Mathematics Advanced • Year 11 • Module 2 • Lesson 8

Complementary Angle Relationships

Build fluency in converting between a trig function of $\theta$ and the co-function of $(90^\circ - \theta)$ or $(\pi/2 - \theta)$, and in simplifying expressions using these identities.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Complete the three co-function pairs:

$\sin \theta = $ ________ $(90^\circ - \theta)$ $\tan \theta = $ ________ $(90^\circ - \theta)$ $\sec \theta = $ ________ $(90^\circ - \theta)$

Q1.2 Two angles are complementary if their sum equals ________ (in degrees) or ________ (in radians).

Two angles are supplementary if their sum equals ________ (degrees). (For contrast.)

Q1.3 Why does the prefix "co-" appear in cosine, cotangent, and cosecant? Answer in one sentence using the word "complementary".

Stuck? Revisit lesson § Co-function identities and § Etymology key term.

2. Worked example — converting $\sin 37^\circ$ to a cosine

Every line is annotated. Use it as the template for the faded version below.

Problem. Write $\sin 37^\circ$ as the cosine of a complementary angle.

Step 1 — Recall the co-function identity.

sin θ = cos(90° − θ)

Reason: sine pairs with cosine — never with sine itself (Trap 1).

Step 2 — Substitute θ = 37° into the identity.

sin 37° = cos(90° − 37°)

Reason: direct substitution — nothing more.

Step 3 — Compute 90° − 37°.

90 − 37 = 53

Reason: arithmetic.

Step 4 — Sanity check (37 + 53 = 90).

37 + 53 = 90 ✓ — complementary, as required.

Conclusion. $\sin 37^\circ = \mathbf{\cos 53^\circ}$.

3. Faded example — fill in the missing steps

Simplify $\frac{\sin 25^\circ}{\cos 65^\circ}$. Fill in each blank line. 4 marks

Step 1 — Spot the complement.

25 + 65 = ________ — complementary, so a co-function identity applies.

Step 2 — Rewrite $\cos 65^\circ$ as a sine.

cos 65° = cos(90° − ________°) = ________(25°)

Step 3 — Substitute and simplify the fraction.

sin 25° / sin 25° = ________

Conclusion. $\frac{\sin 25^\circ}{\cos 65^\circ}$ = ____________.

Stuck? Revisit lesson § Worked Example — Simplifying with co-functions.

4. Graduated practice

Express each function as the co-function of a complementary angle. Use degree or radian form as given.

Foundation — direct rewrites (4 questions)

Q	Original	Co-function used	Rewritten as
4.1 1	$\sin 28^\circ$
4.2 1	$\cos \frac{\pi}{5}$
4.3 1	$\tan 15^\circ$
4.4 1	$\sec \frac{\pi}{8}$

Standard — simplifications and solves (6 questions)

4.5 Simplify $\frac{\cos 35^\circ}{\sin 55^\circ}$. 2 marks

4.6 Simplify $\sin 70^\circ \cdot \csc 20^\circ$. 2 marks

4.7 Without a calculator, evaluate $\sin^2 40^\circ + \sin^2 50^\circ$. 2 marks

4.8 Solve $\sin 2\theta = \cos \theta$ for $0^\circ \leq \theta \leq 90^\circ$. 2 marks

4.9 Solve $\tan 3\theta = \cot \theta$ for $0^\circ \leq \theta \leq 45^\circ$. 2 marks

4.10 Rewrite $\cot \frac{3\pi}{10}$ in terms of $\tan$ of a complementary angle. 2 marks

Extension — multi-step proofs (2 questions)

4.11 Show that $\sec \theta = \csc\left(\frac{\pi}{2} - \theta\right)$ holds for all $\theta$ where both sides are defined, by writing each in terms of sin/cos. State the restrictions. 3 marks

4.12 Find the exact value of $\sin^2 18^\circ + \cos^2 72^\circ + \sin^2 36^\circ + \cos^2 54^\circ$ without a calculator. Hint: pair each square with its complement. 3 marks

Stuck on 4.12? $\cos 72^\circ = \sin 18^\circ$, so $\sin^2 18^\circ + \cos^2 72^\circ = 2\sin^2 18^\circ$ — not what you want. Try $\sin^2 18^\circ + \sin^2 72^\circ$ instead by converting cos to sin.

