QCE Mathematical Methods - Unit 3 - Differentiation of trigonometric functions and differentiation rules
Calculus of Trigonometric Functions | QCE Mathematical Methods
Revise QCE Mathematical Methods trigonometric derivatives, including sine, cosine, radians mode and chain rule applications.
Updated 2026-05-18 - 4 min read
QCAA official coverage - Mathematical Methods 2025 v1.3
Exact syllabus points covered
- Use the rules $\frac{d}{dx}\sin(x)=\cos(x)$ and $\frac{d}{dx}\sin(f(x))=f'(x)\cos(f(x))$.
- Use the rules $\frac{d}{dx}\cos(x)=-\sin(x)$ and $\frac{d}{dx}\cos(f(x))=-f'(x)\sin(f(x))$.
- Model and solve problems that involve derivatives of trigonometric functions, with and without technology.
Trigonometric calculus is about rates of change in periodic motion. If a graph repeats in waves, the derivative tells you how quickly it is moving upward or downward at each point in the cycle.
In QCE Mathematical Methods, the two core derivative rules are:
$ \frac{d}{dx}\sin(x)=\cos(x) $
$ \frac{d}{dx}\cos(x)=-\sin(x) $
These rules assume $x$ is measured in radians. The derivative formulas do not work in the same clean way if your calculator is thinking in degrees.
The radian condition is not a cosmetic setting. The first-principles derivation of $\frac{d}{dx}\sin x$ relies on the limits:
$ \lim_{h\to0}\frac{\sin h}{h}=1 $
and:
$ \lim_{h\to0}\frac{1-\cos h}{h}=0. $
Those limits are true when $h$ is measured in radians. If degrees are used, an extra conversion factor appears, so the clean formulas above no longer match.
Chain rule with trig
Most questions use a function inside the sine or cosine. The QCAA rules are:
$ \frac{d}{dx}\sin(f(x))=f'(x)\cos(f(x)) $
$ \frac{d}{dx}\cos(f(x))=-f'(x)\sin(f(x)) $
The inside function remains inside the trig function. You do not simplify $\cos(3x)$ into $3\cos(x)$; that is not a valid identity.
For common linear-inside functions:
| Function | Derivative | | --- | --- | | $\sin(ax+b)$ | $a\cos(ax+b)$ | | $\cos(ax+b)$ | $-a\sin(ax+b)$ | | $A\sin(ax+b)+c$ | $Aa\cos(ax+b)$ | | $A\cos(ax+b)+c$ | $-Aa\sin(ax+b)$ |
The vertical shift $c$ always differentiates to $0$. The amplitude $A$ stays as a multiplier, and the coefficient $a$ from the inside angle also becomes a multiplier. This is why the rate of change of a fast oscillation is larger: increasing $a$ squeezes more cycles into the same horizontal distance.
Worked example
Sine and cosine as rates
It helps to connect the derivatives to graph shape:
- $\sin(x)$ is increasing fastest at $x=0$, and $\cos(0)=1$.
- $\sin(x)$ has a turning point at $x=\frac{\pi}{2}$, and $\cos\left(\frac{\pi}{2}\right)=0$.
- $\cos(x)$ is decreasing fastest at $x=\frac{\pi}{2}$, and $-\sin\left(\frac{\pi}{2}\right)=-1$.
This is a quick way to catch sign errors. If your derivative says $\cos(x)$ is increasing at $x=0$, something is wrong, because $-\sin(0)=0$.
The unit circle gives the same intuition. On the unit circle, $\sin x$ is the vertical coordinate and $\cos x$ is the horizontal coordinate. Near $x=0$, the sine value rises quickly, so its derivative is close to $1$. Near $x=\frac{\pi}{2}$, sine is at the top of the circle, so its instantaneous change is $0$. That matches $\cos\left(\frac{\pi}{2}\right)=0$.
For cosine, the graph starts at a maximum. It is flat at $x=0$, then begins decreasing, so the derivative should start at $0$ and become negative. That matches $-\sin x$.
Common mistake
Modelling with trig derivatives
If a height is modelled by:
$ h(t)=2+3\sin(0.5t) $
then:
$ h'(t)=1.5\cos(0.5t) $
The derivative gives the vertical velocity. A positive value means height is increasing; a negative value means height is decreasing; a zero value means the object is momentarily at a peak or trough.
For a cosine model:
$ h(t)=A\cos(bt+c)+d, $
the derivative is:
$ h'(t)=-Ab\sin(bt+c). $
The vertical translation $d$ disappears because it does not affect rate of change. The phase shift $c$ stays inside the trig function because it changes when the peak, trough and zero-rate points occur.
Stationary points in trig models
Trig derivatives are also used to locate peaks, troughs and moments where motion changes direction. For:
$ h(t)=2+3\sin(0.5t), $
stationary points occur when:
$ h'(t)=1.5\cos(0.5t)=0. $
So:
$ 0.5t=\frac{\pi}{2}+k\pi $
and:
$ t=\pi+2k\pi. $
Those times correspond to alternating maximum and minimum heights. To classify them, look at the original graph, use the sign of $h'(t)$ on either side, or take the second derivative:
$ h''(t)=-0.75\sin(0.5t). $
This is the same logic as ordinary differentiation, but the periodic nature means there are infinitely many solutions unless the context restricts the time interval.