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Year 11 Methods (Unit 1 & 2) Trigonometric Functions

Graphs Of Sine And Cosine

20 practice questions 1 video lesson Theory + worked examples

Understand the graphs of sine and cosine for Queensland Year 11 Mathematical Methods (QCAA). Both curves have amplitude one and period two pi, forming a smooth wave that repeats every full turn.

You will learn to plot the maximums, minimums and intercepts, see how a coefficient changes the amplitude, and how a constant shifts the curve up or down to change its range — the basis for modelling cycles in the QCAA course.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), the graphs of \(y=\sin t\) and \(y=\cos t\) are smooth waves that repeat every \(2\pi\) with values between \(-1\) and \(1\). This page shows their key features — amplitude, period, maximums, minimums and intercepts — and how a coefficient \(a\) or a vertical shift \(b\) changes them.

Both \(y=\sin t\) and \(y=\cos t\) are periodic with period \(2\pi\): the pattern repeats every full turn. Their amplitude is \(1\) (half the distance from the maximum \(1\) to the minimum \(-1\)), so the range is \([-1,\,1]\).

The key points occur at multiples of \(\dfrac{\pi}{2}\). The sine curve starts at \((0,0)\), peaks at \(\left(\dfrac{\pi}{2},1\right)\), and troughs at \(\left(\dfrac{3\pi}{2},-1\right)\); the cosine curve starts at its maximum \((0,1)\).

A coefficient scales the wave: \(y=a\sin t\) and \(y=a\cos t\) have amplitude \(|a|\). Adding a constant shifts it vertically: \(y=\sin t+b\) moves the whole curve up by \(b\), giving range \([b-|a|,\ b+|a|]\).

Amplitude sets the height, period sets the width. For \(y=a\sin t\) or \(y=a\cos t\) the amplitude is \(|a|\) and (with no horizontal scaling) the period stays \(2\pi\).
Graph of y = sin tOne cycle of y equals sine t from 0 to two pi: it starts at 0, peaks at 1 at pi over two, returns to 0 at pi, troughs at minus 1 at three pi over two, and ends at 0. x y π/2 π 3π/2
\(y=\sin t\): starts at \(0\), peak at \(\dfrac{\pi}{2}\), trough at \(\dfrac{3\pi}{2}\), period \(2\pi\).
Graph of y = cos tOne cycle of y equals cosine t from 0 to two pi: it starts at 1, falls to 0 at pi over two, troughs at minus 1 at pi, returns to 0 at three pi over two, and ends at 1. x y π/2 π 3π/2
\(y=\cos t\): starts at its maximum \(1\), trough at \(\pi\), period \(2\pi\).

For the basic curves \(y=\sin t\) and \(y=\cos t\):

\[\text{amplitude}=1,\qquad \text{period}=2\pi,\qquad -1\le y\le 1\]
period=2π

For \(y=a\sin t\) or \(y=a\cos t\):

\[\text{amplitude}=|a|,\qquad \text{range}=[-|a|,\ |a|]\]
amplitude=|a|

For a vertical shift \(y=a\sin t+b\) (or with cosine):

\[\text{range}=[\,b-|a|,\ b+|a|\,],\qquad \text{midline } y=b\]
range=[b-|a|,b+|a|]
From a graph: amplitude \(=\dfrac{\text{max}-\text{min}}{2}\) and the midline (vertical shift) \(b=\dfrac{\text{max}+\text{min}}{2}\).

How to read or sketch a sine/cosine graph

  1. Amplitude: read \(|a|\) from the coefficient, or compute \(\dfrac{\text{max}-\text{min}}{2}\) from a graph.
  2. Midline and range: the vertical shift \(b\) is the midline; the range is \([b-|a|,\ b+|a|]\).
  3. Key points: mark the maximum, minimum and intercepts at multiples of \(\dfrac{\pi}{2}\) across one period \(2\pi\).
Example 1 — Features of y = sin t
State the amplitude, period, maximum and minimum of \(y=\sin t\).
Solution

Amplitude is half the distance from the maximum to the minimum:

\(\text{amplitude}\)\(=\)\(\dfrac{1-(-1)}{2}\)
\(=\)\(1\)

The curve repeats every full turn:

\(\text{period}\)\(=\)\(2\pi\)

The maximum is \(1\) at \(t=\dfrac{\pi}{2}\) and the minimum is \(-1\) at \(t=\dfrac{3\pi}{2}\).

