Resources For Teachers For Tutors For Students & Parents Pricing
Create Free Account
Year 12 Maths - Methods (Unit 3 & Unit 4) Circular (trigonometric) functions

Sketch graphs y=a sin n(t+-e)+-b and y=a cos n(t+-e)+-b

20 practice questions 0 video lessons Theory + worked examples
Practice 20 questions
Practice questions

Every question with a fully worked solution.

Start practising
Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

A sinusoidal graph of the form \(y=a\sin n(t-e)+b\) (or with \(\cos\)) is a transformed sine wave: \(|a|\) is the amplitude, \(\dfrac{2\pi}{n}\) is the period, \(e\) is the horizontal shift, and \(b\) is the vertical translation. The constant \(b\) fixes the mean line \(y=b\), about which the curve oscillates between \(b-|a|\) and \(b+|a|\). All angles are in radians.

Every graph \(y=a\sin n(t-e)+b\) and \(y=a\cos n(t-e)+b\) is the basic sine or cosine wave stretched, shifted and translated. The amplitude \(|a|\) is the vertical distance from the mean line to a maximum (or minimum); it is half the distance between the highest and lowest points. A negative \(a\) also reflects the curve in the mean line.

The coefficient \(n\) controls the period \(\dfrac{2\pi}{n}\) — the horizontal length of one complete cycle. A bigger \(n\) squeezes the wave into shorter cycles; for example \(n=2\) halves the period to \(\pi\). The number \(e\) is the horizontal (phase) shift: in the factored form \(n(t-e)\), a subtraction \(t-e\) slides the graph \(e\) units right, while \(t+e\) slides it \(e\) units left.

Finally \(b\) is a vertical translation that lifts (or lowers) the whole wave so the centre of the oscillation — the mean line — is \(y=b\). The range is therefore \([\,b-|a|,\ b+|a|\,]\), with maximum \(b+|a|\) and minimum \(b-|a|\) placed symmetrically about \(y=b\).

Key idea. For \(y=a\sin n(t-e)+b\): amplitude \(=|a|\), period \(=\dfrac{2\pi}{n}\), mean line \(y=b\), and range \([\,b-|a|,\ b+|a|\,]\). Sketch the mean line first, then build the wave around it.
Transformed sine y=2 sin t + 3One period of y equals 2 sin t plus 3 over t from 0 to 2 pi; the green dashed mean line is y equals 3, the maximum is 5 and the minimum is 1, so the amplitude is 2 and the range is 1 to 5. t y max 5 min 1 y=3 |a|=2
\(y=2\sin t+3\): mean line \(y=3\), amplitude \(2\), so the range is \([1,5]\)
Transformed cosine y=2 cos 2t + 1y equals 2 cos 2t plus 1; over one period from t equals 0 to t equals pi the maximum is 3 and the minimum is minus 1, the mean line is y equals 1, and the period is pi because n equals 2. t y y=1 period = π max 3 min -1
\(y=2\cos 2t+1\): \(n=2\) gives period \(\pi\); mean line \(y=1\), range \([-1,3]\)

The general transformed sine and cosine, written in factored form so the shift is read off directly:

\[y=a\sin n(t-e)+b \qquad y=a\cos n(t-e)+b\]
y=asinn(t-e)+b

The four features read straight from the equation (with \(n>0\)):

\[\text{amplitude}=|a| \qquad \text{period}=\frac{2\pi}{n} \qquad \text{mean line: } y=b\]
period=2πn

The range sits symmetrically about the mean line:

\[\text{range}=[\,b-|a|,\ b+|a|\,] \qquad \text{max}=b+|a|,\quad \text{min}=b-|a|\]
range=[b-|a|,b+|a|]
Reading the shift. The subtraction \(t-e\) shifts the graph \(e\) to the right; \(t+e\) shifts it \(e\) to the left. If the equation is given as \(a\sin(nt-c)\), first factor: \(n\!\left(t-\dfrac{c}{n}\right)\), so the shift is \(\dfrac{c}{n}\), not \(c\).

How to sketch \(y=a\sin n(t-e)+b\)

