What are trigonometric functions in HSC Mathematics Advanced?

Trigonometric functions ($\\sin$, $\\cos$, $\\tan$ and their reciprocals) are functions that describe periodic phenomena. In HSC Mathematics Advanced you learn their definitions on the unit circle, exact values, identities, equations, and how their graphs can be transformed and used to model real-world periodic behaviour.

How do I memorise the trig exact values?

Memorise these for 0, 30, 45, 60, 90 degrees and the corresponding radians. Use the two basic triangles - the 30-60-90 right triangle and the 45-45-90 right triangle. Build a small table you can write from memory in under 30 seconds. Most strong students draw this table at the start of every trig question.

What is radian measure?

Radians measure angles by arc length on the unit circle. One full revolution is $2\\pi$ radians (equivalent to 360 degrees). $\\pi$ radians = 180 degrees; $\\pi/2$ radians = 90 degrees; $\\pi/4$ radians = 45 degrees. HSC Advanced uses radians for calculus-based trig work and degrees for survey-style applications.

What are the key trig identities for HSC Advanced?

The Pythagorean identity ($\\sin^2\\theta + \\cos^2\\theta = 1$), the double-angle identities ($\\sin 2\\theta = 2\\sin\\theta\\cos\\theta$, $\\cos 2\\theta = \\cos^2\\theta - \\sin^2\\theta$), the sum and difference identities, and the relationships between $\\sin$, $\\cos$, and $\\tan$. Strong students can use these to simplify expressions and solve equations on sight.

How are trig functions graphed?

The basic graphs of $y = \\sin x$ and $y = \\cos x$ are sinusoidal waves with amplitude 1 and period $2\\pi$. $y = \\tan x$ has period $\\pi$ and vertical asymptotes at $x = \\pi/2 + n\\pi$. Transformations like $y = A\\sin(Bx + C) + D$ change amplitude ($A$), period ($2\\pi/B$), horizontal shift ($-C/B$), and vertical shift ($D$).

How are trig functions examined in the HSC?

Trig questions appear in both Section I (multiple choice) and Section II (extended response). Common patterns - solve a trig equation in a given interval, find the period and amplitude of a transformed sinusoid, prove a trig identity, model a real-world periodic phenomenon (tides, temperature, oscillation) with a sinusoid, and use trig in calculus problems.

NSWMaths Advanced

HSC Mathematics Advanced trigonometric functions (2026 guide)

A complete guide to trigonometric functions in HSC Mathematics Advanced. Definitions, exact values, identities, equations, graphs, transformations, modelling, step-by-step worked examples, and embedded practice questions, plus the exam patterns that repeat year to year.

Generated by Claude Opus 4.820 min readNESA Mathematics Advanced Stage 6 Syllabus (2017), Year 12 Trigonometric Functions: MA-T3 Trigonometric Functions and GraphsUpdated 2026-05-25

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

Why trigonometric functions matter
Radian measure
Exact values
Key trigonometric identities
Solving trig equations
Graphs of trig functions
Modelling with sinusoids
Calculus with trig (HSC favourite)
Common HSC trig traps
How trig is examined
Practice strategy
Check your knowledge

Why trigonometric functions matter

Trigonometric functions appear in roughly 20% of the HSC Mathematics Advanced exam. They show up directly (in pure trig questions) and indirectly (in calculus problems with $\sin$ or $\cos$ , or in modelling questions where the variable is a sinusoid).

Students who treat trig as a separate, memorisable topic tend to do well. Students who try to derive everything from scratch under exam pressure run out of time. This is one HSC topic where rote recall pays.

Radian measure

Radians measure angles by arc length on the unit circle.

One full revolution = $2\pi$ radians = 360°.
One radian = the angle subtended by an arc of length 1 on a unit circle.
$\pi$ radians = 180°.
Common values: $\frac{\pi}{6} = 30°$ , $\frac{\pi}{4} = 45°$ , $\frac{\pi}{3} = 60°$ , $\frac{\pi}{2} = 90°$ .

HSC Advanced predominantly uses radians for calculus-based trig work. Always check which mode your calculator is in before computing.

Arc length and sector area

For a sector of radius $r$ subtending angle $\theta$ radians at the centre:

Arc length: $\ell = r\theta$
Sector area: $A = \frac{1}{2} r^2 \theta$

These formulas only work with $\theta$ in radians.

