Pure 5.9 - Trigonometric functions in context | Edexcel A-Level Mathematics

Read and build sinusoidal models for repeating motion and changing quantities, solve their equations in meaningful time intervals, and resolve vectors, velocities and forces. You will connect each parameter to its units and judge what a model can reasonably predict.

Read a trig model

A wheel's lowest point is $1\,\mathrm m$ above the ground. How can a sine or cosine model tell us when a marked point rises above $10\,\mathrm m$ ? We need its centre height, radius, period and starting position. A formula becomes useful when each number has a physical meaning.

Begin with a simple hypothetical daylight model,

H=12+4\cos\left(\frac\pi6(t-6)\right),

where $H$ is daylight duration in hours and $t$ is time in months. This is a mathematical model, rather than measured data for a named location. At $t=6$ , cosine is $1$ , giving $16$ hours; at $t=0$ , it is $-1$ , giving $8$ hours.

You already know

Sine and cosine range from $-1$ to $1$ . Each repeats after $2\pi$ radians, or $360^\circ$ . A negative coefficient reflects a wave in its midline.

A wave's amplitude measures distance from its middle, rather than its total height from bottom to top.

Amplitude, midline and period

The amplitude is half the difference between a sinusoid's maximum and minimum. The midline is their mean. The period is the shortest positive input increase over which the entire non-constant pattern repeats. For a range from $8$ to $16$ , the amplitude is $4$ and the midline is $12$ , not $8$ .

The coefficient inside the trig function is a rate of change of angle, not a time period.

Reading a sinusoidal model

For $y=D+A\sin(B(t-C))$ or $y=D+A\cos(B(t-C))$ , where $A\ne0$ , $B\ne0$ ,

\text{amplitude}=|A|, \quad \text{midline}=D, \quad \text{range}=[D-|A|, D+|A|].

The period is $2\pi/|B|$ for a radian input, or $360/|B|$ when the input's numerical value is in degrees. The period's unit is the unit of $t$ .

Not in the formula booklet: learn it.

A shift of $C$ tells you where a standard cosine has input zero. Whether that is a maximum or minimum also depends on the sign of $A$ .

[DIAGRAM: asset_slug: p1_5_9_use_trigonometric_functions_to_solve_problems_in_context__diagram_01; asset_name: Reading a daylight-duration model; caption: Midline 12, amplitude 4 and period 12 months describe different features; description: Graph of H equal to 12 plus 4 cos pi over 6 times t minus 6, from 0 to 12 months. The maximum 16 occurs at 6 months, minima 8 at both ends, and guides identify the midline and the amplitude.]

Use those features together rather than identifying one coefficient in isolation.

Worked example 1: read the daylight model

Question. Find the range and period of $H=12+4\cos(\frac\pi6(t-6))$ , and state when its maximum first occurs for $0\le t\le12$ .

Plan. Read the vertical shift and amplitude, then make the cosine input zero.

\begin{aligned} H_{\min}&=12-4=8, \quad H_{\max}=12+4=16 && \text{use cosine's range}\\ T&=\frac{2\pi}{\pi/6}=12 && \text{one full input turn}\\ \frac\pi6(t-6)&=0\quad\Rightarrow\quad t=6 && \text{cosine at its maximum} \end{aligned}

Why use radians? The model explicitly uses $\pi$ in its angle; a full turn is $2\pi$ , not $360$ .

Answer. The range is $8\le H\le16$ hours, the period is $12$ months, and the maximum occurs at $t=6$ months.

Check. Inputs at $t=0$ and $12$ are $-\pi$ and $\pi$ , both giving the minimum $8$ .

Change only the amplitude coefficient from $4$ to $5$ . The timing stays the same.

Worked example 2: change the amplitude

Question. Give the same features for $H=12+5\cos(\frac\pi6(t-6))$ .

Plan. Recalculate the extreme values and retain the unchanged input's period and phase.

\begin{aligned} H_{\min}&=12-5=7, \quad H_{\max}=12+5=17 && \text{change the vertical excursion}\\ T&=12, \quad t_{\max}=6 && \text{the angle expression is unchanged} \end{aligned}

Why does the maximum time stay fixed? Changing the amplitude alters height, not the angle at which cosine is $1$ .

Answer. The range is $7\le H\le17$ hours; the period is $12$ months and the maximum is at $t=6$ months.

Check. At $t=3$ , the cosine input is $-\pi/2$ ; both models are on the same midline $H=12$ .

Always attach output units to amplitude and midline, and input units to period.

Watch out

Wrong: the period is $\pi/6$ months. Right: $\pi/6$ is the angular rate in radians per month, so the period is $2\pi/(\pi/6)=12$ months.

Read a pressure model with a different time scale.

Build a periodic model

For a wheel, the centre height is the midline and the radius is the amplitude. If it rotates uniformly, its angle changes at a constant rate. Starting at the bottom suggests negative cosine; starting at the top suggests positive cosine.

Building from extrema and a period

If the maximum is $M$ , the minimum is $m$ , and the period is $T>0$ , then

D=\frac{M+m}{2}, \qquad |A|=\frac{M-m}{2}, \qquad B=\frac{2\pi}{T}

for a radian model. Choose the function, phase and coefficient sign to match the starting position and direction.

Not in the formula booklet: learn it.

You can derive the wheel's height directly: a radius at angle $\theta$ above the horizontal has vertical component $r\sin\theta$ , so $H=c+r\sin\theta$ . Measuring time from a lowest point gives the equivalent negative-cosine model.

