-\dfrac{\sqrt{3}}{3}\) expands \(\cos(x + 60^\circ)\) with a plus sign (the cosine sign must flip). \(3\sqrt{3}\) is \(\dfrac{\cos x}{\sin x}\), the reciprocal, and 0 comes from dropping the factor 2 on the right.
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.