This note covers the graphs of sine, cosine and tangent, and how to stretch, shift and reflect them to model repeating patterns such as tides, sound, daylight hours and rotating wheels. You will learn to read amplitude, period and midline from an equation and to build an equation from a description.
Key ideas
- The graphs of \(y = \sin x\) and \(y = \cos x\) are periodic: they repeat every \(2\pi\) (that is, every \(360^\circ\)). Both have domain all real numbers and range \(-1 \leq y \leq 1\).
- The graph of \(y = \cos x\) is the graph of \(y = \sin x\) shifted left by \(\frac{\pi}{2}\). Both look like the same wave starting at different places.
- The graph of \(y = \sin x\) passes through the origin. The graph of \(y = \cos x\) has a maximum at \(x = 0\).
- The graph of \(y = \tan x\) repeats every \(\pi\), has range all real numbers, and has vertical asymptotes at \(x = \frac{\pi}{2} + k\pi\) for whole numbers \(k\).
- Both \(\sin\) and \(\cos\) are given a general form: \(y = a\sin(b(x - c)) + d\) and \(y = a\cos(b(x - c)) + d\).
- In the general form, \(|a|\) is the amplitude, the distance from the midline to a maximum. The period is \(\frac{2\pi}{b}\) (for \(b \gt 0\)). The value \(c\) is the horizontal shift and \(d\) is the vertical shift, so the midline is \(y = d\).
- The maximum value is \(d + |a|\) and the minimum value is \(d - |a|\).
Transformations. Starting from \(y = \sin x\):
- \(a\) is a vertical stretch with scale factor \(|a|\). If \(a \lt 0\), the graph is also reflected in the \(x\)-axis.
- \(b\) is a horizontal stretch with scale factor \(\frac{1}{b}\). A bigger \(b\) squeezes the graph and makes the period shorter.
- \(c\) translates the graph to the right by \(c\) (to the left if \(c\) is negative).
- \(d\) translates the graph up by \(d\) (down if \(d\) is negative).
When you read a graph to find an equation, find the midline and amplitude from the highest and lowest points: \(d = \frac{\max + \min}{2}\) and \(a = \frac{\max - \min}{2}\). Then find the period to get \(b = \frac{2\pi}{\text{period}}\). Finally choose sine or cosine, and use a clear point such as a maximum to fix the shift. A model that starts at its minimum is \(-\cos\) with no shift.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Period | The length of one full cycle. For \(\sin(bx)\) and \(\cos(bx)\) it is \(\frac{2\pi}{b}\). For \(\tan(bx)\) it is \(\frac{\pi}{b}\). |
| Amplitude \(|a|\) | Half the distance from minimum to maximum. |
| Midline \(y = d\) | The horizontal line halfway between the maximum and minimum. |
| \(y = a\sin(b(x - c)) + d\) | General sine model. |
| \(y = a\cos(b(x - c)) + d\) | General cosine model. |
| \(\cos x = \sin\left(x + \frac{\pi}{2}\right)\) | Links the two graphs through a horizontal shift. |
| Asymptotes of \(\tan x\) | \(x = \frac{\pi}{2} + k\pi\). The function is undefined there. |
Worked examples
Example 1 For the function \(f(x) = 3\cos\left(2\left(x - \dfrac{\pi}{6}\right)\right) + 1\), find the amplitude, period, midline, range, and the position of the first maximum for \(x \geq 0\).
Compare with \(a\cos(b(x - c)) + d\): \(a = 3\), \(b = 2\), \(c = \frac{\pi}{6}\), \(d = 1\).
Amplitude: \(3\). Period: \(\frac{2\pi}{2} = \pi\). Midline: \(y = 1\).
Maximum value \(= 1 + 3 = 4\) and minimum value \(= 1 - 3 = -2\), so the range is \(-2 \leq f(x) \leq 4\).
A cosine graph has a maximum when its bracket is zero, which happens at \(x = c = \frac{\pi}{6}\).
Answer: amplitude 3, period \(\pi\), midline \(y = 1\), range \([-2, 4]\), first maximum at \(x = \frac{\pi}{6}\).
Example 2 A Ferris wheel has its lowest point 2 m above the ground and its highest point 32 m above the ground. It completes one revolution every 40 seconds. A passenger boards at the lowest point at time \(t = 0\). Find a model \(h(t)\) for the height in metres, and use it to find the height at \(t = 12\) and the first time the height is 25 m.
