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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\sin, cos\cos and tan\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.

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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(sinx)=cosx\frac{d}{dx}(\sin x) = \cos x

ddx(cosx)=sinx\frac{d}{dx}(\cos x) = -\sin x

ddx(tanx)=sec2x=1cos2x\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))=ksin(kx+c)\frac{d}{dx}(\cos(k x + c)) = -k \sin(k x + c)

ddx(tan(kx+c))=ksec2(kx+c)\frac{d}{dx}(\tan(k x + c)) = k \sec^2(k x + c)

Integrals

sinxdx=cosx+C\int \sin x \, dx = -\cos x + C

cosxdx=sinx+C\int \cos x \, dx = \sin x + C

sec2xdx=tanx+C\int \sec^2 x \, dx = \tan x + C

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

sin(kx+c)dx=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(ωt+ϕ)+Dy(t) = A \sin(\omega t + \phi) + D

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

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

Differentiating gives the velocity, y(t)=Aωcos(ωt+ϕ)y'(t) = A \omega \cos(\omega t + \phi), and differentiating again gives the acceleration, y(t)=Aω2sin(ωt+ϕ)=ω2(yD)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.

Why radians are compulsory

The derivative rules ddx(sinx)=cosx\frac{d}{dx}(\sin x) = \cos x and ddx(cosx)=sinx\frac{d}{dx}(\cos x) = -\sin x only hold when xx is measured in radians, because they rest on the limit limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1, which is true only in radian measure. If your calculator is in degree mode the geometry of the limit changes and every derivative is off by a factor of π180\frac{\pi}{180}. Always set angles in radians for any calculus involving trigonometric functions, and convert any degree information in the question to radians before differentiating or integrating.

Why the derivative of sinx\sin x is cosx\cos x

The rule ddxsinx=cosx\frac{d}{dx}\sin x = \cos x is worth seeing, not just memorising. The derivative of a function is its gradient at each point, so if we read the gradient of sinx\sin x off the graph and plot those gradient values, we should recover cosx\cos x. We do.

Stage 1, plot y=sinxy = \sin x. Draw one full cycle of sinx\sin x from 00 to 2π2\pi: up to a peak at π2\frac{\pi}{2}, back through zero at π\pi, down to a trough at 3π2\frac{3\pi}{2}, and back to zero at 2π2\pi.

Stage 1: plot y = sin x The curve y equals sin x drawn from x equals 0 to 2 pi, starting at the origin, peaking at x equals pi over 2, crossing zero at pi, reaching a trough at 3 pi over 2, and returning to zero at 2 pi. x π/2π3π/2 y = sin x 1

Stage 2, measure the gradient at sample points. Draw the tangent at a few key points and read its gradient. At x=0x = 0 the curve rises steeply, gradient +1+1; at the peak x=π2x = \frac{\pi}{2} the tangent is flat, gradient 00; at x=πx = \pi the curve falls steeply, gradient 1-1; at the trough it is flat again; at 2π2\pi it is rising, gradient +1+1.

Stage 2: measure the gradient at sample points The same sine curve. Short accent tangent segments are drawn at x equals 0, pi over 2, pi, 3 pi over 2 and 2 pi. The tangent is steepest and positive at x equals 0, flat at the peak x equals pi over 2, steepest and negative at x equals pi, flat at the trough, and positive again at 2 pi. x π/2π3π/2 slope 1 slope 0 slope -1 2

Stage 3, plot each gradient against xx. Take those gradient values, 1,0,1,0,11, 0, -1, 0, 1, and plot them at the same xx-positions. The points sit exactly where cosx\cos x would: cos0=1\cos 0 = 1, cosπ2=0\cos\frac{\pi}{2} = 0, cosπ=1\cos\pi = -1, and so on.

Stage 3: plot each gradient against x The faint sine curve with the measured gradients plotted as accent points: 1 at x equals 0, 0 at pi over 2, negative 1 at pi, 0 at 3 pi over 2 and 1 at 2 pi. These gradient points trace the shape of the cosine curve. x π/2π3π/2 gradient values 3

Stage 4, the gradient function is cosx\cos x. Join the gradient points and the curve y=cosxy = \cos x appears. The gradient of sinx\sin x at every point is cosx\cos x, which is exactly what ddxsinx=cosx\frac{d}{dx}\sin x = \cos x says. The same picture, shifted, shows ddxcosx=sinx\frac{d}{dx}\cos x = -\sin x.

Stage 4: the gradient function is y = cos x The faint sine curve with the gradient curve drawn through the plotted points in the accent colour. The gradient of sin x at every point is cos x, so the derivative of sin x is cos x. x π/2π3π/2 y = cos x sin x (faint) 4

Useful identities for integration

Some trigonometric integrals have no direct antiderivative until you rewrite them with an identity. The most useful in Maths Advanced are the double-angle forms sin2x=1cos2x2\sin^2 x = \frac{1 - \cos 2x}{2} and cos2x=1+cos2x2\cos^2 x = \frac{1 + \cos 2x}{2}. These convert a squared trig function (which you cannot integrate directly) into a constant plus a cosine of a double angle (which you can). For example, cos2xdx=1+cos2x2dx=x2+sin2x4+C\int \cos^2 x\,dx = \int \frac{1 + \cos 2x}{2}\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C. Recognising when to deploy an identity is a key marker of fluency.

How exam questions ask about trigonometric calculus

  • "Differentiate" a trig expression. Apply the standard derivative and the chain-rule factor kk for a linear inside argument; watch the minus sign on cos\cos.
  • "Find dx\int \ldots \, dx" or "evaluate abdx\int_a^b \ldots \, dx". Use the standard integral, divide by the inside coefficient, add +C+C for an indefinite integral, and substitute the limits for a definite one.
  • "Find the rate at which ... is changing." A modelling question: differentiate the given function and substitute the value, exact then with units.
  • "Find the maximum / minimum value" or "the period / amplitude." Read the parameters of Asin(ωt+ϕ)+DA\sin(\omega t + \phi) + D: amplitude AA, period 2πω\frac{2\pi}{\omega}, centre DD; the extreme values are D±AD \pm A.
  • "Show that the motion is simple harmonic." Differentiate twice and show y=ω2(yD)y'' = -\omega^2(y - D): acceleration proportional to displacement and directed back to the centre.
  • An integral of sin2x\sin^2 x or cos2x\cos^2 x. Rewrite with the double-angle identity first; the squared form has no direct antiderivative.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC Q143 marksFind 0π/2cos(2x)dx\int_0^{\pi/2} \cos(2 x) \, dx.
Show worked answer →

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

0π/2cos(2x)dx=[sin(2x)2]0π/2=sinπ2sin02=00=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 π/4\pi/4).

2021 HSC Q123 marksThe height of water in a harbour is modelled by h(t)=3+1.5sin(πt6)h(t) = 3 + 1.5 \sin\left( \frac{\pi t}{6} \right) metres, where tt is in hours after midnight. Find the rate at which the height is changing at t=2t = 2 hours.
Show worked answer →

Differentiate using the chain rule.

h(t)=1.5cos(πt6)π6=π4cos(πt6)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: h(2)=π4cos(π3)=π412=π80.393h'(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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