Exact values

Memorise this table:

$\theta$	0	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$
$\sin\theta$	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
$\cos\theta$	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0
$\tan\theta$	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	undef

A useful memory aid: for $\sin$ at $0, \pi/6, \pi/4, \pi/3, \pi/2$ , the values are $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$ . For $\cos$ , reverse the order.

For angles in other quadrants, use the ASTC rule (All Stations To Central):

Quadrant 1 (0 to $\pi/2$ ): all positive.
Quadrant 2 ( $\pi/2$ to $\pi$ ): $\sin$ positive.
Quadrant 3 ( $\pi$ to $3\pi/2$ ): $\tan$ positive.
Quadrant 4 ( $3\pi/2$ to $2\pi$ ): $\cos$ positive.

Key trigonometric identities

Pythagorean identity

\sin^2\theta + \cos^2\theta = 1

Derived directly from the unit circle. Rearrange to get $\sin^2\theta = 1 - \cos^2\theta$ or $\cos^2\theta = 1 - \sin^2\theta$ .

Dividing through

Dividing the Pythagorean identity by $\cos^2\theta$ :

\tan^2\theta + 1 = \sec^2\theta

By $\sin^2\theta$ :

1 + \cot^2\theta = \csc^2\theta

Double-angle identities

$\sin 2\theta = 2 \sin\theta \cos\theta$
$\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$
$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$

The three forms of $\cos 2\theta$ are interchangeable. Pick whichever simplifies your specific problem.

Sum and difference identities

$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$

(Note the sign flip in the $\cos$ formula.)

Solving trig equations

The standard pattern: rearrange to $\sin\theta = k$ or $\cos\theta = k$ or $\tan\theta = k$ , then find all solutions in the given interval.

Solve

\sin 2\theta = \cos\theta

for

0 \leq \theta \leq 2\pi

Step 1. Replace the double-angle expression using $\sin 2\theta = 2\sin\theta\cos\theta$ .

2\sin\theta\cos\theta = \cos\theta

Step 2. Move everything to one side and factor. Never divide by $\cos\theta$ here - that would erase solutions where $\cos\theta = 0$ .

2\sin\theta\cos\theta - \cos\theta = 0

\cos\theta(2\sin\theta - 1) = 0

Step 3. Solve each factor separately: Either $\cos\theta = 0$ or $\sin\theta = \dfrac{1}{2}$ .
Step 4. Solve $\cos\theta = 0$ on $[0, 2\pi]$: $\theta = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$ .
Step 5. Solve $\sin\theta = \dfrac{1}{2}$ on $[0, 2\pi]$: $\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$ .
Step 6. Combine and order all solutions
Answer: $\theta = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$ .

Critical trap. Do not divide both sides by $\cos\theta$ in step 2 above. That would lose the solutions where $\cos\theta = 0$ . Always factor instead.

Prove the identity

\dfrac{1 - \cos 2\theta}{\sin 2\theta} = \tan\theta

Step 1. Choose the side to manipulate. The left-hand side has more structure, so simplify it.

Step 2. Replace $\cos 2\theta$ using $\cos 2\theta = 1 - 2\sin^2\theta$ . That gives $1 - \cos 2\theta = 1 - (1 - 2\sin^2\theta) = 2\sin^2\theta$ in the numerator.

Step 3. Replace $\sin 2\theta$ using $\sin 2\theta = 2\sin\theta\cos\theta$ in the denominator.

\text{LHS} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta}

Step 4. Cancel the common factor $2\sin\theta$ .

\text{LHS} = \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS}

Answer. Identity proved.

Graphs of trig functions

Basic graphs

$y = \sin x$ : amplitude 1, period $2\pi$ , oscillates between $-1$ and $1$ .
$y = \cos x$ : same shape, shifted left by $\pi/2$ .
$y = \tan x$ : period $\pi$ , vertical asymptotes at $x = \frac{\pi}{2} + n\pi$ .

Transformations

For $y = A\sin(Bx + C) + D$ :

$A$ is the amplitude. $|A|$ is the vertical stretch; the graph oscillates between $-|A| + D$ and $|A| + D$ .
$B$ affects period: $\text{period} = \frac{2\pi}{B}$ . Larger $B$ = shorter period.
$C$ is the phase shift. The graph shifts left by $\frac{C}{B}$ (or right by $-\frac{C}{B}$ ).
$D$ is the vertical shift. The midline moves to $y = D$ .

