Complementary Angle Relationships
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.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
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.
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.
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
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
Skills
- Convert between trig functions of an angle and its complement
- Simplify expressions using co-function identities
- Solve equations involving complementary angles
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:
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)
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.
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.
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.
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 $\frac{\sin 25^\circ}{\cos 65^\circ}$.
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.
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 $\sin(90°-\theta)$, equate angles, then solve for $\theta$.
Solve $\sin 2\theta = \cos \theta$ for $0^\circ \leq \theta \leq 90^\circ$.
Strategy for co-function equations: convert so both sides use the same trig function, then equate angles; $\sin 2\theta = \cos \theta \Rightarrow \sin 2\theta = \sin(90^\circ - \theta) \Rightarrow 2\theta = 90^\circ - \theta \Rightarrow \theta = 30^\circ$
Pause — copy the co-function equation strategy (convert to matching trig functions, equate angles) and the worked solution $\sin 2\theta = \cos\theta \Rightarrow 2\theta = 90°-\theta \Rightarrow \theta = 30°$ into your book.
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.
Students sometimes use $180^\circ - \theta$ instead of $90^\circ - \theta$. Supplementary angles add to $180^\circ$; complementary angles add to $90^\circ$.
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.
Express each trig ratio as a co-function of a complementary angle.
$\sin 28^\circ$
$\sin 28^\circ = \cos(90^\circ - 28^\circ) = \cos 62^\circ$.
$\cos \frac{\pi}{5}$
$\cos \frac{\pi}{5} = \sin\left(\frac{\pi}{2} - \frac{\pi}{5}\right) = \sin \frac{3\pi}{10}$.
$\tan 15^\circ$
$\tan 15^\circ = \cot(90^\circ - 15^\circ) = \cot 75^\circ$.
$\sec \frac{\pi}{8}$
$\sec \frac{\pi}{8} = \csc\left(\frac{\pi}{2} - \frac{\pi}{8}\right) = \csc \frac{3\pi}{8}$.
Simplify $\frac{\cos 35^\circ}{\sin 55^\circ}$.
$\sin 55^\circ = \cos 35^\circ$, so the fraction equals 1.
- sine
- tangent
- secant
- cosecant
- cotangent
- cosine
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:
This is the geometric origin of all co-function identities. Did your explanation capture the key idea of swapping opposite and adjacent?
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.
View comprehensive answer
(a) $\sin 68^\circ = \cos(90^\circ - 68^\circ) = \mathbf{\cos 22^\circ}$.
(b) $\sin^2 68^\circ + \sin^2 22^\circ = \cos^2 22^\circ + \sin^2 22^\circ = \mathbf{1}$.
Solve a co-function equation
Solve $\tan 3\theta = \cot \theta$ for $0^\circ \leq \theta \leq 45^\circ$. Show all working.
View comprehensive answer
Working:
Use $\cot \theta = \tan(90^\circ - \theta)$:
Equate angles:
Answer: $\theta = \mathbf{22.5^\circ}$
Check: $22.5^\circ$ is within $[0^\circ, 45^\circ]$ ✓
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.
View comprehensive answer
Proof:
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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