The shape of the lesson
By the end of the lesson, learners can
- The special triangles: In the lab, step the angle through \(30^\circ\), \(45^\circ\) and \(60^\circ\) and compare the exact values it shows with the sides of the triangles in Fig. 1.
- Related angles of any size: Every angle has a basic acute angle (its angle to the \(x\)-axis) and a sign that depends on the quadrant.
- The two identities: An identity (symbol \(\equiv\)) is true for every value of \(\theta\).
- Exact values: rebuild them from the two special triangles.
- Related angles: acute angle to the \(x\)-axis, then the sign of that quadrant.
- One ratio to another: \(\sin^2\theta + \cos^2\theta = 1\), then choose the sign from the quadrant.
- Proofs: start from one side, combine fractions, replace \(1 - \cos^2\theta\) by \(\sin^2\theta\) (or the reverse), finish on the other side.
- Exact answers: rationalise surd denominators when asked.
One picture every idea returns to
Learn \(\sin 30^\circ = \tfrac{1}{2}\), \(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\), \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\) (cosine runs the same list backwards) and get every related angle from its sign in that quadrant. For any \(\theta\): \(\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}\) and \(\sin^2\theta + \cos^2\theta \equiv 1\).
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, Surds, not decimals.. Two small triangles give you the exact sine, cosine and tangent of \(30^\circ\), \(45^\circ\) and \(60^\circ\); symmetry extends them to angles of any size. 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 special triangles | In the lab, step the angle through \(30^\circ\), \(45^\circ\) and \(60^\circ\) and compare the exact values it shows with the sides of the triangles in Fig. 1. Project tab 2 of the lab (Trigonometric graphs lab, exact values) and let a student drive it while the class predicts. Model the worked example Combining exact values on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Related angles of any size | Every angle has a basic acute angle (its angle to the \(x\)-axis) and a sign that depends on the quadrant. Model the worked example Three related angles on the board, one line at a time. Your turn: Find the exact value of \(\tan\dfrac{5\pi}{4} - 2\cos\dfrac{7\pi}{6}\). | Pairs |
| 21 to 30 min | Teach and check: The two identities | An identity (symbol \(\equiv\)) is true for every value of \(\theta\). Model the worked example One ratio gives the others on the board, one line at a time. Your turn: Prove that \(\tan\theta + \dfrac{1}{\tan\theta} \equiv \dfrac{1}{\sin\theta\cos\theta}\). | 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, 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 two 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 |
|---|---|
| \(\tan^2 30^\circ\) is \((\tan 30^\circ)^2 = \dfrac{1}{3}\), not \(\tan(900^\circ)\) and not \(\tan 30^\circ\) doubled. | Return to The special triangles: ask a pair to state the correct version and show it on the lab before moving on. |
| Find the acute angle to the \(x\)-axis, not to the \(y\)-axis. \(150^\circ\) is \(30^\circ\) from the axis (\(180^\circ - 150^\circ\)), not \(60^\circ\) (\(150^\circ - 90^\circ\)). | Return to Related angles of any size: ask a pair to state the correct version and show it on the lab before moving on. |
| After \(\cos^2\theta = \dfrac{144}{169}\), the square root has two signs. | The quadrant decides which one; dropping the minus sign for an obtuse angle loses the marks. |
Exact values and the two identities: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
Without a calculator, find the exact values of (a) \(\sin 240^\circ\) (b) \(\cos\dfrac{7\pi}{4}\) (c) \(\tan\dfrac{5\pi}{6}\).
Without a calculator, find the exact value of \(3\tan^2\dfrac{\pi}{6} + 4\sin\dfrac{3\pi}{4}\cos\dfrac{\pi}{4}\).
Given that \(\cos\theta = -\dfrac{3}{4}\) and \(180^\circ < \theta < 270^\circ\), find the exact values of \(\sin\theta\) and \(\tan\theta\).
Prove the identity \(\dfrac{1 + \sin\theta}{\cos\theta} + \dfrac{\cos\theta}{1 + \sin\theta} \equiv \dfrac{2}{\cos\theta}\).
Prove the identity \(\tan\theta - \dfrac{1}{\tan\theta} \equiv \dfrac{2\sin^2\theta - 1}{\sin\theta\cos\theta}\).
A Ferris wheel has radius 20 m and its centre is 22 m above the ground. When a car has turned through an angle \(\theta\) anticlockwise from the horizontal line through the centre (on the right), its height is \(h = 22 + 20\sin\theta\) metres and its horizontal displacement to the right of the centre is \(x = 20\cos\theta\) metres.
(a) A car is 34 m above the ground on the left-hand side of the wheel. Find the exact values of \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\), and the value of \(x\).
(b) Find the exact height of a car when \(\theta = \dfrac{7\pi}{6}\) and when \(\theta = \dfrac{5\pi}{4}\).
