1.05q - Trigonometric models in context | OCR A-Level Mathematics A

Use sine and cosine to describe repeating changes, predict when a condition is met, and resolve directed quantities into components. Turn observations into parameters, then interpret the answers in hours, metres or newtons. A model's neat arithmetic is useful only while its assumptions remain sensible.

Building a periodic model from observations

When does a harbour model predict enough depth for a boat needing more than $3.6\,\mathrm m$ ? Suppose the depth varies between $1.8$ and $4.2\,\mathrm m$ , repeats every $12$ hours and first reaches a maximum at 02:00. Build the cycle first; then we can answer the boat's question precisely.

You already know

Sine and cosine take values from $-1$ to $1$ and repeat after $2\pi$ radians. Cosine starts at a maximum when its input is zero. A horizontal shift changes when a feature occurs, without changing its height.

The trigonometric input here is a phase measuring progress through a cycle. Time is not itself a geometrical angle: multiplying hours by radians per hour gives the radian input the function needs.

Parameters of a periodic model

The midline is the centre value; the amplitude is the maximum departure from it. The period is the time for one complete repetition. A phase shift fixes the timing of a reference event, such as a maximum.

A cosine model is convenient when you know a maximum. A sine model with the appropriate shift can describe exactly the same function.

Cosine model with a reference maximum

For $A>0$ , $T>0$ ,

h(t)=d+A\cos\left(\frac{2\pi}{T}(t-t_0)\right)

The maximum is $d+A$ , the minimum is $d-A$ , and maxima occur at $t=t_0+kT$ , $k\in\mathbb Z$ , within the chosen time domain. The angular frequency is $\omega=2\pi/T$ .

Not in the formula sheet: learn it.

Check that the chosen domain contains both a maximum and a minimum before reporting the full range of the cosine model.

Worked example 1: fitting the harbour cycle

Question. Construct the stated harbour-depth model for $0\le t\le24$ , where $t$ is hours since midnight and depth is in metres. State its range.

Plan. Average the extreme depths, halve their difference, then put the observed maximum at the zero of a cosine phase.

\begin{aligned} d&=(4.2+1.8)/2=3 && \text{centre depth}\\ A&=(4.2-1.8)/2=1.2 && \text{half the range}\\ \omega&=2\pi/12=\pi/6 && \text{one turn per cycle}\\ h(t)&=3+1.2\cos\left(\frac\pi6(t-2)\right) && \text{maximum at }t=2 \end{aligned}

Why this step? Cosine equals $1$ when $t=2$ , so the phase directly matches the observed high water.

Answer. The displayed model has domain $[0,24]$ hours and range $[1.8,4.2]$ metres.

Check. It gives high water at $t=2,14$ and low water at $t=8,20$ , with the required heights and $12$ -hour separation.

The graph shows the whole day, including its endpoints; those endpoints will matter in the threshold problem.

Harbour-depth model with its midline, maxima, minima and 3.6 metre threshold

Change only the time of the first maximum. The centre, amplitude and period stay fixed.

Worked example 2: changing the timing

Question. A second harbour has the same extreme depths and period, but its first maximum is at 05:00. Construct its model on $0\le t\le24$ .

Plan. Keep the first model's vertical parameters and replace its reference time.

\begin{aligned} d&=3,\quad A=1.2,\quad\omega=\pi/6 && \text{unchanged observations}\\ h(t)&=3+1.2\cos\left(\frac\pi6(t-5)\right) && \text{shift the maximum} \end{aligned}

Why this step? The whole phase $\omega(t-t_0)$ must be zero at the reference time; subtracting $5$ outside the cosine would change depth instead.

Answer. The model is $h(t)=3+1.2\cos(\pi(t-5)/6)$ , for $0\le t\le24$ .

Check. Maxima occur at $5,17$ hours and minima at $11,23$ hours; the range is still $[1.8,4.2]$ metres.

The angle setting is part of using the model correctly.

Watch out

Wrong: enter $\pi(t-2)/6$ in degree mode. Right: use radians for this phase. Its frequency is $\pi/6$ radians per hour, not $\pi/6$ hours.

Keep the second harbour's timing but change its cycle length.

Times and intervals that satisfy a condition

A threshold asks you to solve over the model's actual time domain, rather than return one inverse-cosine value.

Use the first harbour's model $h(t)=3+1.2\cos(\pi(t-2)/6)$ , $0\le t\le24$ . Its phase runs from $-\pi/3$ to $11\pi/3$ . Listing all boundary crossings and checking between them is safer than guessing a time window from one root.

Worked example 3: a strict depth requirement

Question. Find every time in the stated day when $h(t)>3.6\,\mathrm m$ .

Plan. Solve the threshold equality, then use a maximum to select the correct sides of its crossings.

\begin{aligned} h(t)>3.6 &\iff\cos\left(\frac\pi6(t-2)\right)>\frac12 && \text{subtract and divide by }1.2\\ \frac\pi6(t-2)&=\pm\pi/3+2k\pi && \text{boundary phases}\\ t&=0,4,12,16,24 && \text{boundaries in }[0,24] \end{aligned}

Why this step? The inverse-cosine output gives only one boundary phase; periodicity and the second branch give all crossings.

