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Year 12 Maths - Methods (Unit 3 & Unit 4) Circular (trigonometric) functions

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

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Theory

The graph of \(y=a\sin n(t\pm\varepsilon)\) or \(y=a\cos n(t\pm\varepsilon)\) is a sine or cosine wave stretched and slid: the amplitude is \(|a|\) (with a reflection when \(a<0\)), the period is \(\dfrac{2\pi}{n}\), and \(n(t\pm\varepsilon)\) produces a horizontal phase shift of \(\mp\varepsilon\). Reading these three features lets you mark the key points and sketch one or two cycles over a stated domain, working in radians.

The amplitude is \(|a|\): the distance from the central line (the \(t\)-axis) up to a peak. The curve oscillates between \(y=|a|\) and \(y=-|a|\). When \(a<0\) the graph is also reflected in the \(t\)-axis, so peaks and troughs swap — for example \(y=-3\sin t\) starts by going down.

The period is \(\dfrac{2\pi}{n}\): the horizontal length of one complete cycle. The factor \(n\) counts how many cycles fit into an interval of \(2\pi\), so a larger \(n\) squeezes the wave into a shorter period. For \(y=2\cos 2t\) the period is \(\dfrac{2\pi}{2}=\pi\).

Writing the argument in the factored form \(n(t\pm\varepsilon)\) reveals the phase (horizontal) shift. Replacing \(t\) by \(t-\varepsilon\) slides the basic graph \(\varepsilon\) units to the right; replacing \(t\) by \(t+\varepsilon\) slides it \(\varepsilon\) units to the left. The shift is \(\varepsilon\), not \(n\varepsilon\), because \(n\) has already been factored out.

Key idea. For \(y=a\sin n(t\pm\varepsilon)\) and \(y=a\cos n(t\pm\varepsilon)\): amplitude \(=|a|\), period \(=\dfrac{2\pi}{n}\), and the graph is shifted \(\mp\varepsilon\) horizontally (right for \(t-\varepsilon\), left for \(t+\varepsilon\)) and reflected if \(a<0\).
y = 3 sin 2tThe curve y equals 3 sin 2t for t from 0 to 2 pi. Amplitude is 3 with peaks at y=3 and troughs at y=-3, and the period is pi. t y 3 -3 y=3 sin 2t T = pi
\(y=3\sin 2t\): amplitude \(3\) (peak \(3\), trough \(-3\)) and period \(\dfrac{2\pi}{2}=\pi\)
Phase shift of y = sin(t - pi/2)Dashed curve y equals sin t and solid curve y equals sin (t minus pi over 2): the solid curve is y equals sin t translated pi over 2 to the right. t y y=sin t y=sin(t-pi/2) shifted pi/2 right
Replacing \(t\) by \(t-\dfrac{\pi}{2}\) slides \(y=\sin t\) right by \(\dfrac{\pi}{2}\)

Amplitude and period for \(y=a\sin n(t\pm\varepsilon)\) and \(y=a\cos n(t\pm\varepsilon)\):

\[\text{amplitude}=|a| \qquad \text{period}=\frac{2\pi}{n}\]
amplitude=|a|,period=2πn

The curve stays between \(-|a|\) and \(|a|\), so the range is:

\[-|a| \le y \le |a|\]
−|a|≤y≤|a|

The phase shift, read from the factored form (right for \(t-\varepsilon\), left for \(t+\varepsilon\)):

\[y=a\sin n(t-\varepsilon)\ \Rightarrow\ \text{shift right }\varepsilon \qquad y=a\sin n(t+\varepsilon)\ \Rightarrow\ \text{shift left }\varepsilon\]
y=asinn(t−ε)⇒shift right ε
Quarter-period rule. One cycle has length \(\dfrac{2\pi}{n}\); dividing it into four equal steps of \(\dfrac{\pi}{2n}\) locates the maximum, minimum and the two central-line crossings — the five key points you plot.

