Calculators are banned in many exam questions for a reason: mathematicians want you to know the exact values. The good news is that you only need two special triangles to find the sine, cosine, and tangent of $30^\circ$, $45^\circ$, and $60^\circ$ exactly. Master these two triangles, and you have unlocked half of trigonometry.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Consider an isosceles right-angled triangle with two sides of length 1. What is the length of the hypotenuse? And what are the angles in this triangle? Use this to predict the exact values of $\sin 45^\circ$ and $\cos 45^\circ$.
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: √(a + b) = √a + √b.
Right: The square root of a sum is not the sum of square roots; √(a+b) cannot be simplified this way.
📚 Core Content
Take a square with side length 1 and cut it diagonally. You get two isosceles right-angled triangles. By Pythagoras' theorem, the hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$.
| Angle | $\sin$ | $\cos$ | $\tan$ |
|---|---|---|---|
| $45^\circ$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $1$ |
Take an equilateral triangle with side length 2 and cut it in half. You get two right-angled triangles with sides 1, $\sqrt{3}$, and 2.
| Angle | $\sin$ | $\cos$ | $\tan$ |
|---|---|---|---|
| $30^\circ$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{3}$ |
| $60^\circ$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |
Once you know the exact values for $30^\circ$, $45^\circ$, and $60^\circ$, you can find the exact value for any angle that has one of these as a reference angle. Just follow these steps:
🧮 Worked Examples
🧪 Activities
1 $\sin 60^\circ$
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Answer in your workbook.
2 $\cos \frac{\pi}{4}$
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3 $\tan \frac{\pi}{6}$
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4 $\sin^2 45^\circ + \cos^2 45^\circ$
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Answer in your workbook.
5 $\cos 30^\circ \tan 60^\circ$
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Answer in your workbook.
1 $\sin 135^\circ$
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Answer in your workbook.
2 $\cos \frac{7\pi}{6}$
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Answer in your workbook.
3 $\tan 300^\circ$
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Answer in your workbook.
4 $\cos\left(-\frac{\pi}{3}\right)$
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Answer in your workbook.
Earlier you were asked: An isosceles right triangle with legs of 1 has what hypotenuse, and what are $\sin 45^\circ$ and $\cos 45^\circ$?
By Pythagoras, the hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$. Since the two legs are equal:
This is the only acute angle where sine and cosine are exactly equal.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. Find the exact values of: (a) $\sin \frac{5\pi}{4}$ (b) $\cos 210^\circ$ (c) $\tan \frac{11\pi}{6}$. Show reference angles and ASTC reasoning for each. 3 MARKS
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Answer in your workbook.
9. Simplify the expression $\frac{\sin 60^\circ}{\cos 30^\circ} + \tan 45^\circ$ to a single exact value. 2 MARKS
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Answer in your workbook.
10. A student writes: $\cos 150^\circ = -\cos 30^\circ$. Explain why this is true using the unit circle and the concept of reference angles. 3 MARKS
Type your answer below:
Answer in your workbook.
1. $\frac{\sqrt{3}}{2}$
2. $\frac{\sqrt{2}}{2}$
3. $\frac{\sqrt{3}}{3}$
4. $1$ (Pythagorean identity)
5. $\frac{\sqrt{3}}{2} \times \sqrt{3} = \frac{3}{2}$
1. $\sin 135^\circ = \frac{\sqrt{2}}{2}$ (QII, reference $45^\circ$)
2. $\cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$ (QIII, reference $\frac{\pi}{6}$)
3. $\tan 300^\circ = -\sqrt{3}$ (QIV, reference $60^\circ$)
4. $\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}$ (QI, reference $\frac{\pi}{3}$)
1. A — $\frac{\sqrt{2}}{2}$.
2. A — $\frac{\sqrt{3}}{2}$.
3. A — $\sqrt{3}$.
4. A — $\frac{1}{2}$.
5. A — Identity equals 1.
Q8 (3 marks): (a) QIII, reference $\frac{\pi}{4}$, $\sin = -\frac{\sqrt{2}}{2}$ [1]. (b) QIII, reference $30^\circ$, $\cos = -\frac{\sqrt{3}}{2}$ [1]. (c) QIV, reference $30^\circ$, $\tan = -\frac{\sqrt{3}}{3}$ [1].
Q9 (2 marks): $\frac{\sin 60^\circ}{\cos 30^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} = 1$ [1]. $1 + \tan 45^\circ = 1 + 1 = 2$ [1].
Q10 (3 marks): $150^\circ$ is in Quadrant II [0.5]. Its reference angle is $180^\circ - 150^\circ = 30^\circ$ [0.5]. On the unit circle, the $x$-coordinate (cosine) for $150^\circ$ has the same magnitude as for $30^\circ$ but is reflected across the $y$-axis, making it negative [1.5]. Therefore $\cos 150^\circ = -\cos 30^\circ$ [0.5].
Face the boss using your knowledge of exact trig values and special triangles. Pool: lessons 1–5.
Tick when you've finished all activities and checked your answers.