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.
Today's hook, 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.
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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.
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Co-function identities
Work through the core explanation before applying it.
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Co-function identities
+5 XP to read
The prefix "co-" in cosine, cotangent, and cosecant stands for "complementary". These functions are exactly the original functions evaluated at the complementary angle. This is not a coincidence, it is built into the geometry of right triangles.
In a right-angled triangle with acute angles $\theta$ and $(90^\circ - \theta)$, the side opposite $\theta$ is adjacent to $(90^\circ - \theta)$. Therefore $\sin \theta = \cos(90^\circ - \theta)$. The same logic extends to all six trig ratios.
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
The complementary angle identities for sine, cosine, and tangent
The corresponding identities for secant, cosecant, and cotangent
That "co-function" means the function of the complement
Understand
Concepts
Why the co-function identities follow from swapping opposite and adjacent sides
How the unit circle reflects these symmetries
Why the prefix "co-" appears in cosine, cotangent, and cosecant
Can do
Skills
Convert between trig functions of an angle and its complement
Simplify expressions using co-function identities
Solve equations involving complementary angles
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Key terms
Complementary anglesTwo angles that add to $90^\circ$ (or $\frac{\pi}{2}$ radians).
Co-functionA trig function evaluated at the complementary angle: $\cos \theta = \sin(90^\circ - \theta)$.
Co-function identityAn equation relating a trig function to its "co-" partner at the complement.
Reflection symmetryThe transformation across $y = x$ that swaps $x$- and $y$-coordinates on the unit circle.
Supplementary anglesTwo angles that add to $180^\circ$. Not the same as complementary.
Etymology"Cosine" was originally "complementi sinus", the sine of the complementary angle.
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Why co-functions work, triangle and circle
Work through the core explanation before applying it.
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Why co-functions work, triangle and circle
Consider a right-angled triangle with acute angles $\theta$ and $(90^\circ - \theta)$. Looking 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:
$$\sin \theta = \cos(90^\circ - \theta)$$
On the unit circle, the angle $\frac{\pi}{2} - \theta$ represents a reflection across the line $y = x$. This swaps the $x$- and $y$-coordinates:
$\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta$ (the new $x$-coordinate is the old $y$-coordinate)
$\sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta$ (the new $y$-coordinate is the old $x$-coordinate)
Why "cosine" has "co-" in its name. The word "sine" comes from the Latin "sinus," meaning curve or fold. "Cosine" was originally called "complementi sinus", the sine of the complementary angle. Over time, this was shortened to "cosinus" and eventually "cosine." The same pattern applies to cotangent and cosecant.
Six co-function pairs: $\sin \theta = \cos(90^\circ{-}\theta)$, $\tan \theta = \cot(90^\circ{-}\theta)$, $\sec \theta = \csc(90^\circ{-}\theta)$; Geometric reason: swapping opposite and adjacent sides swaps the trig ratio with its "co-" partner
Pause, copy the three co-function pairs ($\sin\theta = \cos(90°-\theta)$; $\tan\theta = \cot(90°-\theta)$; $\sec\theta = \csc(90°-\theta)$) and the geometric reason (swapping opp/adj swaps the ratio with its co-partner) into your book.
True or false: $\sin 40^\circ = \cos 50^\circ$.
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Worked Example, Converting to a co-function
+5 XP for trying first
We just saw that $\sin\theta = \cos(90°-\theta)$ because swapping opp and adj in a right triangle swaps the ratio with its co-partner. That raises a question: how does this look in practice when we need to rewrite a specific angle? This card answers it → subtract the angle from 90°: $\sin 37° = \cos 53°$ because $37 + 53 = 90$.
Write $\sin 37^\circ$ as the cosine of a complementary angle.
Your turn first. Try it yourself before viewing the solution.
Template: $\sin \theta = \cos(90^\circ - \theta)$, subtract the angle from $90^\circ$; Example: $\sin 37^\circ = \cos 53^\circ$ because $37 + 53 = 90$
Pause, copy the conversion template ($\sin\theta = \cos(90°-\theta)$: subtract the angle from 90°) and the sum-to-90° check into your book.
Write $\cos 72^\circ$ as the sine of its complementary angle. Type your answer in the form shown by the placeholder (for example, $\sin 40^\circ$ would be typed sin 40).
sin °
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Worked Example, Simplifying with co-functions
+5 XP for trying first
We just saw that $\sin 37° = \cos 53°$ because the two angles sum to 90°, we convert by subtracting from 90°. That raises a question: what happens when a fraction has co-functions of complementary angles in numerator and denominator? This card answers it → the fraction collapses to 1 because numerator and denominator become identical after conversion.
