Worked examples — Complementary angle relationships — sin(90−θ) = cos θ etc.

This page is the "no surprises" workout for complementary angle identities. We build a scenario matrix — every kind of case this topic can throw at you — and then knock out each cell with a fully worked example. By the end, no exam question can show you a case you haven't already seen.

Before we start, one reminder of the only tools we use, each already earned in the parent note:

Recall The six cofunction identities (tap to reveal)

$\sin (90°-\theta )=\cos \theta$, $\cos (90°-\theta )=\sin \theta$ $\tan (90°-\theta )=\cot \theta$, $\cot (90°-\theta )=\tan \theta$ $\sec (90°-\theta )=\csc \theta$, $\csc (90°-\theta )=\sec \theta$ Complementary = adds to $90°$ (a corner). Not $180°$ (that's supplementary).

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The scenario matrix

Every problem this topic produces falls into one of these case classes. Think of it as a checklist: each row is a way the world can be, and the last column names the example that handles it.

#	Case class	What makes it tricky	Handled by
A	Evaluate a value using its complement	Recognise $90°-\theta$ before touching a calculator	Example 1
B	Both acute, solve for the angle ($\sin A=\cos B$)	Convert one function, set arguments complementary	Example 2
C	Radian form ($\pi /2-\theta$)	Never mix degrees and radians	Example 3
D	Prove an expression $=$ constant	Chain cofunction $+$ reciprocal identities	Example 4
E	Degenerate / boundary input ($\theta =0°$ or $\theta =90°$)	A function blows up or a ratio hits $0$	Example 5
F	Limiting behaviour ($\theta \to 90°$)	Watch which side of infinity you approach	Example 6
G	Real-world word problem (angle of elevation)	Translate geometry into the identity	Example 7
H	Exam twist (multiple solutions / general angle)	Acute assumption fails — count all answers	Example 8

We now walk cell by cell. Guess each answer before reading on — that "Forecast" habit is how the pattern sinks in.

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Example 1 — Cell A · Evaluate using a complement

Step 1. Rewrite $72°$ as a complement. $72°=90°-18°$ Why this step? The identities only fire when the angle wears the costume $90°-\theta$. We hunt for a partner that adds to $90°$: $72°+18°=90°$. ✓

Step 2. Apply $\cos (90°-\theta )=\sin \theta$ with $\theta =18°$. $\cos (72°)=\cos (90°-18°)=\sin (18°)=0.309$ Why this step? This is the second cofunction identity, read left to right.

Answer: $\cos (72°)=0.309$.

Verify: $72°$ is close to $90°$, where cosine is small, so a value near $0.3$ is sensible (not near $1$). Forecast confirmed: smaller than $0.5$. Numeric check in VERIFY.

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Example 2 — Cell B · Both acute, solve for the angle

Step 1. Convert the cosine into a sine so both sides match. $\cos (x+10°)=\sin (90°-(x+10°))=\sin (80°-x)$ Why this step? $\cos \theta =\sin (90°-\theta )$. Matching functions lets us compare arguments directly (Solving Trigonometric Equations).

Step 2. Equate the two sines. $\sin (4x)=\sin (80°-x)$

Step 3. For acute angles, equal sines mean equal arguments. $4x=80°-x$ Why this step? On $(0°,90°)$ sine is increasing — one-to-one — so $\sin \alpha =\sin \beta \Rightarrow \alpha =\beta$.

Step 4. Solve algebraically. $5x=80°\Rightarrow x=16°$

Answer: $x=16°$.

Verify: $\sin (64°)$ vs $\cos (26°)$. Since $64°+26°=90°$ they are cofunctions and equal. ✓ Forecast: $16°<20°$. Numeric check in VERIFY.

