Two angles that add to $90^\circ$ are called complementary. In a right-angled triangle, the two non-right angles are always complementary — and this creates a beautiful symmetry between sine and cosine, tangent and cotangent, secant and cosecant. In this lesson, you will learn these co-function relationships and how to use them to simplify calculations.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
In a right-angled triangle, the two acute angles add up to $90^\circ$. If one angle is $\theta$, the other is $90^\circ - \theta$. How do you think the sine of one angle relates to the cosine of the other? Try to explain why this relationship exists using the definitions of sine and cosine in a right triangle.
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: tan(θ) = sin(θ) + cos(θ).
Right: tan(θ) = sin(θ)/cos(θ); it is a ratio, not a sum.
📚 Core Content
Consider a right-angled triangle with acute angles $\theta$ and $(90^\circ - \theta)$. If you look at $\sin \theta$, it is $\frac{\text{opposite}}{\text{hypotenuse}}$ relative to $\theta$. But that same side is the adjacent side relative to $(90^\circ - \theta)$. Therefore:
$$\sin \theta = \cos(90^\circ - \theta)$$
This is the essence of the co-function identities. The same logic applies to all trig ratios:
| Identity | In degrees | In radians |
|---|---|---|
| $\sin \theta = \cos(90^\circ - \theta)$ | $\sin \theta = \cos(90^\circ - \theta)$ | $\sin \theta = \cos\left(\frac{\pi}{2} - \theta\right)$ |
| $\cos \theta = \sin(90^\circ - \theta)$ | $\cos \theta = \sin(90^\circ - \theta)$ | $\cos \theta = \sin\left(\frac{\pi}{2} - \theta\right)$ |
| $\tan \theta = \cot(90^\circ - \theta)$ | $\tan \theta = \cot(90^\circ - \theta)$ | $\tan \theta = \cot\left(\frac{\pi}{2} - \theta\right)$ |
| $\sec \theta = \csc(90^\circ - \theta)$ | $\sec \theta = \csc(90^\circ - \theta)$ | $\sec \theta = \csc\left(\frac{\pi}{2} - \theta\right)$ |
The angle $\frac{\pi}{2} - \theta$ represents a reflection of $\theta$ across the line $y = x$. On the unit circle, this swaps the $x$- and $y$-coordinates. That is why:
🧮 Worked Examples
🧪 Activities
1 $\sin 28^\circ$
Type your answer:
Answer in your workbook.
2 $\cos \frac{\pi}{5}$
Type your answer:
Answer in your workbook.
3 $\tan 15^\circ$
Type your answer:
Answer in your workbook.
4 $\sec \frac{\pi}{8}$
Type your answer:
Answer in your workbook.
1 $\frac{\cos 35^\circ}{\sin 55^\circ}$
Type your answer:
Answer in your workbook.
2 $\sin^2 20^\circ + \sin^2 70^\circ$
Type your answer:
Answer in your workbook.
3 Solve $\cos 2\theta = \sin \theta$ for $0^\circ \leq \theta \leq 90^\circ$.
Type your answer:
Answer in your workbook.
Earlier you were asked: How does the sine of one acute angle in a right triangle relate to the cosine of the other?
In a right-angled triangle, the two acute angles are complementary: $\theta + (90^\circ - \theta) = 90^\circ$. The side that is opposite $\theta$ is adjacent to $(90^\circ - \theta)$. Therefore:
$$\sin \theta = \frac{\text{opposite to } \theta}{\text{hypotenuse}} = \frac{\text{adjacent to } (90^\circ - \theta)}{\text{hypotenuse}} = \cos(90^\circ - \theta)$$
This is the geometric origin of all co-function 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. (a) Express $\sin 68^\circ$ as a cosine of a complementary angle. (b) Hence, evaluate $\sin^2 68^\circ + \sin^2 22^\circ$ without a calculator. 3 MARKS
Type your answer below:
Answer in your workbook.
9. Solve $\tan 3\theta = \cot \theta$ for $0^\circ \leq \theta \leq 45^\circ$. Show all working. 3 MARKS
Type your answer below:
Answer in your workbook.
10. A student claims that $\sec(90^\circ - \theta) = \csc \theta$ for all values of $\theta$ where both sides are defined. Prove this identity and explain why the restriction "where both sides are defined" is necessary. 3 MARKS
Type your answer below:
Answer in your workbook.
1. $\cos 62^\circ$
2. $\sin \frac{3\pi}{10}$ (or $\sin 54^\circ$)
3. $\cot 75^\circ$
4. $\csc \frac{3\pi}{8}$
1. $\frac{\cos 35^\circ}{\sin 55^\circ} = \frac{\cos 35^\circ}{\cos 35^\circ} = 1$
2. $\sin^2 20^\circ + \cos^2 20^\circ = 1$
3. $\cos 2\theta = \cos(90^\circ - \theta) \Rightarrow 2\theta = 90^\circ - \theta \Rightarrow \theta = 30^\circ$
1. A — Definition of co-function.
2. A — $\cos 30^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}$.
3. A — $\tan \alpha = \cot \beta$ when $\alpha + \beta = 90^\circ$.
4. A — $\sin 15^\circ = \cos 75^\circ$.
5. A — $\sec(90^\circ - \theta) = \frac{1}{\cos(90^\circ - \theta)} = \frac{1}{\sin \theta} = \csc \theta$.
Q8 (3 marks): (a) $\sin 68^\circ = \cos 22^\circ$ [1]. (b) $\sin^2 68^\circ + \sin^2 22^\circ = \cos^2 22^\circ + \sin^2 22^\circ = 1$ [2].
Q9 (3 marks): $\cot \theta = \tan(90^\circ - \theta)$ [1]. So $\tan 3\theta = \tan(90^\circ - \theta)$, giving $3\theta = 90^\circ - \theta$ [1]. Hence $\theta = 22.5^\circ$ [1].
Q10 (3 marks): LHS $= \sec(90^\circ - \theta) = \frac{1}{\cos(90^\circ - \theta)} = \frac{1}{\sin \theta} = \csc \theta$ = RHS [2]. The restriction is needed because $\sec(90^\circ - \theta)$ is undefined when $\cos(90^\circ - \theta) = 0$ (i.e., $\theta = 0^\circ, 180^\circ, \\dots$), and $\csc \theta$ is undefined when $\sin \theta = 0$ (same values) [1].
Defend your ship by blasting the correct answers for Complementary Angle Relationships. Scores count toward the Asteroid Blaster leaderboard.
Play Asteroid Blaster →Complementary Angle Relationships
Tick when you've finished all activities and checked your answers.