Identify the amplitude, period, phase shift and vertical shift of

y = 3\sin\left(2x + \dfrac{\pi}{3}\right) - 1

Step 1. Compare to the general form $y = A\sin(Bx + C) + D$: Here $A = 3$ , $B = 2$ , $C = \dfrac{\pi}{3}$ , $D = -1$ .
Step 2. Read off the amplitude: $|A| = 3$ , so the graph oscillates 3 units above and below the midline.
Step 3. Compute the period using $\dfrac{2\pi}{B}$: Period $= \dfrac{2\pi}{2} = \pi$ .
Step 4. Compute the phase shift using $-\dfrac{C}{B}$: Phase shift $= -\dfrac{\pi/3}{2} = -\dfrac{\pi}{6}$ . The graph shifts left by $\dfrac{\pi}{6}$ .
Step 5. Read off the vertical shift: $D = -1$ , so the midline is $y = -1$ .
Step 6. State the range: The graph oscillates between $D - |A| = -4$ and $D + |A| = 2$ .
Answer: Amplitude $3$ , period $\pi$ , phase shift $\dfrac{\pi}{6}$ to the left, midline $y = -1$ , range $[-4, 2]$ .

Modelling with sinusoids

A common HSC question pattern: model a real-world periodic phenomenon (tide height, temperature, ferris wheel position) with a sinusoid.

Tide modelling - peak depth 8 m at 6 am, minimum 2 m at 12 pm

The water depth at a port varies sinusoidally with time. Write an equation for the depth $h$ (m) as a function of time $t$ (hours after midnight).

Step 1. Find the midline: Average of max and min: $\dfrac{8 + 2}{2} = 5$ m. So $D = 5$ .
Step 2. Find the amplitude: Half the difference: $\dfrac{8 - 2}{2} = 3$ m. So $A = 3$ .
Step 3. Find the period: Maximum to next minimum is half a period, and that takes $12 - 6 = 6$ hours. So the period is $12$ hours.

Step 4. Convert the period to $B$ using $\dfrac{2\pi}{B} = 12$ . Hence $B = \dfrac{\pi}{6}$ .

Step 5. Choose a phase so that the peak lands at $t = 6$ . A sine peak occurs when the argument equals $\dfrac{\pi}{2}$ . Setting $\dfrac{\pi}{6}(t - k) = \dfrac{\pi}{2}$ at $t = 6$ gives $k = 3$ .

Step 6. Write the model.

h(t) = 5 + 3\sin\left(\dfrac{\pi}{6}(t - 3)\right)

Step 7. Verify at $t = 6$ . $\sin\left(\dfrac{\pi}{6} \cdot 3\right) = \sin\dfrac{\pi}{2} = 1$ , so $h(6) = 5 + 3 = 8$ m as required.

Answer. $h(t) = 5 + 3\sin\left(\dfrac{\pi}{6}(t - 3)\right)$ metres.

Find the times during the day when the tide depth is below 3.5 m (using the model above)

Step 1. Set up the inequality.

5 + 3\sin\left(\dfrac{\pi}{6}(t - 3)\right) < 3.5

Step 2. Isolate the sine term. Subtract 5, then divide by 3:

\sin\left(\dfrac{\pi}{6}(t - 3)\right) < -\dfrac{1}{2}

Step 3. Find boundary angles: Let $u = \dfrac{\pi}{6}(t - 3)$ . Then $\sin u = -\dfrac{1}{2}$ at $u = -\dfrac{\pi}{6} + 2\pi n$ (quadrant 4 reference) and $u = -\dfrac{5\pi}{6} + 2\pi n$ (quadrant 3 reference). On one period the inequality $\sin u < -\dfrac{1}{2}$ holds for $u \in \left( -\dfrac{5\pi}{6}, -\dfrac{\pi}{6} \right)$ , or equivalently $u \in \left( \pi + \dfrac{\pi}{6}, 2\pi - \dfrac{\pi}{6} \right) = \left( \dfrac{7\pi}{6}, \dfrac{11\pi}{6} \right)$ in $[0, 2\pi]$ .
Step 4. Convert $u$ back to $t$: From $u = \dfrac{\pi}{6}(t - 3)$ , $t = 3 + \dfrac{6u}{\pi}$ . At $u = \dfrac{7\pi}{6}$ , $t = 3 + 7 = 10$ . At $u = \dfrac{11\pi}{6}$ , $t = 3 + 11 = 14$ .
Answer: The depth is below 3.5 m between $t = 10$ and $t = 14$ , that is, from 10 am to 2 pm.