Answer key and marking notes
- (a) \(240^\circ = 180^\circ + 60^\circ\), third quadrant: \(-\sin 60^\circ = -\dfrac{\sqrt{3}}{2}\)
- (b) \(2\pi - \dfrac{\pi}{4}\), fourth quadrant: \(\cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}\)
- (c) \(\pi - \dfrac{\pi}{6}\), second quadrant: \(-\tan\dfrac{\pi}{6} = -\dfrac{1}{\sqrt{3}} = -\dfrac{\sqrt{3}}{3}\)
- \(3 \times \left(\dfrac{1}{\sqrt{3}}\right)^2 = 3 \times \dfrac{1}{3} = 1\)
- \(\sin\dfrac{3\pi}{4} = \dfrac{\sqrt{2}}{2}\), so \(4 \times \dfrac{\sqrt{2}}{2} \times \dfrac{\sqrt{2}}{2} = 4 \times \dfrac{1}{2} = 2\)
- Total \(= 1 + 2 = 3\)
- \(\sin^2\theta = 1 - \dfrac{9}{16} = \dfrac{7}{16}\), so \(\sin\theta = \pm\dfrac{\sqrt{7}}{4}\).
- Third quadrant: sine is negative, so \(\sin\theta = -\dfrac{\sqrt{7}}{4}\).
- \(\tan\theta = \dfrac{-\sqrt{7}/4}{-3/4} = \dfrac{\sqrt{7}}{3}\) (positive, as expected in the third quadrant).
- LHS \(= \dfrac{(1 + \sin\theta)^2 + \cos^2\theta}{\cos\theta(1 + \sin\theta)}\)
- Numerator \(= 1 + 2\sin\theta + \sin^2\theta + \cos^2\theta = 2 + 2\sin\theta = 2(1 + \sin\theta)\)
- LHS \(= \dfrac{2(1 + \sin\theta)}{\cos\theta(1 + \sin\theta)} = \dfrac{2}{\cos\theta}\) = RHS
- LHS \(= \dfrac{\sin\theta}{\cos\theta} - \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta - \cos^2\theta}{\sin\theta\cos\theta}\)
- Replace \(\cos^2\theta\) by \(1 - \sin^2\theta\): numerator \(= \sin^2\theta - 1 + \sin^2\theta = 2\sin^2\theta - 1\)
- So LHS \(= \dfrac{2\sin^2\theta - 1}{\sin\theta\cos\theta}\) = RHS
- (a) \(22 + 20\sin\theta = 34\) gives \(\sin\theta = \dfrac{12}{20} = \dfrac{3}{5}\).
- Left-hand side means \(\theta\) is obtuse, so \(\cos\theta = -\sqrt{1 - \dfrac{9}{25}} = -\dfrac{4}{5}\) and \(\tan\theta = -\dfrac{3}{4}\).
- \(x = 20 \times \left(-\dfrac{4}{5}\right) = -16\): the car is 16 m to the left of the centre.
- (b) \(\sin\dfrac{7\pi}{6} = -\tfrac{1}{2}\), so \(h = 22 - 10 = 12\) m.
- \(\sin\dfrac{5\pi}{4} = -\dfrac{\sqrt{2}}{2}\), so \(h = 22 - 10\sqrt{2}\) m.
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 \(\tan 225^\circ + \cos 300^\circ\).
- A\(\dfrac{3}{2}\)
- B\(-\dfrac{1}{2}\)
- C\(\dfrac{1}{2}\)
- D\(-\dfrac{3}{2}\)
Find the exact value of \(\sin\left(-\dfrac{2\pi}{3}\right)\).
- A\(-\dfrac{\sqrt{3}}{2}\)
- B\(\dfrac{\sqrt{3}}{2}\)
- C\(-\dfrac{1}{2}\)
- D\(\dfrac{1}{2}\)
Given that \(\tan\theta = 2\) and \(180^\circ < \theta < 270^\circ\), find the exact value of \(\sin\theta\).
- A\(-\dfrac{2}{\sqrt{5}}\)
- B\(\dfrac{2}{\sqrt{5}}\)
- C\(-\dfrac{1}{\sqrt{5}}\)
- D\(\dfrac{1}{\sqrt{5}}\)
Which expression is identical to \(\dfrac{1 - \cos^2\theta}{\cos\theta}\)?
- A\(\sin\theta\tan\theta\)
- B\(\tan\theta\)
- C\(\dfrac{1}{\cos\theta} - 1\)
- D\(\sin^2\theta\)
Given that \(\sin\theta = -\dfrac{7}{25}\) and \(\dfrac{3\pi}{2} < \theta < 2\pi\), find the exact value of \(\tan\theta\).
- A\(-\dfrac{7}{24}\)
- B\(\dfrac{7}{24}\)
- C\(-\dfrac{24}{7}\)
- D\(-\dfrac{7}{25}\)
Simplify \((\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2\).
- A\(2\)
- B\(0\)
- C\(1\)
- D\(4\sin\theta\cos\theta\)
Which expression is identical to \(\tan\theta + \dfrac{1}{\tan\theta}\)?