  1. Read off \(a,\ n,\ e,\ b\). If needed, factor the argument into the form \(n(t-e)\) so the shift \(e\) is visible.
  2. Amplitude and mean line. The amplitude is \(|a|\); draw the mean line \(y=b\) as a dashed guide, with the maximum \(b+|a|\) and minimum \(b-|a|\).
  3. Period. Compute the period \(\dfrac{2\pi}{n}\) and divide one cycle into four equal quarters.
  4. Apply the shift. Start the cycle at \(t=e\): a sine curve leaves the mean line rising, a cosine curve starts at a maximum (or a minimum if \(a<0\)).
  5. Plot and join. Mark the five key points across one period and draw a smooth wave; then state the range \([\,b-|a|,\ b+|a|\,]\).
Five-point tip. Over one sine cycle the key heights are mean, max, mean, min, mean; for cosine they are max, mean, min, mean, max. Spacing them a quarter-period apart makes the sketch quick and accurate.
Example 1 — amplitude, period, range
For \(y=3\sin t+2\), state the amplitude, period, mean line and range.
Solution
Read off \(a=3\), \(n=1\), \(b=2\):
amplitude\(=\)\(|a|=3\)
period\(=\)\(\dfrac{2\pi}{1}=2\pi\)
mean line\(:\)\(y=2\)
Range \(=[\,b-|a|,\ b+|a|\,]\):
range\(=\)\([\,2-3,\ 2+3\,]\)
\(=\)\([-1,5]\)
\(\therefore\) amplitude \(3\), period \(2\pi\), mean \(y=2\), range \([-1,5]\)
[-1,5]
Example 2 — a cosine with \(n\ne 1\)
For \(y=2\cos 3t-1\), find the amplitude, period, mean line and range.
Solution
Read off \(a=2\), \(n=3\), \(b=-1\):
amplitude\(=\)\(|2|=2\)
period\(=\)\(\dfrac{2\pi}{3}\)
mean line\(:\)\(y=-1\)
Range about the mean line \(y=-1\):
range\(=\)\([\,-1-2,\ -1+2\,]\)
\(=\)\([-3,1]\)
\(\therefore\) amplitude \(2\), period \(\dfrac{2\pi}{3}\), mean \(y=-1\), range \([-3,1]\)
period=2π3
Example 3 — full transformation
For \(y=4\sin 2\!\left(t-\dfrac{\pi}{6}\right)+1\), state the amplitude, period, horizontal shift, mean line and range.
Solution
Read off \(a=4\), \(n=2\), \(e=\dfrac{\pi}{6}\), \(b=1\):
amplitude\(=\)\(|4|=4\)
period\(=\)\(\dfrac{2\pi}{2}=\pi\)
shift\(:\)\(\dfrac{\pi}{6}\text{ right}\)
Mean line and range about it:
mean line\(:\)\(y=1\)
range\(=\)\([\,1-4,\ 1+4\,]=[-3,5]\)
\(\therefore\) amplitude \(4\), period \(\pi\), shift \(\dfrac{\pi}{6}\) right, mean \(y=1\), range \([-3,5]\)
[-3,5]
Example 4 — sketch a complete cycle
Sketch one complete cycle of \(y=2\sin\!\left(t-\dfrac{\pi}{3}\right)+1\), showing the mean line and key points.
Solution
Read off the features — \(a=2\), \(n=1\), \(e=\dfrac{\pi}{3}\), \(b=1\):
amplitude\(=\)\(2\)
period\(=\)\(2\pi\)
mean line\(:\)\(y=1\)
The cycle starts at \(t=e=\dfrac{\pi}{3}\) (sine leaves the mean rising). Quarter-period \(=\dfrac{\pi}{2}\), so the five key points are:
\(t=\dfrac{\pi}{3}\)\(:\)\(y=1\) (mean)
\(t=\dfrac{5\pi}{6}\)\(:\)\(y=3\) (max)
\(t=\dfrac{4\pi}{3}\)\(:\)\(y=1\) (mean)
\(t=\dfrac{11\pi}{6}\)\(:\)\(y=-1\) (min)
\(t=\dfrac{7\pi}{3}\)\(:\)\(y=1\) (mean)
\(\therefore\) range \([-1,3]\); the cycle runs from \(t=\dfrac{\pi}{3}\) to \(t=\dfrac{7\pi}{3}\)
One cycle of y=2 sin(t - pi/3) + 1One complete cycle of y equals 2 sin of (t minus pi on 3) plus 1; the mean line is y equals 1, the curve rises through the mean, reaches a maximum of 3, returns to the mean, falls to a minimum of minus 1, and returns to the mean after one period of 2 pi. t y y=1 max 3 min -1
y=2sin(t-π3)+1

Common pitfalls

Factor before reading the shift. For \(y=\sin(2t-\pi)\) the shift is not \(\pi\). Factor first: \(\sin 2\!\left(t-\dfrac{\pi}{2}\right)\), so the graph shifts \(\dfrac{\pi}{2}\) to the right, not \(\pi\).
Period is \(\dfrac{2\pi}{n}\), not \(2\pi n\). A larger \(n\) makes the cycle shorter. For \(y=2\cos 3t-1\) the period is \(\dfrac{2\pi}{3}\), well under \(2\pi\).
The mean line is \(y=b\), not \(y=0\). After a vertical translation the max and min are \(b+|a|\) and \(b-|a|\). For \(y=3\sin t+2\) the maximum is \(5\) and the minimum is \(-1\), never \(\pm 3\).

Frequently asked questions

What is the amplitude of y = a sin n(t - e) + b?

The amplitude is \(|a|\), the distance from the mean line up to a maximum or down to a minimum — half the distance between the highest and lowest points. For example \(y=3\sin t+2\) has amplitude \(3\).

How do you find the period of y = a sin nt + b?

The period is \(\dfrac{2\pi}{n}\), where \(n\) is the coefficient of \(t\). A larger \(n\) gives shorter cycles: \(y=2\cos 3t-1\) has period \(\dfrac{2\pi}{3}\), and \(y=2\cos 2t+1\) has period \(\pi\).

What does b do to a sine or cosine graph?

\(b\) is a vertical translation, so the mean line becomes \(y=b\) instead of \(y=0\) and the range becomes \([\,b-|a|,\ b+|a|\,]\).

Which way does the graph shift for t - e compared with t + e?

In the factored form \(a\sin n(t-e)+b\), the subtraction \(t-e\) shifts the graph \(e\) units right, while \(t+e\) shifts it \(e\) units left. Here \(e\) is the horizontal (phase) shift.

What is the range of y = a sin n(t - e) + b?

The range is \([\,b-|a|,\ b+|a|\,]\), with maximum \(b+|a|\) and minimum \(b-|a|\) placed symmetrically about the mean line \(y=b\).

How do you sketch one cycle of a transformed sine graph?

Draw the mean line \(y=b\), find the amplitude \(|a|\) and period \(\dfrac{2\pi}{n}\), divide the period into quarters, apply the shift \(e\), then plot the five key points (mean, max, mean, min, mean for sine) and join them smoothly.

More in Circular (trigonometric) functions
Further symmetry properties and the Pythagorean identity
20 questions
Graphs of sine and cosine
20 questions
Sketch graphs y=a sin n(t+-e) and y=a cos n(t+-e)
20 questions
Determining rules for graphs of trig functions
20 questions
The tangent function
20 questions