HSC Mathematics Advanced trigonometric functions (2026 guide)

Step 1. Isolate the trig function. Divide both sides by 2:

Step 2. Find the reference (first-quadrant) angle

Since $\sin\dfrac{\pi}{6} = \dfrac{1}{2}$, the reference angle is $\dfrac{\pi}{6}$.

Step 3. Identify all quadrants where the function is positive

$\sin$ is positive in quadrants 1 and 2.

Step 4. Write the general solutions in those quadrants on $[0, 2\pi]$

Quadrant 1: $\theta = \dfrac{\pi}{6}$. Quadrant 2: $\theta = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$.

Answer

$\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$.

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

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

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

$\theta = \dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$.

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.

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

Answer. Identity proved.

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 $[-4, 2]$.

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.

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.

Step 1. Set up the inequality.

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

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.

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 $\left( \dfrac{5\pi}{4}, -\sqrt{2} \right)$.

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.

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.

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.

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.

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 $\sin(3x)$.

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.

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 $k = 9$.

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.

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