[DIAGRAM: asset_slug: p1_5_9_use_trigonometric_functions_to_solve_problems_in_context__diagram_02; asset_name: A wheel and its height-time curve; caption: The vertical radius component creates a sinusoid about the centre height; description: A wheel geometry panel shows H equal to c plus r sin theta. A separate example graph H equal to 9 minus 7 cos pi t over 20 starts at 2 metres, reaches 16 metres at 20 seconds and completes its cycle at 40 seconds.]
The vertical radius component creates a sinusoid about the centre height

Here is a useful geometrical surprise: diametrically opposite points always have heights adding to $2c$ . Their vertical displacements are equal and opposite. For a wheel centred at $6\,\mathrm m$ , the opposite heights add to $12\,\mathrm m$ at every angle.

Worked example 3: start at the bottom

Question. A point is on a wheel of radius $5\,\mathrm m$ with centre $6\,\mathrm m$ above the ground. It starts lowest at $t=0$ , and one revolution takes $24\,\mathrm s$ . Find its height model.

Plan. Use the centre height as midline, the radius as amplitude, and negative cosine for the lowest start.

\begin{aligned} H&=6-5\cos(Bt) && \text{match the starting position}\\ \frac{2\pi}{B}&=24\quad\Rightarrow\quad B=\frac\pi{12} && \text{match the period} \end{aligned}

Why negative cosine? At $t=0$ , it subtracts the full radius from the centre height.

Answer. $H=6-5\cos(\pi t/12)$ , with $H$ in metres and $t$ in seconds.

Check. $H(0)=1$ , $H(12)=11$ and $H(24)=1$ , matching bottom, top and bottom again.

Change only the starting position from bottom to top. The wheel's size and speed are unchanged.

Worked example 4: start at the top

Question. Find a model for the same wheel point if it starts at its highest position at $t=0$ .

Plan. Keep the midline and angular rate, but use positive cosine.

\begin{aligned} H&=6+5\cos(Bt) && \text{highest initial position}\\ B&=\frac\pi{12} && \text{the period remains }24\,\mathrm s \end{aligned}

Why change just the sign? It swaps maxima and minima without changing the period or radius.

Answer. $H=6+5\cos(\pi t/12)$ .

Check. $H(0)=11$ , $H(12)=1$ , and $H(24)=11$ , as required.

A point starting level with the centre and moving upwards is often simpler in sine form. For example, $H=8+6\sin(\pi t/15)$ has radius $6\,\mathrm m$ and period $30\,\mathrm s$ . Its first arrival at $H=11$ solves $\sin(\pi t/15)=1/2$ , giving $\pi t/15=\pi/6$ , so $t=2.5\,\mathrm s$ .

Watch out

Wrong: choose cosine simply because the motion is circular. Right: choose the phase and sign from the starting height and direction. Sine and cosine can describe the same motion with different time origins.

Build a model after changing the wheel radius.

Use single-function form

If a model combines sine and cosine of the same input, their coefficients are not separate amplitudes. Rewrite their sum as one function to expose its actual range.

You already know

Expand the requested shifted form and match coefficients. For $a\cos x+b\sin x$ , the amplitude is $r=\sqrt{a^2+b^2}$ . This result is derived from supplied compound-angle formulae but is not itself printed in the formula booklet.

A constant added outside the wave changes the midline, leaving that amplitude unchanged.

Worked example 5: recover an unknown midline

Question. A population index is modelled by $N=A+12\cos((30t)^\circ)-5\sin((30t)^\circ)$ , where $t$ is in months. Its maximum during a full year is $93$ . Find $A$ , a single-function form, and the minimum.

Plan. Combine the oscillating terms, then use maximum equals midline plus amplitude.

With $x=(30t)^\circ$ , expand $r\cos(x+\alpha)$ :

\begin{aligned} r\cos\alpha&=12, \quad r\sin\alpha=5 && \text{match the two coefficients}\\ r&=\sqrt{12^2+5^2}=13 && \text{square and add}\\ \alpha&=\tan^{-1}(5/12) && 0^\circ<\alpha<90^\circ\\ A+13&=93\quad\Rightarrow\quad A=80 && \text{use the given maximum} \end{aligned}

Why can the wave reach both extremes during a year? Its input advances by $360^\circ$ in $12$ months, one complete cycle.

Answer. $N=80+13\cos((30t)^\circ+\alpha)$ , with exact $\alpha=\tan^{-1}(5/12)\approx22.62^\circ$ . The minimum is $80-13=67$ .

Check. At $t=0$ , the original is $92$ , and the new form gives $80+13(12/13)=92$ .

Change only the sine coefficient's sign. The range stays fixed, while the peak occurs at a different time.

Worked example 6: the phase changes, the range stays

Question. Replace the negative sine term by a positive one, keeping the full-year maximum $93$ . Find the single-function model and minimum.

Plan. Expand cosine difference to match the positive sine coefficient.

\begin{aligned} 12\cos x+5\sin x&=13\cos(x-\alpha) && \cos\alpha=12/13, \ \sin\alpha=5/13\\ A+13&=93\quad\Rightarrow\quad A=80 && \text{the amplitude is unchanged}\\ N_{\min}&=80-13=67 && \text{one full cycle is included} \end{aligned}

Why does the sign affect timing? It changes the phase from $+\alpha$ to $-\alpha$ , rather than the size of the vertical excursion.

Answer. $N=80+13\cos((30t)^\circ-\alpha)$ , with the same exact $\alpha$ ; the minimum remains $67$ .

Check. At $t=3$ , the input is $90^\circ$ : the new model gives $85$ , matching $80+5$ .

The full-cycle condition matters. On a shorter time interval, the wave may never reach its theoretical maximum.

Watch out

Wrong: $20-6\cos u$ has maximum $20$ . Right: its full-cycle maximum is $26$ , because $\cos u=-1$ makes the added term positive.

Match a negative cosine form and interpret its range.

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Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 5.9 - Trigonometric functions in context

Read a trig model

Build a periodic model

Use single-function form

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