HSC Mathematics Advanced trigonometric functions (2026 guide)

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.
• IMATH_9 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: IMATH_17

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

Exact values

Memorise this table:

IMATH_19	0	IMATH_20	IMATH_21	IMATH_22	IMATH_23
IMATH_24	0	IMATH_25	IMATH_26	IMATH_27	1
IMATH_28	1	IMATH_29	IMATH_30	IMATH_31	0
IMATH_32	0	IMATH_33	1	IMATH_34	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 Students Take Calculus):

• 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

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 $\cos^2\theta$:

By $\sin^2\theta$:

Double-angle identities

• IMATH_53
• IMATH_54
• IMATH_55

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

Sum and difference identities

• IMATH_57
• IMATH_58

(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.

A worked example

Solve $2\sin\theta = 1$ for $0 \leq \theta \leq 2\pi$.

Step 1: $\sin\theta = \frac{1}{2}$.

Step 2: The reference angle is $\frac{\pi}{6}$.

Step 3: $\sin$ is positive in quadrants 1 and 2. So $\theta = \frac{\pi}{6}$ or $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

Step 4: Both are in $[0, 2\pi]$, so the solutions are $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$.

A more complex example

Solve $\sin 2\theta = \cos\theta$ for $0 \leq \theta \leq 2\pi$.

Use the double-angle identity: $\sin 2\theta = 2\sin\theta\cos\theta$.

So either $\cos\theta = 0$ (giving $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$) or $\sin\theta = \frac{1}{2}$ (giving $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$).

Solutions: $\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$.

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

Graphs of trig functions

Basic graphs

• IMATH_82 : amplitude 1, period $2\pi$, oscillates between $-1$ and $1$.
• IMATH_86 : same shape, shifted left by $\pi/2$.
• IMATH_88 : 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 $y = D$.

A worked transformation

Sketch $y = 3\sin\left(2x + \frac{\pi}{3}\right) - 1$ for $0 \leq x \leq 2\pi$.

• Amplitude: 3 (oscillates between $-4$ and $2$).
• Period: $\frac{2\pi}{2} = \pi$.
• Phase shift: $-\frac{\pi/3}{2} = -\frac{\pi}{6}$ (shifted left by $\pi/6$).
• Vertical shift: $-1$ (midline at $y = -1$).

So the graph is a sinusoid with two full cycles over $[0, 2\pi]$, oscillating between $-4$ and $2$, midline at $-1$, with the first peak slightly left of $x = 0$.

Modelling with sinusoids

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

A typical question

The water depth at a port varies sinusoidally with time. The maximum depth is 8 metres at 6:00am and the minimum depth is 2 metres at 12:00pm. Write an equation for the depth $h$ as a function of time $t$ in hours after midnight.

The midline is $\frac{8 + 2}{2} = 5$ metres.

The amplitude is $\frac{8 - 2}{2} = 3$ metres.

The period is 12 hours (from one max to the next is 12 hours, since max to min is half a period = 6 hours).

So $\frac{2\pi}{B} = 12$, giving $B = \frac{\pi}{6}$.

At $t = 6$, we want max. So we need a sine function offset to peak at $t = 6$.

(Check: at $t = 6$, $\sin\left(\frac{\pi}{6} \cdot 3\right) = \sin\left(\frac{\pi}{2}\right) = 1$, so $h = 5 + 3 = 8$ ✓.)

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).

In one sentence

HSC Mathematics Advanced trigonometric functions reward rule memorisation (exact values, identities, transformations) plus pattern recognition on equations and modelling questions. Drill the exact-value table; never divide by $\cos$ or $\sin$ when you should factor; always check calculator mode.