Answer. The predicted times are $0<t<4$ or $12<t<16$ hours since midnight.

Check. At $t=2,14$ , depth is $4.2$ , so the selected intervals are above threshold. Each listed boundary has depth exactly $3.6$ and is excluded.

Change only “more than” to “at least”. The equality cases now belong in the answer.

Worked example 4: an included endpoint

Question. Find all times in the same day when $h(t)\ge3.6\,\mathrm m$ .

Plan. Include every threshold boundary as well as the above-threshold intervals.

\begin{aligned} h(t)\ge3.6 &\iff\cos\left(\frac\pi6(t-2)\right)\ge\frac12 && \text{same threshold phase}\\ t&\in[0,4]\cup[12,16]\cup\{24\} && \text{retain domain endpoints} \end{aligned}

Why this step? At $t=24$ equality holds, although the following above-threshold interval lies outside the day. A single point can belong to an inequality solution.

Answer. From midnight to 04:00 inclusive, from 12:00 to 16:00 inclusive, and at 24:00.

Check. The next maximum would be at $t=26$ , outside the domain; no interval beyond $24$ is included.

That isolated endpoint is a useful surprise: an inequality need not produce intervals alone when you restrict the domain. Also check the range first. Since $4.2$ is the maximum, $h(t)\ge4.2$ holds only at $t=2,14$ ; $h(t)>4.2$ has no solutions.

Exam tip

Report the full time intervals, their endpoint inclusions and their units. A list of equality times alone does not answer a “greater than” condition.

Lower the threshold to the midline, keeping the non-strict relation.

Combining effects with one frequency

Contributions with the same frequency can combine into one wave, but their amplitudes do not usually add.

Combining cosine and sine contributions

For $a,b>0$ ,

a\cos u+b\sin u=R\cos(u-\alpha),\quad R=\sqrt{a^2+b^2},\quad \cos\alpha=\frac aR,\quad\sin\alpha=\frac bR

Here $0<\alpha<\pi/2$ . Expand the cosine to derive and check the coefficient matching; handle other signs with a suitable phase.

Not in the formula sheet: learn it.

The compound-angle identity is supplied, but this harmonic-form result is derived from it. A phase in radians becomes a time only after division by angular frequency.

Worked example 5: two positive contributions

Question. A depth is modelled by $h(t)=3+0.6\cos(\omega t)+0.8\sin(\omega t)$ , where $\omega=\pi/6$ radians per hour and $t\ge0$ . Find its range and first maximum time.

Plan. Combine the equal-frequency terms and set the new cosine phase to zero.

\begin{aligned} R&=\sqrt{0.6^2+0.8^2}=1 && \text{amplitude}\\ \alpha&=\arctan(4/3) && \text{first-quadrant phase}\\ h(t)&=3+\cos(\omega t-\alpha) && \text{match coefficients}\\ t_{\max}&=\alpha/\omega=\frac{6\arctan(4/3)}\pi\approx1.77 && \text{convert phase to hours} \end{aligned}

Why this step? The two contributions reach their own maxima at different times, so $0.6+0.8=1.4$ is not the combined amplitude.

Answer. The range is $[2,4]$ metres; the first maximum is about $1.77$ hours after $t=0$ , to $3$ significant figures.

Check. At that time, cosine and sine of $\omega t$ are $0.6$ and $0.8$ ; the depth is $3+0.36+0.64=4$ .

Change only the sign of the sine contribution. The amplitude is unchanged, but the first non-negative maximum moves.

Worked example 6: a reversed sine contribution

Question. Find the range and first maximum time for $h(t)=3+0.6\cos(\omega t)-0.8\sin(\omega t)$ , with the same $\omega$ and $t\ge0$ .

Plan. Use $\cos(\omega t+\alpha)$ and choose the first admissible maximum phase.

\begin{aligned} h(t)&=3+\cos(\omega t+\alpha) && \alpha=\arctan(4/3)\\ \omega t+\alpha&=2\pi && \text{first maximum with }t\ge0\\ t&=\frac{2\pi-\alpha}{\pi/6}\approx10.2 && \text{hours, to }3\text{ s.f.} \end{aligned}

Why this step? Setting the phase to zero gives a negative time; the next cosine maximum is one full turn later.

Answer. Range $[2,4]$ metres; first maximum at approximately $10.2$ hours.

Check. At the exact time, $\cos(\omega t)=0.6$ and $\sin(\omega t)=-0.8$ , giving depth $4$ . Maxima repeat every $12$ hours.

A single-wave combination requires the same frequency. Different-frequency contributions can make a more complicated pattern; coefficient matching into one cosine would be unjustified.

Watch out

Wrong: the phase $\alpha\approx0.927$ means a delay of $0.927$ hours. Right: divide the phase by $\omega$ , with units, to get the corresponding time shift.

Change the ratio of the two contributions while retaining their signs and frequency.

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1.05q - Trigonometric models in context

Building a periodic model from observations

Times and intervals that satisfy a condition

Combining effects with one frequency

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