Simplify $\frac{\sin 25^\circ}{\cos 65^\circ}$.
Your turn first. Try it yourself before viewing the solution.
When numerator and denominator are co-functions of complementary angles, the fraction simplifies to 1; Strategy: check if the two angles sum to $90^\circ$, then apply the identity to cancel
Pause, copy the cancellation rule (co-functions of complementary angles in a fraction → simplifies to 1) and the two-step check (sum to 90°, then cancel) into your book.
What is the simplified value of $\dfrac{\cos 35^\circ}{\sin 55^\circ}$?
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Worked Example, Solving a complementary angle equation
+5 XP for trying first
We just saw that checking whether two angles sum to 90° lets us collapse a fraction to 1. That raises a question: what if we have an equation like $\sin 2\theta = \cos\theta$, can we solve it using the same co-function idea? This card answers it → convert $\cos\theta$ to $\sin(90°-\theta)$, then use BOTH angle families of $\sin A=\sin B$, because sine takes the same value at two angles in a revolution.
Your turn first. Try it yourself before viewing the solution.
SOLUTION
2 marks · HSC band 4+
Step 1: Convert cosine to sine of complement.
$$\cos \theta = \sin(90^\circ - \theta)$$
Step 2: Rewrite the equation.
$$\sin 2\theta = \sin(90^\circ - \theta)$$
Step 3: Equate the angles, both ways. $\sin A = \sin B$ does not force $A=B$: sine repeats, so either $A = B$ or $A = 180^\circ - B$ (plus multiples of $360^\circ$).
Step 5: Keep the solutions inside $0^\circ \leq \theta \leq 90^\circ$. Both survive. Check $\theta = 90^\circ$: $\sin 180^\circ = 0$ and $\cos 90^\circ = 0$, so it is genuine.
Answer: $\theta = \mathbf{30^\circ}$ or $\theta = \mathbf{90^\circ}$
Strategy for co-function equations: convert so both sides use the same trig function, then take BOTH angle families, since $\sin A = \sin B$ gives $A = B$ or $A = 180^\circ - B$; $\sin 2\theta = \cos \theta \Rightarrow \sin 2\theta = \sin(90^\circ - \theta) \Rightarrow \theta = 30^\circ$ or $\theta = 90^\circ$
Pause, copy the co-function equation strategy (convert to matching trig functions, then take both angle families) and the worked solution $\sin 2\theta = \cos\theta \Rightarrow \theta = 30°$ or $\theta = 90°$ into your book.
Which of the following is the odd one out it is NOT a co-function identity?
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Common traps
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This is only true when $\theta = 45^\circ$. In general, $\sin \theta$ equals $\cos(90^\circ - \theta)$, not $\sin(90^\circ - \theta)$. Sine pairs with cosine, tangent pairs with cotangent, secant pairs with cosecant.
Trap 2, Confusing complementary with supplementary
Students sometimes use $180^\circ - \theta$ instead of $90^\circ - \theta$. Supplementary angles add to $180^\circ$; complementary angles add to $90^\circ$.
Trap 3, Not checking that the solution is within the required domain
When solving equations with co-functions, there may be multiple solutions from the general sine equation. Always check which ones fall in the specified interval.
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Drill, build fluency
Work through the core explanation before applying it.
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Drill, build fluency
+5 XP for 5 correct
Express each trig ratio as a co-function of a complementary angle.
1 $\sin 55^\circ = \cos 35^\circ$, so the fraction equals 1.
Match each function to its co-function partner.
sine
tangent
secant
cosecant
cotangent
cosine
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Revisit, the geometric origin
Run the quick drill and copy the summary into your book.
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Revisit, the geometric origin
+5 XP for checking
Return to your original answer from Section 01. 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. Did your explanation capture the key idea of swapping opposite and adjacent?
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Multiple choice
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Practice
Multiple choice
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Short answer
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Short Answer
Structured response questions
ApplyBand 4–5
Co-function with Pythagorean identity
(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.
Check: $22.5^\circ$ is within $[0^\circ, 45^\circ]$ ✓
AnalyseBand 6
Prove a co-function identity
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.
Restriction: $\sec(90^\circ - \theta)$ 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). Both sides must be defined for the identity to hold.
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