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NSWMaths AdvancedSyllabus dot point

How do we differentiate and integrate trigonometric functions and use them to model periodic phenomena?

Find derivatives and integrals of $\sin$, $\cos$ and $\tan$ (with linear inside arguments) and apply them to model simple harmonic and periodic motion

A focused answer to the HSC Maths Advanced dot point on trigonometric calculus. Derivatives and integrals of sin, cos and tan, plus modelling periodic motion such as tides and oscillations.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answer

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What this dot point is asking

NESA wants you to differentiate and integrate the three standard trigonometric functions, including with linear inside arguments such as sin⁑(kx+c)\sin(k x + c), and apply this calculus to model periodic phenomena (tides, oscillating springs, biological cycles).

The answer

Derivatives

The angle is always measured in radians. The derivatives are:

ddx(sin⁑x)=cos⁑x\frac{d}{dx}(\sin x) = \cos x

ddx(cos⁑x)=βˆ’sin⁑x\frac{d}{dx}(\cos x) = -\sin x

ddx(tan⁑x)=sec⁑2x=1cos⁑2x\frac{d}{dx}(\tan x) = \sec^2 x = \frac{1}{\cos^2 x}

With a linear inside argument, apply the chain rule:

ddx(sin⁑(kx+c))=kcos⁑(kx+c)\frac{d}{dx}(\sin(k x + c)) = k \cos(k x + c)

ddx(cos⁑(kx+c))=βˆ’ksin⁑(kx+c)\frac{d}{dx}(\cos(k x + c)) = -k \sin(k x + c)

ddx(tan⁑(kx+c))=ksec⁑2(kx+c)\frac{d}{dx}(\tan(k x + c)) = k \sec^2(k x + c)

Integrals

∫sin⁑x dx=βˆ’cos⁑x+C\int \sin x \, dx = -\cos x + C

∫cos⁑x dx=sin⁑x+C\int \cos x \, dx = \sin x + C

∫sec⁑2x dx=tan⁑x+C\int \sec^2 x \, dx = \tan x + C

With a linear inside argument, divide by the inside coefficient:

∫sin⁑(kx+c) dx=βˆ’1kcos⁑(kx+c)+C\int \sin(k x + c) \, dx = -\frac{1}{k} \cos(k x + c) + C

∫cos⁑(kx+c) dx=1ksin⁑(kx+c)+C\int \cos(k x + c) \, dx = \frac{1}{k} \sin(k x + c) + C

Modelling periodic motion

A function of the form

y(t)=Asin⁑(Ο‰t+Ο•)+Dy(t) = A \sin(\omega t + \phi) + D

describes simple harmonic motion (or any sinusoidal cycle). The parameters are:

  • IMATH_13 is the amplitude (half the peak-to-trough range).
  • IMATH_14 is the angular frequency in radians per unit time. The period is T=2πωT = \frac{2 \pi}{\omega}.
  • IMATH_16 is the phase shift.
  • IMATH_17 is the vertical shift (the centre of oscillation).

Differentiating gives the velocity, yβ€²(t)=AΟ‰cos⁑(Ο‰t+Ο•)y'(t) = A \omega \cos(\omega t + \phi), and differentiating again gives the acceleration, yβ€²β€²(t)=βˆ’AΟ‰2sin⁑(Ο‰t+Ο•)=βˆ’Ο‰2(yβˆ’D)y''(t) = -A \omega^2 \sin(\omega t + \phi) = -\omega^2 (y - D). The acceleration is proportional to the displacement from the centre and points back towards it.

Worked example: derivative of a composite

Differentiate y=sin⁑2xy = \sin^2 x.

Rewrite as y=(sin⁑x)2y = (\sin x)^2 and apply the chain rule with u=sin⁑xu = \sin x.

dydx=2sin⁑xβ‹…cos⁑x=sin⁑(2x)\frac{dy}{dx} = 2 \sin x \cdot \cos x = \sin(2 x),

using the double-angle identity.

Worked example: integral via substitution

Evaluate ∫sin⁑xcos⁑x dx\int \sin x \cos x \, dx.

Let u=sin⁑xu = \sin x, so du=cos⁑x dxdu = \cos x \, dx.

∫u du=u22+C=sin⁑2x2+C\int u \, du = \frac{u^2}{2} + C = \frac{\sin^2 x}{2} + C.

Equivalently, you could rewrite using sin⁑xcos⁑x=12sin⁑(2x)\sin x \cos x = \frac{1}{2} \sin(2 x) and integrate directly, giving βˆ’14cos⁑(2x)+C-\frac{1}{4} \cos(2 x) + C. The two forms differ by a constant.