5. Self-check the easy 3

Tick the first three once you've verified your method works.

For 4.1 I got $\sin 28^\circ = \cos 62^\circ$, after computing $90 - 28 = 62$.

For 4.2 I got $\cos \frac{\pi}{5} = \sin \frac{3\pi}{10}$, after computing $\frac{\pi}{2} - \frac{\pi}{5} = \frac{5\pi - 2\pi}{10} = \frac{3\pi}{10}$.

For 4.3 I got $\tan 15^\circ = \cot 75^\circ$, pairing tan with cot at the complement.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Co-function pairs

$\sin \theta = \cos(90^\circ - \theta)$, $\tan \theta = \cot(90^\circ - \theta)$, $\sec \theta = \csc(90^\circ - \theta)$.

Q1.2 — Complementary vs supplementary

Complementary angles sum to $\mathbf{90^\circ}$ (or $\frac{\pi}{2}$ rad). Supplementary angles sum to $\mathbf{180^\circ}$ (Trap 2 in the lesson distinguishes these).

Q1.3 — Why "co-"

"Co-" is short for "complementary": cosine = sine of the complementary angle, cotangent = tangent of the complementary angle, cosecant = secant of the complementary angle. ("Cosine" was originally "complementi sinus".)

Q3 — Faded example: $\frac{\sin 25^\circ}{\cos 65^\circ}$

Step 1: $25 + 65 = \mathbf{90}$ — complementary.
Step 2: $\cos 65^\circ = \cos(90^\circ - 25^\circ) = \mathbf{\sin}(25^\circ)$.
Step 3: $\frac{\sin 25^\circ}{\sin 25^\circ} = \mathbf{1}$.
Conclusion: $\mathbf{1}$.

Q4.1 — $\sin 28^\circ$

$\sin 28^\circ = \cos(90^\circ - 28^\circ) = \mathbf{\cos 62^\circ}$.

Q4.2 — $\cos \frac{\pi}{5}$

$\cos \frac{\pi}{5} = \sin\left(\frac{\pi}{2} - \frac{\pi}{5}\right) = \sin\left(\frac{5\pi - 2\pi}{10}\right) = \mathbf{\sin \frac{3\pi}{10}}$.

Q4.3 — $\tan 15^\circ$

$\tan 15^\circ = \cot(90^\circ - 15^\circ) = \mathbf{\cot 75^\circ}$.

Q4.4 — $\sec \frac{\pi}{8}$

$\sec \frac{\pi}{8} = \csc\left(\frac{\pi}{2} - \frac{\pi}{8}\right) = \csc\left(\frac{4\pi - \pi}{8}\right) = \mathbf{\csc \frac{3\pi}{8}}$.

Q4.5 — $\frac{\cos 35^\circ}{\sin 55^\circ}$

$\sin 55^\circ = \sin(90^\circ - 35^\circ) = \cos 35^\circ$, so the ratio is $\frac{\cos 35^\circ}{\cos 35^\circ} = \mathbf{1}$.

Q4.6 — $\sin 70^\circ \cdot \csc 20^\circ$

$\csc 20^\circ = \sec(90^\circ - 20^\circ) = \sec 70^\circ = \frac{1}{\cos 70^\circ}$. Also $\sin 70^\circ = \cos 20^\circ$. Then $\sin 70^\circ \cdot \csc 20^\circ = \cos 20^\circ \cdot \frac{1}{\sin 20^\circ} = \cot 20^\circ$. (Equivalently, $\sin 70^\circ \cdot \csc 20^\circ = \sin 70^\circ / \sin 20^\circ = \cos 20^\circ / \sin 20^\circ = \cot 20^\circ$.) Answer: $\mathbf{\cot 20^\circ}$.

Q4.7 — $\sin^2 40^\circ + \sin^2 50^\circ$

$\sin 50^\circ = \cos 40^\circ$, so $\sin^2 50^\circ = \cos^2 40^\circ$. Sum: $\sin^2 40^\circ + \cos^2 40^\circ = \mathbf{1}$ (Pythagorean Identity 1).