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Example 3 — Cell C · Radian form (never mix units)

Step 1. Apply the identity in radian dress: $\sin (\frac{\pi }{2}-\theta )=\cos \theta$. $\frac{\sin \left(\frac{\pi }{2}-\theta \right)}{\cos \theta }=\frac{\cos \theta }{\cos \theta }$ Why this step? Here $90°$ becomes $\pi /2$ because $\theta$ is in radians. Mixing $90°-\frac{\pi }{6}$ would be nonsense — a degree minus a radian.

Step 2. Cancel (valid whenever $\cos \theta \ne 0$). $\frac{\cos \theta }{\cos \theta }=1$ Why this step? Any non-zero quantity over itself is $1$.

Answer: The expression is $1$ for all $\theta$ with $\cos \theta \ne 0$; at $\theta =\frac{\pi }{6}$ it is $1$.

Verify: At $\theta =\frac{\pi }{6}$: $\sin (\frac{\pi }{2}-\frac{\pi }{6})=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$ and $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$; ratio $=1$. ✓ Numeric check in VERIFY.

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Example 4 — Cell D · Prove an expression equals a constant

Step 1. Apply $\tan (90°-\theta )=\cot \theta$. $\tan (90°-\theta )\cdot \tan \theta =\cot \theta \cdot \tan \theta$ Why this step? Third cofunction identity turns the awkward $90°-\theta$ into the clean $\cot \theta$.

Step 2. Rewrite $\cot \theta =\frac{1}{\tan \theta }$. $\cot \theta \cdot \tan \theta =\frac{1}{\tan \theta }\cdot \tan \theta$ Why this step? Cotangent is the reciprocal of tangent — this exposes the cancellation.

Step 3. Multiply. $\frac{1}{\tan \theta }\cdot \tan \theta =1$ Why this step? Non-zero $\times$ its reciprocal $=1$. This needs $\tan \theta \ne 0$ and $\tan \theta$ finite — i.e. $\theta \ne 0°,90°$.

Answer: $\tan (90°-\theta )\tan \theta =1$. $■$

Verify: $\theta =30°$: $\tan 60°\cdot \tan 30°=\sqrt{3}\cdot \frac{1}{\sqrt{3}}=1$. ✓ Numeric check in VERIFY.

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Example 5 — Cell E · Degenerate / boundary inputs

Boundary cases are where identities either stay graceful or explode. We check both ends. See the figure: the right triangle collapses as $\theta \to 0°$.

Step 1. Evaluate $\cos (90°-0°)$ using $\cos (90°-\theta )=\sin \theta$ with $\theta =0°$. $\cos (90°-0°)=\sin (0°)=0$ Why this step? $\theta =0°$ is a legal input for sine — no division involved, so it's safe. Geometrically (figure) the opposite side has shrunk to length $0$.

Step 2. Evaluate $\cot (90°-90°)$ using $\cot (90°-\theta )=\tan \theta$ with $\theta =90°$. $\cot (90°-90°)=\tan (90°)=\text{undefined}$ Why this step? $\tan 90°=\frac{\sin 90°}{\cos 90°}=\frac{1}{0}$ — division by zero. The identity is only valid where both sides exist; at $\theta =90°$ it breaks.

Answer: $\cos (90°-0°)=0$ (defined); $\cot (90°-90°)$ is undefined.

Verify: $\sin (0°)=0$ exactly; $\cos (90°)=0$ so $\tan (90°)$ has zero denominator. ✓ Check in VERIFY.

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Example 6 — Cell F · Limiting behaviour

Step 1. Rewrite using the cofunction identity $\sec (90°-\theta )=\csc \theta$. $\sec (90°-\theta )=\csc \theta =\frac{1}{\sin \theta }$ Why this step? This converts the limit into a familiar reciprocal of sine, whose small-angle behaviour we understand.

Step 2. Send $\theta \to 0{°}^{+}$. $\sin \theta \to 0^{+}\Rightarrow \frac{1}{\sin \theta }\to +\infty$ Why this step? As the sine shrinks to a tiny positive number, its reciprocal grows without bound. On the Unit Circle, $\sin \theta$ is the height, and near $0°$ that height is nearly zero.