Worked example: harmonic motion

A spring oscillates so that its displacement from equilibrium is x(t)=0.1cos⁑(4t)x(t) = 0.1 \cos(4 t) metres, with tt in seconds.

Velocity. v(t)=xβ€²(t)=βˆ’0.4sin⁑(4t)v(t) = x'(t) = -0.4 \sin(4 t) m/s.

Acceleration. a(t)=vβ€²(t)=βˆ’1.6cos⁑(4t)=βˆ’16 x(t)a(t) = v'(t) = -1.6 \cos(4 t) = -16 \, x(t) m/s2^2.

The maximum speed is 0.40.4 m/s, the period is T=2Ο€4=Ο€2T = \frac{2 \pi}{4} = \frac{\pi}{2} seconds, and the acceleration is always directed back toward equilibrium with magnitude proportional to displacement.

Common traps

Working in degrees. The derivative rules assume radians. If you differentiate sin⁑x\sin x in degrees, you get Ο€180cos⁑x\frac{\pi}{180} \cos x, which is not what NESA expects.

Sign error on cos⁑\cos. ddx(cos⁑x)=βˆ’sin⁑x\frac{d}{dx}(\cos x) = -\sin x and ∫sin⁑x dx=βˆ’cos⁑x+C\int \sin x \, dx = -\cos x + C. Forgetting the minus is the single most common arithmetic slip.

Forgetting to divide by the inside coefficient when integrating. ∫cos⁑(3x) dx=13sin⁑(3x)+C\int \cos(3 x) \, dx = \frac{1}{3} \sin(3 x) + C, not sin⁑(3x)+C\sin(3 x) + C.

Treating sin⁑2x\sin^2 x like sin⁑(x2)\sin(x^2). sin⁑2x=(sin⁑x)2\sin^2 x = (\sin x)^2 has derivative 2sin⁑xcos⁑x2 \sin x \cos x. sin⁑(x2)\sin(x^2) has derivative 2xcos⁑(x2)2 x \cos(x^2).

Confusing period and angular frequency. In sin⁑(Ο‰t)\sin(\omega t), the angular frequency is Ο‰\omega and the period is 2πω\frac{2 \pi}{\omega}.

In one sentence

Trigonometric calculus differentiates and integrates sin⁑\sin, cos⁑\cos and tan⁑\tan in radians, with chain-rule and reverse-chain-rule adjustments for linear inside arguments, and underpins the modelling of every periodic phenomenon in Maths Advanced.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q143 marksFind $\int_0^{\pi/2} \cos(2 x) \, dx$.
Show worked answer β†’

Use ∫cos⁑(kx) dx=sin⁑(kx)k\int \cos(k x) \, dx = \frac{\sin(k x)}{k} with k=2k = 2.

∫0Ο€/2cos⁑(2x) dx=[sin⁑(2x)2]0Ο€/2=sin⁑π2βˆ’sin⁑02=0βˆ’0=0\int_0^{\pi/2} \cos(2 x) \, dx = \left[ \frac{\sin(2 x)}{2} \right]_0^{\pi/2} = \frac{\sin \pi}{2} - \frac{\sin 0}{2} = 0 - 0 = 0.

Markers reward the correct antiderivative, the bracket notation, and an evaluated answer (here exactly zero, since the integrand is symmetric about Ο€/4\pi/4).

2021 HSC Q123 marksThe height of water in a harbour is modelled by $h(t) = 3 + 1.5 \sin\left( \frac{\pi t}{6} \right)$ metres, where $t$ is in hours after midnight. Find the rate at which the height is changing at $t = 2$ hours.
Show worked answer β†’

Differentiate using the chain rule.

hβ€²(t)=1.5cos⁑(Ο€t6)β‹…Ο€6=Ο€4cos⁑(Ο€t6)h'(t) = 1.5 \cos\left( \frac{\pi t}{6} \right) \cdot \frac{\pi}{6} = \frac{\pi}{4} \cos\left( \frac{\pi t}{6} \right).

At t=2t = 2: hβ€²(2)=Ο€4cos⁑(Ο€3)=Ο€4β‹…12=Ο€8β‰ˆ0.393h'(2) = \frac{\pi}{4} \cos\left( \frac{\pi}{3} \right) = \frac{\pi}{4} \cdot \frac{1}{2} = \frac{\pi}{8} \approx 0.393 m/hr.

The water is rising at about 0.390.39 metres per hour at t=2t = 2 hours.

Markers expect the chain rule applied correctly, the standard angle evaluated exactly, and units in the final answer.

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