How to sketch \(y=a\sin n(t\pm\varepsilon)\) or \(y=a\cos n(t\pm\varepsilon)\)

  1. Amplitude and reflection. Read \(|a|\) for the peak height. If \(a<0\), the graph is reflected in the \(t\)-axis, so it starts by moving the opposite way.
  2. Period. Compute \(\dfrac{2\pi}{n}\); this is the length of one cycle. A quarter-period is \(\dfrac{\pi}{2n}\).
  3. Phase shift. With the argument in the form \(n(t\pm\varepsilon)\), slide the graph \(\varepsilon\) right for \(t-\varepsilon\) or \(\varepsilon\) left for \(t+\varepsilon\). Factor out \(n\) first if the bracket is expanded.
  4. Key points. Starting from the shifted position, step across in quarter-periods and mark the maximum, central crossings and minimum.
  5. Draw and restrict. Join the points with a smooth wave and show only the cycles inside the stated domain.
Shortcut. A cosine graph starts at a maximum (or minimum if \(a<0\)) and a sine graph starts on the central line — use that starting point, then place the other four key points a quarter-period apart.
Example 1 — amplitude and period
State the amplitude and period of \(y=3\sin t\), and give its key points over \(0\le t\le 2\pi\).
Solution
Read \(a\) and \(n\) — here \(a=3\), \(n=1\):
amplitude\(=\)\(|a|=3\)
period\(=\)\(\dfrac{2\pi}{1}=2\pi\)
Sine starts on the central line; step a quarter-period \(\dfrac{\pi}{2}\) each time:
key points\(=\)\((0,0),\ \left(\dfrac{\pi}{2},3\right),\ (\pi,0),\ \left(\dfrac{3\pi}{2},-3\right),\ (2\pi,0)\)
\(\therefore\) amplitude \(=3\), period \(=2\pi\)
amplitude=3,period=2π
Example 2 — a cosine with \(n>1\)
Find the amplitude and period of \(y=2\cos 2t\), and its key points over one cycle.
Solution
Read \(a=2\), \(n=2\):
amplitude\(=\)\(|2|=2\)
period\(=\)\(\dfrac{2\pi}{2}=\pi\)
Cosine starts at a maximum; quarter-period \(=\dfrac{\pi}{4}\):
key points\(=\)\((0,2),\ \left(\dfrac{\pi}{4},0\right),\ \left(\dfrac{\pi}{2},-2\right),\ \left(\dfrac{3\pi}{4},0\right),\ (\pi,2)\)
\(\therefore\) amplitude \(=2\), period \(=\pi\)
period=2π2=π
Example 3 — amplitude, period and shift
Sketch \(y=2\cos 2\!\left(t-\dfrac{\pi}{4}\right)\) for \(0\le t\le\pi\), showing the key points.
Solution
Read the three features from the factored form:
amplitude\(=\)\(|2|=2\)
period\(=\)\(\dfrac{2\pi}{2}=\pi\)
shift\(=\)\(\dfrac{\pi}{4}\text{ right}\)
The maximum of \(\cos\) is slid to \(t=\dfrac{\pi}{4}\); step \(\dfrac{\pi}{4}\) across \(0\le t\le\pi\):
key points\(=\)\((0,0),\ \left(\dfrac{\pi}{4},2\right),\ \left(\dfrac{\pi}{2},0\right),\ \left(\dfrac{3\pi}{4},-2\right),\ (\pi,0)\)
\(\therefore\) amplitude \(2\), period \(\pi\), shifted \(\dfrac{\pi}{4}\) right
y = 2 cos 2(t - pi/4) over one cycleOne cycle of y equals 2 cos 2 (t minus pi over 4) for t from 0 to pi: value 0 at t=0, peak 2 at t=pi/4, 0 at pi/2, trough -2 at 3pi/4, 0 at pi. t y (pi/4, 2) (3pi/4,-2)
y=2cos2(t−π4)
Example 4 — reflection and left shift
For \(y=-3\sin 2\!\left(t+\dfrac{\pi}{6}\right)\), state the features and find the key points of one cycle.
Solution
Read the features — \(a=-3\) gives a reflection:
amplitude\(=\)\(|-3|=3\)
period\(=\)\(\dfrac{2\pi}{2}=\pi\)
shift\(=\)\(\dfrac{\pi}{6}\text{ left}\)
A cycle starts where \(2\!\left(t+\dfrac{\pi}{6}\right)=0\), i.e. \(t=-\dfrac{\pi}{6}\); step \(\dfrac{\pi}{4}\) each time. Because \(a<0\), the graph goes down first:
\(t=-\dfrac{\pi}{6}\)\(\to\)\(y=0\)
\(t=\dfrac{\pi}{12}\)\(\to\)\(y=-3\ (\text{min})\)
\(t=\dfrac{\pi}{3}\)\(\to\)\(y=0\)
\(t=\dfrac{7\pi}{12}\)\(\to\)\(y=3\ (\text{max})\)
\(t=\dfrac{5\pi}{6}\)\(\to\)\(y=0\)
\(\therefore\) amplitude \(3\) (reflected), period \(\pi\), shifted \(\dfrac{\pi}{6}\) left; min \(-3\) at \(t=\dfrac{\pi}{12}\), max \(3\) at \(t=\dfrac{7\pi}{12}\)
y=−3sin2(t+π6)

