The Pythagorean theorem is one of the most famous results in mathematics. But did you know it hides inside every trigonometric function? In this lesson, you will discover the three Pythagorean identities that connect sine, cosine, tangent, and their reciprocals — identities so powerful they appear in almost every trigonometry problem you will ever solve.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
You already know that $\sin^2 \theta + \cos^2 \theta = 1$. What do you think happens if you divide every term in this equation by $\sin^2 \theta$? And what happens if you divide every term by $\cos^2 \theta$? Try to predict the results before reading on.
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: The sine and cosine functions have a period of 360° for all transformations.
Right: The period is affected by horizontal dilation; y = sin(nx) has period 360°/n, not 360°.
📚 Core Content
This is the fundamental identity, derived directly from the unit circle equation $x^2 + y^2 = 1$:
$$\sin^2 \theta + \cos^2 \theta = 1$$
Divide Identity 1 by $\cos^2 \theta$ (provided $\cos \theta \neq 0$):
$$\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}$$
Which simplifies to:
$$\tan^2 \theta + 1 = \sec^2 \theta$$
Divide Identity 1 by $\sin^2 \theta$ (provided $\sin \theta \neq 0$):
$$\frac{\sin^2 \theta}{\sin^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}$$
Which simplifies to:
$$1 + \cot^2 \theta = \csc^2 \theta$$
Each identity is useful in different situations:
| From | Rearrangements |
|---|---|
| $\sin^2 \theta + \cos^2 \theta = 1$ | $\sin^2 \theta = 1 - \cos^2 \theta$ $\cos^2 \theta = 1 - \sin^2 \theta$ |
| $1 + \tan^2 \theta = \sec^2 \theta$ | $\tan^2 \theta = \sec^2 \theta - 1$ $\sec^2 \theta - \tan^2 \theta = 1$ |
| $1 + \cot^2 \theta = \csc^2 \theta$ | $\cot^2 \theta = \csc^2 \theta - 1$ $\csc^2 \theta - \cot^2 \theta = 1$ |
🧮 Worked Examples
🧪 Activities
1 If $\tan \theta = 2$ and $\theta$ is acute, find $\sec \theta$.
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2 If $\csc \theta = 3$ and $\frac{\pi}{2} < \theta < \pi$, find $\cot \theta$.
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3 If $\cos \theta = -\frac{2}{3}$ and $\pi < \theta < \frac{3\pi}{2}$, find $\sin \theta$.
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4 If $\sec \theta = -\frac{5}{3}$ and $\frac{\pi}{2} < \theta < \pi$, find $\tan \theta$.
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1 $\sin^2 \theta + \sin^2 \theta \cot^2 \theta$
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2 $\frac{\csc^2 \theta - 1}{\cot \theta}$
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3 $(\sec \theta + 1)(\sec \theta - 1)$
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Earlier you were asked: What happens if you divide $\sin^2 \theta + \cos^2 \theta = 1$ by $\sin^2 \theta$ or $\cos^2 \theta$?
Dividing by $\cos^2 \theta$:
$$\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} \Rightarrow \tan^2 \theta + 1 = \sec^2 \theta$$
Dividing by $\sin^2 \theta$:
$$\frac{\sin^2 \theta}{\sin^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta} \Rightarrow 1 + \cot^2 \theta = \csc^2 \theta$$
One simple identity, two divisions, three powerful results. This is the elegance of the Pythagorean identities.
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. Prove that $1 + \cot^2 \theta = \csc^2 \theta$ starting from $\sin^2 \theta + \cos^2 \theta = 1$. State any restrictions. 3 MARKS
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9. If $\tan \theta = -\frac{4}{3}$ and $\frac{3\pi}{2} < \theta < 2\pi$, find the exact values of $\sin \theta$ and $\cos \theta$. Show all working. 4 MARKS
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10. Simplify $\frac{\sin^2 \theta}{1 - \cos \theta}$ to a single trigonometric expression. Justify each step. 3 MARKS
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Answer in your workbook.
1. $1 + 4 = \sec^2 \theta \Rightarrow \sec \theta = \sqrt{5}$ (positive, acute)
2. $1 + \cot^2 \theta = 9 \Rightarrow \cot^2 \theta = 8$. In QII, cot is negative, so $\cot \theta = -2\sqrt{2}$.
3. $\sin^2 \theta = 1 - \frac{4}{9} = \frac{5}{9}$. In QIII, sine is negative, so $\sin \theta = -\frac{\sqrt{5}}{3}$.
4. $\tan^2 \theta = \frac{25}{9} - 1 = \frac{16}{9}$. In QII, tan is negative, so $\tan \theta = -\frac{4}{3}$.
1. $\sin^2 \theta(1 + \cot^2 \theta) = \sin^2 \theta \cdot \csc^2 \theta = \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 1$
2. $\frac{\cot^2 \theta}{\cot \theta} = \cot \theta$
3. $\sec^2 \theta - 1 = \tan^2 \theta$ (difference of squares)
1. A — Fundamental Pythagorean identity.
2. A — $1 + \tan^2 \theta = \sec^2 \theta$.
3. A — $1 + \cot^2 \theta = \csc^2 \theta$.
4. A — $\cos^2 \theta = 1 - \frac{4}{9} = \frac{5}{9}$.
5. A — $\sec^2 \theta - \tan^2 \theta = 1$.
Q8 (3 marks): Start with $\sin^2 \theta + \cos^2 \theta = 1$ [0.5]. Divide by $\sin^2 \theta$ (provided $\sin \theta \neq 0$) [0.5]: $1 + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}$ [1]. This gives $1 + \cot^2 \theta = \csc^2 \theta$ [1].
Q9 (4 marks): Using $1 + \tan^2 \theta = \sec^2 \theta$: $\sec^2 \theta = 1 + \frac{16}{9} = \frac{25}{9}$ [1]. In QIV, $\sec \theta > 0$, so $\sec \theta = \frac{5}{3}$, giving $\cos \theta = \frac{3}{5}$ [1]. Using $\tan \theta = \frac{\sin \theta}{\cos \theta}$: $\sin \theta = -\frac{4}{3} \times \frac{3}{5} = -\frac{4}{5}$ [1]. Check with identity: $\frac{16}{25} + \frac{9}{25} = 1$ [1].
Q10 (3 marks): Replace $\sin^2 \theta$ with $1 - \cos^2 \theta$ [1]: $\frac{1 - \cos^2 \theta}{1 - \cos \theta} = \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 - \cos \theta}$ [1]. Cancel to get $1 + \cos \theta$ [1] (where $\cos \theta \neq 1$).
Answer questions on applying and proving Pythagorean trig identities. Pool: lessons 1–7.
Tick when you've finished all activities and checked your answers.