Calculus with trig (HSC favourite)

Find the stationary points of

y = \sin x + \cos x

[0, 2\pi]

Step 1. Differentiate: $\dfrac{dy}{dx} = \cos x - \sin x$ .
Step 2. Set the derivative to zero: $\cos x - \sin x = 0$ , so $\sin x = \cos x$ , giving $\tan x = 1$ .
Step 3. Solve $\tan x = 1$ on $[0, 2\pi]$: $\tan$ is positive in quadrants 1 and 3, so $x = \dfrac{\pi}{4}$ or $x = \dfrac{5\pi}{4}$ .
Step 4. Find the y-coordinate at each: At $x = \dfrac{\pi}{4}$ : $y = \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} = \sqrt{2}$ . At $x = \dfrac{5\pi}{4}$ : $y = -\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} = -\sqrt{2}$ .
Step 5. Classify with the second derivative: $y'' = -\sin x - \cos x$ . At $x = \dfrac{\pi}{4}$ : $y'' = -\dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} = -\sqrt{2} < 0$ , so local maximum at $\left( \dfrac{\pi}{4}, \sqrt{2} \right)$ . At $x = \dfrac{5\pi}{4}$ : $y'' = \sqrt{2} > 0$ , so local minimum at $\left( \dfrac{5\pi}{4}, -\sqrt{2} \right)$ .
Answer: Local maximum at $\left( \dfrac{\pi}{4}, \sqrt{2} \right)$ ; local minimum at $\left( \dfrac{5\pi}{4}, -\sqrt{2} \right)$ .

Common HSC trig traps

Dividing by $\cos\theta$ or $\sin\theta$: Loses solutions. Always factor instead.
Mixing degrees and radians: Always check calculator mode. HSC Advanced is primarily radians; questions using arc length or sector area formulas REQUIRE radians.
Forgetting all solutions: $\sin\theta = \frac{1}{2}$ has solutions in both quadrants 1 and 2; $\cos\theta = \frac{1}{2}$ has solutions in 1 and 4; $\tan\theta = 1$ has solutions in 1 and 3. Missing one = lost mark.
Sign confusion in the cosine sum identity: $\cos(A + B) = \cos A \cos B - \sin A \sin B$ (note the minus). $\cos(A - B) = \cos A \cos B + \sin A \sin B$ (note the plus). Easy to flip.
Period of $\tan$: Many students apply the $\sin$ / $\cos$ period $2\pi$ to $\tan$ . Wrong: $\tan$ has period $\pi$ .

How trig is examined

In Mathematics Advanced HSC paper:

Multiple choice. Exact value problems. Identifying transformations from a graph. Solving simple equations.
Section II short questions. Solving equations in a given interval. Computing arc length or sector area.
Section II medium questions. Proving identities. Multi-step equations using double-angle or factor identities.
Section II extended questions. Modelling with sinusoids. Calculus problems involving trig functions (e.g. optimisation with a $\sin$ -based volume).

Practice strategy

For HSC Mathematics Advanced trigonometric functions:

Term 2-3 of Year 12. Drill exact values and the basic identities until they are automatic.
Term 3. Solving equations. Aim to solve any equation involving $\sin$ , $\cos$ , $\tan$ at first glance.
Term 4. Modelling questions. Past papers. Look at the last 5 years of HSC papers and identify the recurring modelling patterns (tides, oscillators, biological cycles).

Check your knowledge

A sector of a circle has radius $8$ cm and angle $\dfrac{3\pi}{4}$ radians. Find the arc length and the sector area.
Solve $2\cos\theta - \sqrt{3} = 0$ for $0 \leq \theta \leq 2\pi$ .
Solve $\tan\theta = -1$ for $0 \leq \theta \leq 2\pi$ .
Prove that $\dfrac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$ .
State the amplitude, period, phase shift and range of $y = -2\cos\left(\dfrac{x}{2} - \dfrac{\pi}{4}\right) + 3$ .
Differentiate $y = \sin^2(3x)$ .
The temperature in a city varies sinusoidally with a maximum of $28^\circ$ C at 3 pm and a minimum of $16^\circ$ C at 3 am. Write a sinusoidal model $T(t)$ where $t$ is hours after midnight.
Solve $2\sin^2\theta - \sin\theta - 1 = 0$ for $0 \leq \theta \leq 2\pi$ .

