- Any expression \(a\sin x + b\cos x\) can be written as a single wave, \(R\cos(x – \alpha)\), called the harmonic form.
- Find \(R = \sqrt{a^2 + b^2}\) and \(\alpha\) from \(\tan\alpha = \dfrac{a}{b}\), checking the quadrant from the signs of \(R\sin\alpha = a\) and \(R\cos\alpha = b\).
- The maximum value is \(R\) (when \(x = \alpha\)) and the minimum is \(-R\) (when \(x = \alpha + 180°\)). Equations become a single \(\cos\) equation.
- In our case, a tide model \(h = 5 + 3\sin 30t° + 4\cos 30t°\) becomes \(5 + 5\cos(30t – 36.87)°\): high water of 10 m at about 1:14 a.m., and depth of at least 8 m for about 3.54 hours around each high tide.
- Exam point: expand \(R\cos(x – \alpha) = R\cos x\cos\alpha + R\sin x\sin\alpha\) and compare coefficients; do not guess which of \(a\) and \(b\) goes with \(\cos\alpha\).
The case: when can the ship enter the harbour?
A cargo ship needs at least 8 metres of water to cross the entrance channel of a small harbour. The harbour master models the depth of water in the channel, in metres, \(t\) hours after midnight, as
The angle \(30t\) is in degrees, so the pattern repeats every \(360 \div 30 = 12\) hours, close to the real gap between high tides in places with two tides a day. (The figures are illustrative; real tide tables come from harmonic analysis of many more terms.)
The captain asks three questions. How deep does the water get, and when? How shallow does it get? And for how long is it at least 8 m deep? With the sine and cosine terms separate, none of these is easy to read off. The two waves have the same period but peak at different times, so their sum peaks somewhere in between. The harmonic form combines them into one wave with a clear height and a clear timing.
The concept: two waves make one
Graph \(y = 3\sin x\) and \(y = 4\cos x\) and add them. The result looks like a single cosine wave that is taller than either part and shifted to the right. That is not a coincidence. The compound angle formula says
So if we choose \(R\) and \(\alpha\) to make \(R\sin\alpha = 3\) and \(R\cos\alpha = 4\), the right-hand side is exactly \(3\sin x + 4\cos x\). Squaring and adding gives \(R^2(\sin^2\alpha + \cos^2\alpha) = 9 + 16\), so \(R^2 = 25\) and \(R = 5\). Dividing gives \(\tan\alpha = \frac{3}{4}\), so \(\alpha = 36.87°\).[1]
Here \(a\) and \(b\) are the coefficients of \(\sin x\) and \(\cos x\). \(R\) is the amplitude of the combined wave: it can never be more than \(|a| + |b|\), and it is never less than the larger of \(|a|\) and \(|b|\). \(\alpha\) is how far the combined wave is shifted to the right compared with \(\cos x\). The two lines \(R\sin\alpha = a\) and \(R\cos\alpha = b\) come from matching coefficients, and they fix \(\alpha\) uniquely, including its quadrant.[4]
Mathematics Made Simple explains the harmonic form R cos(x − α) with worked examples and graphs.Get the textbook · $12
Which of the four forms to use depends on the signs. A useful rule: pick the form whose expansion has the same signs as your expression. For \(a\sin x + b\cos x\) with both positive, \(R\cos(x – \alpha)\) or \(R\sin(x + \alpha)\) both work with an acute \(\alpha\). For \(b\cos x – a\sin x\), use \(R\cos(x + \alpha)\). For \(a\sin x – b\cos x\), use \(R\sin(x – \alpha)\). Exam questions normally tell you which form to use.[3]
Why this method fits the case
The tide model is a sum of a sine wave and a cosine wave with the same period. That is exactly the shape \(a\sin\theta + b\cos\theta\) with \(\theta = 30t\). The harmonic form turns it into \(5 + R\cos(\theta – \alpha)\), where every number has a physical meaning: 5 m is the mean depth, \(R\) is how far the tide rises and falls from that mean, and \(\alpha\) sets the time of high water.
It also turns the captain’s question into a standard equation. “When is \(h = 8\)?” becomes \(\cos(\theta – \alpha) = \text{constant}\), which has a principal value and a partner, just like any basic trig equation. Without the harmonic form you would have a sum of sine and cosine on one side, which a calculator cannot invert directly.