Amplitude \(1\), period \(2\pi\); max \(1\) at \(\dfrac{\pi}{2}\), min \(-1\) at \(\dfrac{3\pi}{2}\).

y = sin t: amplitude 1, period 2piy equals sine t with maximum 1 at pi over two and minimum minus 1 at three pi over two. x y π/2 π 3π/2
amplitude=1,period=2π
Example 2 — A scaled cosine, y = 4 cos t
For \(y=4\cos t\), state the amplitude, period and range, and the coordinates of the maximum and minimum on \([0,\,2\pi]\).
Solution

The amplitude is \(|a|\) with \(a=4\):

\(\text{amplitude}\)\(=\)\(|4|\)
\(=\)\(4\)

There is no horizontal scaling, so the period is unchanged, and the range follows:

\(\text{period}\)\(=\)\(2\pi\)
\(\text{range}\)\(=\)\([-4,\ 4]\)

Cosine starts at its maximum: max \(4\) at \(t=0\) and \(t=2\pi\); min \(-4\) at \(t=\pi\).

Amplitude \(4\), period \(2\pi\), range \([-4,\,4]\); max \((0,4)\), min \((\pi,-4)\).

y = 4 cos t: amplitude 4y equals four cosine t with maximum 4 at 0 and two pi and minimum minus 4 at pi. x y π/2 π 3π/2
amplitude=4
Example 3 — A vertical shift, y = 2 sin t - 1
For \(y=2\sin t-1\), state the amplitude, the vertical shift and the range, and the maximum and minimum values.
Solution

Amplitude is \(|a|\) with \(a=2\); the constant \(b=-1\) is the vertical shift (midline):

\(\text{amplitude}\)\(=\)\(|2|=2\)
\(\text{midline}\)\(=\)\(y=-1\)

The range runs from \(b-|a|\) to \(b+|a|\):

\(\text{range}\)\(=\)\([\,-1-2,\ -1+2\,]\)
\(=\)\([-3,\ 1]\)

So the maximum value is \(1\) (at \(t=\dfrac{\pi}{2}\)) and the minimum is \(-3\) (at \(t=\dfrac{3\pi}{2}\)).

Amplitude \(2\), shift \(-1\), range \([-3,\,1]\); max \(1\), min \(-3\).

y = 2 sin t - 1: shifted down 1y equals two sine t minus one with midline at minus one, maximum 1 at pi over two and minimum minus 3 at three pi over two. x y π/2 π 3π/2
range=[-3,1]
Example 4 — Reading a rule off a graph
A cosine graph \(y=a\cos t+b\) has a maximum value of \(4\) and a minimum value of \(-2\). Find \(a\) and \(b\), and write the rule.
Solution

The amplitude is half the max-to-min distance:

\(a\)\(=\)\(\dfrac{\text{max}-\text{min}}{2}\)
\(=\)\(\dfrac{4-(-2)}{2}\)
\(=\)\(3\)

The vertical shift is the midline, halfway between them:

\(b\)\(=\)\(\dfrac{\text{max}+\text{min}}{2}\)
\(=\)\(\dfrac{4+(-2)}{2}\)
\(=\)\(1\)

A cosine starts at its maximum, so the rule is \(y=3\cos t+1\).

\(a=3,\ b=1\), so \(y=3\cos t+1\).

y = 3 cos t + 1 from its maximum and minimumA cosine graph with maximum 4 and minimum minus 2, midline 1; the amplitude is 3 and the vertical shift is 1. x y π/2 π 3π/2
y=3cost+1

Common pitfalls

Confusing amplitude with the maximum. For \(y=2\sin t-1\) the amplitude is \(2\), but the maximum value is \(1\) because of the shift; amplitude is half the max-to-min distance.
Changing the period when only \(a\) changes. \(y=4\cos t\) still has period \(2\pi\); only a coefficient of \(t\) inside the function would change the period.
Starting sine and cosine at the same point. \(y=\sin t\) starts at \((0,0)\), but \(y=\cos t\) starts at its maximum \((0,1)\).

Frequently asked questions

What is the period of y = sin t and y = cos t?

Both have period \(2\pi\): the graph repeats every full turn.

What is the amplitude of y = a sin t?

The amplitude is \(|a|\), the absolute value of the coefficient. For example \(y=4\cos t\) has amplitude \(4\).

How do the graphs of sine and cosine differ?

They are the same shape, but shifted: \(y=\sin t\) starts at \((0,0)\) rising, while \(y=\cos t\) starts at its maximum \((0,1)\).

How does adding a constant change the graph?

\(y=\sin t+b\) shifts the whole curve up by \(b\), moving the midline to \(y=b\) and the range to \([b-|a|,\ b+|a|]\).

How do you read amplitude and vertical shift off a graph?

Amplitude \(=\dfrac{\text{max}-\text{min}}{2}\) and the vertical shift \(b=\dfrac{\text{max}+\text{min}}{2}\) (the midline).