Solutions to check-your-knowledge questions

These eight questions cover arc length and sector area, solving trig equations, proving identities, reading graph features, differentiating with the chain rule, modelling with sinusoidal functions, and solving a quadratic-in-trig equation.

Step 1: Q1. Arc length and sector area for $r = 8$ , $\theta = \frac{3\pi}{4}$

The arc length formula is $\ell = r\theta$ and the sector area formula is $A = \frac{1}{2}r^2\theta$ . Substitute the given values directly.

\ell = 8 \cdot \frac{3\pi}{4} = 6\pi \text{ cm}

A = \frac{1}{2} \cdot 64 \cdot \frac{3\pi}{4} = 24\pi \text{ cm}^2

Step 2: Q2. Solve $\cos\theta = \frac{\sqrt{3}}{2}$ for $\theta \in [0, 2\pi]$

The reference angle is $\frac{\pi}{6}$ because $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ . Cosine is positive in quadrants 1 and 4.

\theta = \frac{\pi}{6} \quad \text{or} \quad \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}

Step 3: Q3. Solve $\tan\theta = -1$ for $\theta \in [0, 2\pi]$

The reference angle is $\frac{\pi}{4}$ because $\tan\frac{\pi}{4} = 1$ . Tangent is negative in quadrants 2 and 4.

\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \quad \text{or} \quad \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}

Step 4: Q4. Prove $\dfrac{\sin 2\theta}{1 + \cos 2\theta} = \tan\theta$

Replace the double-angle expressions using $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$ , so $1 + \cos 2\theta = 2\cos^2\theta$ . Then simplify by cancelling the common factor.

\text{LHS} = \frac{2\sin\theta\cos\theta}{2\cos^2\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS}

Identity proved.

Step 5: Q5. Read off features of $y = -2\cos\left(\frac{x}{2} - \frac{\pi}{4}\right) + 3$

Match to the standard form $y = A\cos(Bx + C) + D$ with $A = -2$ , $B = \frac{1}{2}$ , $C = -\frac{\pi}{4}$ , $D = 3$ . Each parameter controls one feature.

Amplitude $= |A| = 2$ . Period $= \dfrac{2\pi}{B} = 4\pi$ . Phase shift $= -\dfrac{C}{B} = \dfrac{\pi}{2}$ to the right. Range $= [D - |A|,\, D + |A|] = [1, 5]$ .

Step 6: Q6. Differentiate $y = \sin^2(3x)$

Write $y = [\sin(3x)]^2$ and apply the chain rule twice: first for the outer square, then for the inner $\sin(3x)$ .

\frac{dy}{dx} = 2\sin(3x) \cdot \cos(3x) \cdot 3 = 6\sin(3x)\cos(3x) = 3\sin(6x)

The last step uses the double-angle identity $\sin 2u = 2\sin u\cos u$ .

Step 7: Q7. Fit a sinusoidal model to daily temperature data

The midline is the average of maximum and minimum; the amplitude is half their difference; the period is 24 hours.

D = \frac{28 + 16}{2} = 22, \quad A = \frac{28 - 16}{2} = 6, \quad B = \frac{2\pi}{24} = \frac{\pi}{12}

The peak occurs at $t = 15$ (3 pm), so shift the sine so its argument equals $\frac{\pi}{2}$ at $t = 15$ : $\frac{\pi}{12}(15 - k) = \frac{\pi}{2}$ gives $k = 9$ .

T(t) = 22 + 6\sin\!\left(\frac{\pi}{12}(t - 9)\right)

Step 8: Q8. Solve $2\sin^2\theta - \sin\theta - 1 = 0$ for $\theta \in [0, 2\pi]$

Substitute $u = \sin\theta$ to turn this into a quadratic, then factor.

2u^2 - u - 1 = (2u + 1)(u - 1) = 0 \implies u = -\frac{1}{2} \text{ or } u = 1

From $\sin\theta = -\dfrac{1}{2}$ : $\theta = \dfrac{7\pi}{6}$ or $\theta = \dfrac{11\pi}{6}$ . From $\sin\theta = 1$ : $\theta = \dfrac{\pi}{2}$ .

Final answer: $\theta = \dfrac{\pi}{2},\ \dfrac{7\pi}{6},\ \dfrac{11\pi}{6}$ .