When is it not the right tool? The method needs both terms to have the same angle. It works for \(3\sin 2x + 4\cos 2x\) but not for \(3\sin x + 4\cos 2x\). It also cannot combine terms with different periods, such as the two main tidal components in places where the sun and moon pull at different rates; real tide prediction sums many such waves.[4] If the expression has \(\sin^2 x\) or \(\sin x\cos x\) terms, use double-angle identities first.
Setting it up
Write \(\theta = 30t\) (in degrees). Here are the quantities in the case.
| Quantity | Symbol | Value |
|---|---|---|
| Hours after midnight | \(t\) | \(0 \le t \le 12\) |
| Angle | \(\theta\) | \(30t\) degrees |
| Mean depth | – | 5 m |
| Coefficient of \(\sin\theta\) | \(a\) | 3 |
| Coefficient of \(\cos\theta\) | \(b\) | 4 |
| Amplitude | \(R\) | to be found |
| Phase angle | \(\alpha\) | to be found |
| Depth the ship needs | – | at least 8 m |
Our target is to write \(3\sin\theta + 4\cos\theta\) as \(R\cos(\theta – \alpha)\), then read off the high water and solve \(h = 8\).
Solving it
We first express a sum in harmonic form, then solve an equation with it, and then answer the captain.
Express \(3\sin x + 4\cos x\) in the form \(R\cos(x – \alpha)\), where \(R > 0\) and \(0° < \alpha < 90°\). Hence state its maximum and minimum values and where they occur for \(0° \le x \le 360°\).
\(3\sin x + 4\cos x = 5\cos(x – 36.87°)\), with maximum 5 at \(x = 36.9°\) and minimum −5 at \(x = 216.9°\).
Solve \(5\cos x – 12\sin x = 6.5\) for \(0° \le x \le 360°\).
\(x = 232.6°\) or \(x = 352.6°\). Check: \(5\cos 232.62° – 12\sin 232.62° = 6.50\).
Using the tide model, find the time and height of high water, the height of low water, and the times when the depth is exactly 8 m.
High water is 10 m at about 1:14 a.m. The depth falls to 8 m at 3:00 a.m. and rises back to 8 m at about 11:27 a.m. Each safe window lasts \(\frac{2 \times 53.13}{30} \approx 3.54\) hours, centred on high water.
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What the answer tells us
The harmonic form turns three loose numbers into a clear story. The water swings 5 m either side of a 5 m mean, so it ranges from 0 m to 10 m. High water comes 1.23 hours after midnight, not at midnight, because the \(3\sin\) term pushes the peak later than the \(4\cos\) term alone would. The ship can cross whenever \(\cos(30t – 36.87)° \ge 0.6\), which is a 3.54-hour window around each high tide: from about 11:27 p.m. the night before until 3:00 a.m., and again from about 11:27 a.m. until 3:00 p.m.
Reasonableness check. At \(t = 0\), \(h = 5 + 0 + 4 = 9\) m, and at \(t = 3\), \(h = 5 + 3\sin 90° + 4\cos 90° = 8\) m. The first agrees with high water being just after midnight; the second matches our solution exactly. The amplitude 5 also passes the test: it is bigger than both 3 and 4, and smaller than their sum, 7.
Sensitivity. If the ship were loaded more lightly and needed only 7 m, the equation would become \(\cos(30t – 36.87)° = 0.4\), with principal value 66.42°. The window would grow to \(\frac{2 \times 66.42}{30} \approx 4.43\) hours, almost an hour longer. Every metre of draught matters near the top of the tide, because the wave is flattest there. The chart below shows why ignoring the sine term is dangerous: the gold curve, \(5 + 4\cos 30t°\), peaks at only 9 m at midnight, while the true combined wave peaks at 10 m over an hour later.
The Mathematics Made Simple textbook covers compound angle formulas, double angles and all four harmonic forms with graphs. For practice choosing the right form, finding \(\alpha\) in the right quadrant and solving equations, the Mathematics Made Simple Practice Questions book has graded sets with full worked answers.
Master trigonometric identities and the harmonic form with both Mathematics Made Simple books for $16.Both books · $16
Practice questions
For each question, write out the expansion of the target form before comparing coefficients.
EasyQ1. Express \(\sqrt{3}\sin x + \cos x\) in the form \(R\cos(x – \alpha)\), where \(R > 0\) and \(0° < \alpha < 90°\). Show worked answer
\(R = \sqrt{1 + 3} = 2\). \(\tan\alpha = \sqrt{3}\), so \(\alpha = 60°\).
