Differentiate trigonometric functions, including compositions of the form $\sin(f(x))$, $\cos(f(x))$ and $\tan(f(x))$, working in radians

What this dot point is asking

QCAA wants you to differentiate trigonometric functions and their compositions in radians, including combinations with the product, quotient and chain rules. Trigonometric derivatives appear in Paper 1 short answer, in Paper 2 modelling questions about oscillating quantities (tides, sound, AC currents, planetary orbits), and in PSMTs that involve periodic phenomena.

The answer

Why radians

In Methods, all calculus on trig functions is done with $x$ in radians. The formula $\frac{d}{dx}(\sin x) = \cos x$ is only true when $x$ is in radians. If you work in degrees the derivative picks up an awkward factor of $\frac{\pi}{180}$, which is why QCAA requires radians for calculus questions and why most calculators default to degree mode on a fresh reset (always check and switch to radians).

The standard derivatives

The minus sign on the derivative of $\cos$ is the single most common Paper 1 trap. Memorise it.

Chain-rule generalisations

For any differentiable inner $f(x)$:

The most common case is a linear inner function, where the chain rule factor is just a constant.

Linear inside argument

For constants $a$ and $b$:

This is the most heavily examined form. Almost every modelling question is of the type $A \sin(\omega t + \phi) + C$.

Worked examples

Direct chain rule

Differentiate $y = \cos(3x - 1)$.

$\dfrac{dy}{dx} = -3 \sin(3x - 1)$.

Inner is a polynomial

Differentiate $y = \sin(x^2)$.

Let $f(x) = x^2$, $f'(x) = 2x$. $\dfrac{dy}{dx} = 2x \cos(x^2)$.

Combining with the product rule

Differentiate $y = x^2 \sin x$.

Product rule: $u = x^2$, $v = \sin x$, $u' = 2x$, $v' = \cos x$.

$\dfrac{dy}{dx} = 2x \sin x + x^2 \cos x.$

Modelling: simple harmonic motion

A particle moves so that its displacement from the origin is $s(t) = 4 \sin(2 t)$ metres, with $t$ in seconds. Find its velocity at $t = \dfrac{\pi}{6}$.

$v(t) = s'(t) = 8 \cos(2 t)$.

$v(\pi / 6) = 8 \cos(\pi / 3) = 8 \cdot \dfrac{1}{2} = 4 \text{ m/s}.$

The exact value $\cos(\pi/3) = \tfrac{1}{2}$ is expected without a calculator on Paper 1.

Tangent example

Differentiate $y = \tan(3x)$.

$\dfrac{dy}{dx} = 3 \sec^2(3x) = \dfrac{3}{\cos^2(3x)}$.

Common traps

Working in degrees. $\frac{d}{dx}(\sin x) = \cos x$ only when $x$ is in radians. Set your calculator to radians for Paper 2. Paper 1 problems are always in radians by convention.

Dropping the minus sign on $\cos$. $\frac{d}{dx}(\cos x) = -\sin x$. Reversing the sign turns a maximum into a minimum and a velocity into its negative.

Missing the chain rule factor. $\frac{d}{dx}(\sin(2x)) = 2 \cos(2x)$, not $\cos(2x)$. The factor of $2$ comes from differentiating the inner $2x$.

Forgetting exact values. Paper 1 expects $\sin(\pi/6) = 1/2$, $\cos(\pi/4) = \sqrt{2}/2$, $\tan(\pi/3) = \sqrt{3}$ and the related angle values without a calculator. Drill the unit circle.

Combining identities incorrectly. $\sin^2 x + \cos^2 x = 1$, not $\sin^2 x - \cos^2 x = 1$. Sign errors here lead to wrong derivatives on optimisation questions.

In one sentence

In radians, $\sin x$, $\cos x$ and $\tan x$ differentiate to $\cos x$, $-\sin x$ and $\sec^2 x$ respectively, and the chain rule extends this to $\frac{d}{dx} \sin(f(x)) = f'(x) \cos(f(x))$ and analogous forms for $\cos$ and $\tan$.