Midline: \(d = \dfrac{32 + 2}{2} = 17\). Amplitude: \(\dfrac{32 - 2}{2} = 15\). Period 40, so \(b = \dfrac{2\pi}{40} = \dfrac{\pi}{20}\).
The passenger starts at the minimum, so use a reflected cosine: \(h(t) = 17 - 15\cos\left(\dfrac{\pi t}{20}\right)\).
At \(t = 12\): \(h = 17 - 15\cos(0.6\pi) = 17 - 15(-0.30902) \approx 21.6\) m (GDC in radian mode).
For \(h = 25\): \(17 - 15\cos\left(\frac{\pi t}{20}\right) = 25\), so \(\cos\left(\frac{\pi t}{20}\right) = -\frac{8}{15}\).
Use the inverse cosine on your GDC: \(\frac{\pi t}{20} = \cos^{-1}\left(-\frac{8}{15}\right) \approx 2.133\), so \(t \approx 13.6\) s. (The graph of \(h\) first reaches 25 on its way up, which is the first solution.)
Answer: \(h(t) = 17 - 15\cos\left(\frac{\pi t}{20}\right)\); \(h(12) \approx 21.6\) m; first reaches 25 m at \(t \approx 13.6\) s.
Example 3 Describe a sequence of transformations that maps \(y = \sin x\) onto \(y = 2\sin(3x) + 1\), and state the period and range of the new function.
Here \(a = 2\), \(b = 3\), \(c = 0\), \(d = 1\).
Step 1: a horizontal stretch with scale factor \(\frac{1}{3}\) gives \(y = \sin(3x)\).
Step 2: a vertical stretch with scale factor 2 gives \(y = 2\sin(3x)\).
Step 3: a translation up by 1 unit, \(\begin{pmatrix} 0 \\ 1 \end{pmatrix}\), gives \(y = 2\sin(3x) + 1\).
Period \(= \frac{2\pi}{3}\). The maximum is \(1 + 2 = 3\) and the minimum is \(1 - 2 = -1\).
Answer: period \(\frac{2\pi}{3}\); range \(-1 \leq y \leq 3\).
Common errors
- Reading the period as \(b\) or \(\frac{b}{2\pi}\). The period of \(\sin(bx)\) is \(\frac{2\pi}{b}\).
- Reading the shift wrongly when the bracket is not factorised. The graph of \(y = \sin(2x - \pi)\) is shifted right by \(\frac{\pi}{2}\), not \(\pi\), because \(2x - \pi = 2\left(x - \frac{\pi}{2}\right)\).
- Using \(\frac{\pi}{2}\) for the period of tangent or \(2\pi\) in place of \(\pi\).
- Giving a negative amplitude. Amplitude is a distance, so it is \(|a|\), and a negative \(a\) means a reflection.
- Using degree mode on the GDC for a model written with \(\pi\), which makes the values look nonsense.
Exam tip: Write down the maximum, minimum and period first. Then calculate \(a\), \(d\) and \(b\) separately. Check your model by substituting one given point.
Check yourself
- State the amplitude, period, midline and range of \(y = -4\sin\left(\dfrac{x}{2}\right) + 3\).
Answer
Amplitude 4. Here \(b = \frac{1}{2}\), so the period is \(\frac{2\pi}{1/2} = 4\pi\). Midline \(y = 3\). Range \(-1 \leq y \leq 7\). - For \(y = \tan(2x)\) with \(0 \lt x \lt \pi\), state the period and the equations of the vertical asymptotes.
Answer
Period \(\frac{\pi}{2}\). Asymptotes occur when \(2x = \frac{\pi}{2}\) or \(\frac{3\pi}{2}\), so \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\). - A cosine function has maximum value 5, minimum value \(-1\) and period 6, and has a maximum at \(x = 0\). Write its equation.
Answer
\(d = \frac{5 + (-1)}{2} = 2\), \(a = \frac{5 - (-1)}{2} = 3\), \(b = \frac{2\pi}{6} = \frac{\pi}{3}\). So \(y = 3\cos\left(\frac{\pi x}{3}\right) + 2\). - Write the equation of the graph of \(y = \sin x\) after a shift of \(\frac{\pi}{4}\) to the left and 2 units down.
Answer
A shift left by \(\frac{\pi}{4}\) replaces \(x\) with \(x + \frac{\pi}{4}\). So \(y = \sin\left(x + \frac{\pi}{4}\right) - 2\).