\(\sqrt{3}\sin x + \cos x = 2\cos(x – 60°)\).
EasyQ2. Find the maximum and minimum values of \(6\sin x + 8\cos x\), and the values of \(x\) in \(0° \le x \le 360°\) where they occur. Show worked answer
Maximum 10 when \(x = 36.9°\).
Minimum −10 when \(x = 36.87° + 180° = 216.9°\).
MediumQ3. Express \(2\sin x – \cos x\) in the form \(R\sin(x – \alpha)\), with \(0° < \alpha < 90°\), and hence solve \(2\sin x – \cos x = 1\) for \(0° \le x \le 360°\). Show worked answer
\(R = \sqrt{5}\), \(\alpha = \tan^{-1}\frac{1}{2} = 26.57°\).
\(\sin(x – 26.57°) = \frac{1}{\sqrt{5}}\), so \(x – 26.57° = 26.57°\) or \(153.43°\).
\(x = 53.1°\) or \(x = 180°\).
MediumQ4. Solve \(\sqrt{3}\sin x + \cos x = \sqrt{2}\) for \(0 \le x \le 2\pi\), giving exact answers. Show worked answer
\(2\sin\left(x + \frac{\pi}{6}\right) = \sqrt{2}\), so \(\sin\left(x + \frac{\pi}{6}\right) = \frac{\sqrt{2}}{2}\).
\(x + \frac{\pi}{6}\) lies in \(\left[\frac{\pi}{6}, \frac{13\pi}{6}\right]\): \(x + \frac{\pi}{6} = \frac{\pi}{4}\) or \(\frac{3\pi}{4}\).
\(x = \frac{\pi}{12}\) or \(x = \frac{7\pi}{12}\).
ExamQ5. (a) Express \(8\cos x – 15\sin x\) in the form \(R\cos(x + \alpha)\), where \(R > 0\) and \(0° < \alpha < 90°\). Give \(\alpha\) to 2 decimal places. [3] (b) Hence solve \(8\cos x – 15\sin x = 10\) for \(0° \le x \le 360°\). [3] (c) State the greatest value of \(\dfrac{1}{20 + 8\cos x – 15\sin x}\). [1] Show worked answer
(b) \(\cos(x + 61.93°) = \frac{10}{17}\) [M1]. Principal value \(53.97°\); range of \(x + 61.93°\) is \(61.93°\) to \(421.93°\), so \(x + 61.93° = 306.03°\) or \(413.97°\) [M1].
\(x = 244.1°\) or \(x = 352.0°\) [A1].
(c) Least denominator \(= 20 – 17 = 3\), so the greatest value is \(\frac{1}{3}\) [B1].
Frequently asked questions
Why is it called the harmonic form?
Because it describes a single pure wave, or harmonic, with one amplitude and one phase. Physicists and engineers use the same idea to describe sound, alternating current and tides, where several waves of the same frequency combine into one.
How do I choose between R cos(x − α) and R sin(x + α)?
Both are correct for \(a\sin x + b\cos x\) with \(a, b > 0\); they give the same \(R\) and different values of \(\alpha\). Use the form the question asks for. If you are free to choose, pick the one whose expansion matches the signs in your expression, so that \(\alpha\) comes out acute.
Is R always equal to √(a² + b²)?
Yes. Squaring \(R\cos\alpha = b\) and \(R\sin\alpha = a\) and adding gives \(R^2 = a^2 + b^2\), whichever form you use. Only \(\alpha\) depends on the form.
Can I use radians instead of degrees?
Yes. The method is identical; give \(\alpha\) in radians (for example \(\alpha = 0.6435\) instead of \(36.87°\)) and use \(\pi\) and \(2\pi\) when shifting the interval. Match whatever unit the question uses.
References
- Abramson, J. et al. (2021) Algebra and Trigonometry 2e. Houston: OpenStax. https://openstax.org/details/books/algebra-and-trigonometry-2e
- Cambridge International (2024) Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. https://www.cambridgeinternational.org/programmes-and-qualifications/cambridge-international-as-and-a-level-mathematics-9709/
- International Baccalaureate Organization (2019) Mathematics: analysis and approaches guide. Geneva: IBO.
- Abramson, J. et al. (2021) Precalculus 2e. Houston: OpenStax. https://openstax.org/details/books/precalculus-2e