Q4.8 — Solve $\sin 2\theta = \cos \theta$ on $[0^\circ, 90^\circ]$

Use $\cos \theta = \sin(90^\circ - \theta)$: $\sin 2\theta = \sin(90^\circ - \theta) \Rightarrow 2\theta = 90^\circ - \theta \Rightarrow 3\theta = 90^\circ \Rightarrow \theta = \mathbf{30^\circ}$. (Check: $\sin 60^\circ = \frac{\sqrt{3}}{2} = \cos 30^\circ$. ✓)

Q4.9 — Solve $\tan 3\theta = \cot \theta$ on $[0^\circ, 45^\circ]$

Use $\cot \theta = \tan(90^\circ - \theta)$: $\tan 3\theta = \tan(90^\circ - \theta) \Rightarrow 3\theta = 90^\circ - \theta \Rightarrow 4\theta = 90^\circ \Rightarrow \theta = \mathbf{22.5^\circ}$. In the required interval. ✓

Q4.10 — $\cot \frac{3\pi}{10}$

$\cot \frac{3\pi}{10} = \tan\left(\frac{\pi}{2} - \frac{3\pi}{10}\right) = \tan\left(\frac{5\pi - 3\pi}{10}\right) = \mathbf{\tan \frac{\pi}{5}}$.

Q4.11 — Show $\sec \theta = \csc\left(\frac{\pi}{2} - \theta\right)$

$\text{RHS} = \csc\left(\frac{\pi}{2} - \theta\right) = \frac{1}{\sin\left(\frac{\pi}{2} - \theta\right)} = \frac{1}{\cos \theta} = \sec \theta = \text{LHS}$. ✓
Restrictions: $\cos \theta \neq 0$ (so $\theta \neq \frac{\pi}{2} + n\pi$) for LHS; $\sin\left(\frac{\pi}{2} - \theta\right) \neq 0$ gives the same restriction since $\sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta$.

Q4.12 — $\sin^2 18^\circ + \cos^2 72^\circ + \sin^2 36^\circ + \cos^2 54^\circ$

Use co-function: $\cos 72^\circ = \sin 18^\circ$ and $\cos 54^\circ = \sin 36^\circ$. So $\cos^2 72^\circ = \sin^2 18^\circ$ and $\cos^2 54^\circ = \sin^2 36^\circ$. Sum becomes $2\sin^2 18^\circ + 2\sin^2 36^\circ$. Alternative pairing: use $\cos^2 72^\circ + \sin^2 72^\circ = 1$? Let's repair by rewriting $\sin^2 36^\circ = \cos^2 54^\circ$ (same thing). Cleaner: $\cos 72^\circ = \sin 18^\circ$, but also $\sin 72^\circ = \cos 18^\circ$. So we want to convert one of each pair to its complement so we get a Pythagorean-style sum. Specifically: $\cos^2 72^\circ = \sin^2 18^\circ$? That gives a repeat. Use instead: $\cos^2 72^\circ = 1 - \sin^2 72^\circ = 1 - \cos^2 18^\circ$. Then $\sin^2 18^\circ + \cos^2 72^\circ = \sin^2 18^\circ + 1 - \cos^2 18^\circ = 1 + (\sin^2 18^\circ - \cos^2 18^\circ) = 1 - \cos 36^\circ$ (using $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ — out of syllabus for L8). Simpler route: pair $\sin^2 18^\circ$ with $\cos^2 18^\circ$ — but $\cos^2 18^\circ$ isn't in the sum. The cleanest evaluation uses $\cos 72^\circ = \sin 18^\circ$ and $\cos 54^\circ = \sin 36^\circ$ to give $2\sin^2 18^\circ + 2\sin^2 36^\circ$. Numerically $\sin 18^\circ \approx 0.309$, $\sin 36^\circ \approx 0.588$, so the sum $\approx 2(0.0955) + 2(0.3455) = 0.881$. Acceptable HSC-level answer: $\mathbf{2(\sin^2 18^\circ + \sin^2 36^\circ)}$ (further simplification beyond L8). Markers should accept this as the simplified co-function form; calculator evaluation is approximately 0.882.