Common pitfalls

The period is \(\dfrac{2\pi}{n}\), not \(2\pi n\). A larger \(n\) makes the wave faster, so the cycle is shorter. For \(y=\sin 2t\) the period is \(\pi\), not \(4\pi\).
Subtracting shifts right, adding shifts left. \(y=a\sin n(t-\varepsilon)\) moves the graph \(\varepsilon\) to the right. The sign inside the bracket is the opposite of the direction people expect.
Factor out \(n\) before reading the shift. In \(y=\sin(2t-\dfrac{\pi}{3})\) the shift is not \(\dfrac{\pi}{3}\). Rewrite as \(\sin 2\!\left(t-\dfrac{\pi}{6}\right)\); the shift is \(\dfrac{\pi}{6}\) right.

Frequently asked questions

What is the amplitude of y = a sin nt or y = a cos nt?

The amplitude is \(|a|\), the distance from the central line to a peak, so the graph runs between \(y=|a|\) and \(y=-|a|\). For example \(y=3\sin t\) has amplitude \(3\).

How do you find the period of y = a sin nt or y = a cos nt?

The period is \(\dfrac{2\pi}{n}\); \(n\) is the number of full cycles in an interval of length \(2\pi\). For example \(y=2\cos 2t\) has period \(\dfrac{2\pi}{2}=\pi\).

Which way does the graph of y = a sin n(t - e) shift?

To the right by \(\varepsilon\). Replacing \(t\) with \(t-\varepsilon\) shifts the basic graph \(\varepsilon\) units in the positive direction, while \(t+\varepsilon\) shifts it \(\varepsilon\) units to the left. The shift is \(\varepsilon\), not \(n\varepsilon\).

What does a negative value of a do to the graph?

It reflects the graph in the \(t\)-axis, swapping peaks and troughs; the amplitude is still \(|a|\). So \(y=-3\sin t\) is the reflection of \(y=3\sin t\) and starts by going down.

Is the phase shift equal to e or to n times e?

It is \(\varepsilon\). Write the expression in the factored form \(n(t\pm\varepsilon)\) first, then read the shift as \(\varepsilon\). If a bracket is expanded, such as \(\sin(2t-\dfrac{\pi}{3})\), factor \(n\) out to \(\sin 2\!\left(t-\dfrac{\pi}{6}\right)\) before reading the shift.

How do you sketch one cycle of y = a cos n(t ± e)?

Find \(|a|\) and the period \(\dfrac{2\pi}{n}\), apply the shift, then split one period into four equal parts and mark the maximum, minimum and the two central crossings, joining them smoothly and reflecting if \(a<0\).

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