Step 3. Cross-check the un-transformed side. As $\theta \to 0°$, $90°-\theta \to 90°$, and $\sec 90°=1/\cos 90°=1/0\to \infty$. Same conclusion. ✓ Why this step? Confirming both sides diverge together shows the identity respects the limit.

Answer: $\sec (90°-\theta )\to +\infty$ as $\theta \to 0{°}^{+}$.

Verify: At $\theta =1°$, $\csc (1°)\approx 57.30$, already large and growing. ✓ Check in VERIFY.

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Example 7 — Cell G · Real-world word problem

A surveyor stands at point $P$. A tower top $T$ is seen at angle of elevation $\theta$ above the horizontal. A drone at the same height as $T$, but on the other side of the vertical line, is seen at angle $90°-\theta$. The figure shows the two right triangles sharing the vertical.

Step 1. From $P$, elevation $\theta$: opposite $=a$ (height), adjacent $=b$ (ground). $\tan \theta =\frac{a}{b}$ Why this step? Tangent $=$ opposite/adjacent in the ground-level right triangle (Right Triangle Trigonometry).

Step 2. Look from the top $T$ down the line $TP$. The angle this line makes with the vertical is $90°-\theta$, because the two acute angles of a right triangle sum to $90°$. $\text{angle at top}=90°-\theta$ Why this step? Same triangle, other acute angle — exactly the parent note's swap-of-perspective idea.

Step 3. From $T$'s corner, the side opposite $(90°-\theta )$ is the ground $b$; the side adjacent is the height $a$. $\tan (90°-\theta )=\frac{b}{a}$

Step 4. Confirm they're reciprocals via the identity. $\tan (90°-\theta )=\cot \theta =\frac{1}{\tan \theta }=\frac{1}{a/b}=\frac{b}{a}$ Why this step? The geometry ($b/a$) must match the algebra ($\cot \theta$) — and it does.

Answer: $\tan \theta =\frac{a}{b}$ and $\tan (90°-\theta )=\frac{b}{a}$; they are reciprocals, as the identity predicts.

Verify: With $a=3\text{ m}$, $b=4\text{ m}$: $\tan \theta =0.75$, $\tan (90°-\theta )=4/3\approx 1.333$, product $=1$. Units cancel (m/m). ✓ Check in VERIFY.

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Example 8 — Cell H · Exam twist (acute assumption fails)

Step 1. Convert cosine to sine. $\cos (x)=\sin (90°-x)\Rightarrow \sin (3x)=\sin (90°-x)$ Why this step? Standard cofunction move to match functions.

Step 2. Use the full rule for equal sines (not just the acute shortcut). If $\sin \alpha =\sin \beta$ then either $\alpha =\beta \text{or}\alpha =180°-\beta (+360°k).$ Why this step? Over a wider range sine is not one-to-one — it repeats and reflects. Using only $\alpha =\beta$ would miss solutions (Solving Trigonometric Equations).

Step 3a. Branch 1: $3x=90°-x$. $4x=90°\Rightarrow x=22.5°$

Step 3b. Branch 2: $3x=180°-(90°-x)=90°+x$. $2x=90°\Rightarrow x=45°$ Why this step? The reflection branch adds a second valid angle inside $[0°,180°]$.

Step 4. Check the range and look for shifted copies. Adding $360°$ to either branch throws $x$ outside $[0°,180°]$, so no more.

Answer: $x=22.5°$ and $x=45°$.

Verify: $x=22.5°$: $\sin 67.5°=\cos 22.5°$ (complementary). $x=45°$: $\sin 135°=\sin 45°=\frac{\sqrt{2}}{2}$ and $\cos 45°=\frac{\sqrt{2}}{2}$. ✓ Both checked in VERIFY.

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Recall check

Which cells did we cover? Try to name the case class for each.