- A\(\dfrac{1}{\sin\theta\cos\theta}\)
- B\(1\)
- C\(\sin\theta\cos\theta\)
- D\(\dfrac{2}{\sin\theta\cos\theta}\)
Given that \(\tan\theta = t\), where \(0 < \theta < \dfrac{\pi}{2}\), express \(\cos\theta\) in terms of \(t\).
- A\(\dfrac{1}{\sqrt{1 + t^2}}\)
- B\(\dfrac{1}{\sqrt{1 - t^2}}\)
- C\(\sqrt{1 + t^2}\)
- D\(\dfrac{t}{\sqrt{1 + t^2}}\)
Quiz key
\(225^\circ\) is in the third quadrant, where tan is positive: \(\tan 225^\circ = \tan 45^\circ = 1\). \(300^\circ\) is in the fourth quadrant, where cos is positive: \(\cos 300^\circ = \cos 60^\circ = \tfrac{1}{2}\). Total \(\tfrac{3}{2}\). The tempting \(-\tfrac{1}{2}\) makes \(\tan 225^\circ\) negative; in the third quadrant sine and cosine are both negative, so their quotient tan is positive. \(\tfrac{1}{2}\) makes the cosine negative, and \(-\tfrac{3}{2}\) makes both negative.
\(-\dfrac{2\pi}{3}\) is the same direction as \(\dfrac{4\pi}{3}\), in the third quadrant, where sine is negative. The related angle is \(\dfrac{\pi}{3}\), so \(\sin\left(-\dfrac{2\pi}{3}\right) = -\sin\dfrac{\pi}{3} = -\dfrac{\sqrt{3}}{2}\). The tempting \(\dfrac{\sqrt{3}}{2}\) ignores the minus sign, but sine is an odd function: \(\sin(-x) = -\sin x\). \(\pm\tfrac{1}{2}\) use the cosine value of \(\dfrac{\pi}{3}\) instead of the sine.
The triangle with opposite 2 and adjacent 1 has hypotenuse \(\sqrt{1^2 + 2^2} = \sqrt{5}\), so the acute value is \(\dfrac{2}{\sqrt{5}}\). In the third quadrant sine is negative, so \(\sin\theta = -\dfrac{2}{\sqrt{5}}\). The tempting \(\dfrac{2}{\sqrt{5}}\) forgets the quadrant: a positive tan does not mean a positive sine. \(\pm\dfrac{1}{\sqrt{5}}\) is the cosine value (adjacent over hypotenuse).
\(\dfrac{1 - \cos^2\theta}{\cos\theta} = \dfrac{\sin^2\theta}{\cos\theta} = \sin\theta \times \dfrac{\sin\theta}{\cos\theta} = \sin\theta\tan\theta\). The tempting \(\dfrac{1}{\cos\theta} - 1\) splits the fraction wrongly: \(\dfrac{\cos^2\theta}{\cos\theta} = \cos\theta\), not 1, so the split is \(\dfrac{1}{\cos\theta} - \cos\theta\). \(\tan\theta\) loses a factor of \(\sin\theta\), and \(\sin^2\theta\) forgets the denominator.
\(\cos^2\theta = 1 - \dfrac{49}{625} = \dfrac{576}{625}\), and cosine is positive in the fourth quadrant, so \(\cos\theta = \dfrac{24}{25}\). Then \(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{-7/25}{24/25} = -\dfrac{7}{24}\). The tempting \(\dfrac{7}{24}\) takes the cosine as negative too; in the fourth quadrant only cosine is positive, so tan is negative. \(-\dfrac{24}{7}\) is \(\dfrac{\cos\theta}{\sin\theta}\), the quotient upside down, and \(-\dfrac{7}{25}\) is just \(\sin\theta\).
\((\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta\) and \((\sin\theta - \cos\theta)^2 = \sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta\). Adding, the cross terms cancel and \(2(\sin^2\theta + \cos^2\theta) = 2\). The tempting 1 uses the identity once and forgets there are two copies. \(4\sin\theta\cos\theta\) is the difference of the two squares, not the sum, and 0 cancels the squares instead of the cross terms.
\(\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta}\). The tempting 1 treats \(\tan\theta \times \dfrac{1}{\tan\theta} = 1\) as if it were the sum. \(\sin\theta\cos\theta\) inverts the answer, and \(\dfrac{2}{\sin\theta\cos\theta}\) counts the numerator as 2 instead of \(\sin^2\theta + \cos^2\theta = 1\).
With opposite \(t\) and adjacent 1, the hypotenuse is \(\sqrt{1 + t^2}\), so \(\cos\theta = \dfrac{1}{\sqrt{1 + t^2}}\) (positive, as \(\theta\) is acute). Check: \(\sin^2\theta + \cos^2\theta = \dfrac{t^2 + 1}{1 + t^2} = 1\). The tempting \(\dfrac{1}{\sqrt{1 - t^2}}\) subtracts as if \(t\) were a hypotenuse-based ratio like sine. \(\sqrt{1 + t^2}\) is hypotenuse over adjacent, upside down, and \(\dfrac{t}{\sqrt{1 + t^2}}\) is \(\sin\theta\).
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.