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Year 12 Maths Advanced (2027) Functions

Transformations of Trigonometric Functions

20 practice questions 0 video lessons Theory + worked examples
NSW · Year 12 Mathematics Advanced · Further graph transformations and modelling

Transformations of trigonometric functions covers stretching, shifting and reflecting the graphs of \(y=\sin x\), \(y=\cos x\) and \(y=\tan x\), and reading amplitude, period, phase shift and vertical shift from \(y=a\sin\!\big(b(x+c)\big)+d\).

Part of the NSW Year 12 Mathematics Advanced course, in the Functions area of study (Further graph transformations and modelling focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.

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Theory

Transformations of trigonometric functions reflect, translate and dilate the graphs of \(\sin\), \(\cos\) and \(\tan\). This Year 12 Mathematics Advanced topic (MAV-12-01) shows how to read the amplitude, period, phase shift, vertical shift and range from \(y=a\sin\!\big(b(x-c)\big)+d\), and how to find the equation of a transformed sine or cosine graph. Angles are in radians.

A transformation of a trigonometric function changes its graph by a reflection, a translation or a dilation, while the graph stays periodic (wave-like). Every transformed sine or cosine can be written as \(y=a\sin\!\big(b(x-c)\big)+d\), and each constant controls one feature.

The amplitude \(|a|\) is the vertical dilation — half the distance from the lowest to the highest point. The number \(b\) is a horizontal dilation that sets the period (the length of one full cycle). The constant \(c\) is the horizontal (phase) shift and \(d\) is the vertical shift that moves the centre line to \(y=d\).

A negative value of \(a\) reflects the graph in the \(x\)-axis. The same ideas apply to \(\cos\) and \(\tan\), except that \(\tan\) has no amplitude and a different period.

Sine graph before and after transformation The graph of y = sin x in grey and y = 3 sin 2x in navy, showing the amplitude stretched to 3 and the period compressed from 2 pi to pi. xy 3 -3 π period = π y = 3 sin 2x y = sin x
Amplitude and period: \(y=\sin x\) becomes \(y=3\sin 2x\).
Vertical dilation and translation of a sine graph The graph of y = 2 sin x + 1 oscillates about the midline y = 1 between a minimum of -1 and a maximum of 3, so its range is -1 to 3. xy y=1 3 -1 y = 2 sin x + 1
Vertical shift: \(y=2\sin x+1\) oscillates about \(y=1\), range \([-1,3]\).

For \(y=a\sin\!\big(b(x-c)\big)+d\) with \(b>0\) (and likewise for \(\cos\)):

\[\text{amplitude}=|a|\]
amplitude=|a|
\[\text{period}=\dfrac{2\pi}{b}\quad(\sin,\cos),\qquad \text{period}=\dfrac{\pi}{b}\quad(\tan)\]
period=2πb
\[\text{range}=[\,d-|a|,\ d+|a|\,]\]
range=[d-|a|,d+|a|]
Factor first. To read the phase shift from a form like \(y=\sin(2x-\pi)\), factor the coefficient of \(x\) out first: \(y=\sin\!\big(2(x-\dfrac{\pi}{2})\big)\), so the shift is \(\dfrac{\pi}{2}\) to the right.

How to analyse a transformed trig function

  1. Write it in the form \(y=a\sin\!\big(b(x-c)\big)+d\) — factor the coefficient of \(x\) out of the bracket.
  2. Read the constants \(a\), \(b\), \(c\) and \(d\).
  3. Apply the rules: amplitude \(=|a|\), period \(=\dfrac{2\pi}{b}\), phase shift \(=c\), vertical shift \(=d\), and range \(=[\,d-|a|,\,d+|a|\,]\).
Example 1 — Amplitude & period
State the amplitude and period of \(y=4\sin 3x\).
Solution
\(\text{amplitude}\)\(=\)\(|4|=4\)
\(\text{period}\)\(=\)\(\dfrac{2\pi}{b}=\dfrac{2\pi}{3}\)
period=2π3
Graph of y = 4 sin 3xThree full cycles of a sine wave of amplitude 4 between x = 0 and x = 2 pi. 4 -4 x

Amplitude \(4\), period \(\dfrac{2\pi}{3}\).

Example 2 — Range after shifts
Find the range of \(y=2\cos x-1\).
Solution

Amplitude \(|a|=2\), vertical shift \(d=-1\).

\(\text{max}\)\(=\)\(d+|a|=-1+2=1\)
\(\text{min}\)\(=\)\(d-|a|=-1-2=-3\)
range=[-3,1]

Range \(-3\le y\le 1\).

Example 3 — Describe the transformations
Describe the transformations taking \(y=\sin x\) to \(y=-\sin\!\big(2(x-\dfrac{\pi}{3})\big)+1\), and state the period.
Solution

Read \(a=-1,\ b=2,\ c=\dfrac{\pi}{3},\ d=1\):

• reflection in the \(x\)-axis (\(a<0\));
• horizontal dilation, factor \(\dfrac{1}{2}\) (\(b=2\));
• translation \(\dfrac{\pi}{3}\) right and \(1\) up.

\(\text{period}\)\(=\)\(\dfrac{2\pi}{2}=\pi\)
Example 4 — Find the equation
The graph below is \(y=a\cos bx\). Find \(a\) and \(b\).
Solution
Graph of y = a cos bx to identifyA cosine curve starting at a maximum of 3, reaching a minimum of -3 and returning, completing one cycle at x = pi. 3 -3 x π/2 π

The height is \(3\), so \(|a|=3\); it starts at a maximum, so \(a=3\). One cycle ends at \(x=\pi\):

\(\pi\)\(=\)\(\dfrac{2\pi}{b}\)
\(b\)\(=\)\(2\)
y=3cos2x

So \(y=3\cos 2x\).

Common pitfalls

Multiplying instead of dividing by \(b\). The period is \(\dfrac{2\pi}{b}\), not \(2\pi b\). A larger \(b\) makes the wave faster, so the period gets shorter.
Treating the amplitude as negative. For \(y=-4\sin x\) the amplitude is \(4\). The minus sign is a reflection in the \(x\)-axis, not a negative amplitude.
Reading the phase shift before factoring. In \(y=\sin(2x-\pi)\) the shift is not \(\pi\). Factor first: \(y=\sin\!\big(2(x-\dfrac{\pi}{2})\big)\), so the shift is \(\dfrac{\pi}{2}\) right.
Using \(2\pi\) for the tangent period. The period of \(y=\tan bx\) is \(\dfrac{\pi}{b}\), and \(\tan\) has no amplitude.

Frequently asked questions

What is the amplitude of a trigonometric function?

The amplitude is how far the graph rises above or falls below its centre line. For y equals a sin(bx) or a cos(bx) the amplitude is the absolute value of a. For example y equals 4 sin 3x has amplitude 4. The amplitude is always positive; a negative value of a just reflects the graph in the x-axis.

How do you find the period of y = sin(bx)?

Divide 2 pi by b. The period of y equals sin(bx) or y equals cos(bx) is 2 pi over b, so a larger b gives a shorter period (a faster wave). For example y equals 4 sin 3x has period 2 pi over 3.

What does the b value do to a sine or cosine graph?

The b value is a horizontal dilation. It squeezes or stretches the graph sideways and sets the period, which is 2 pi over b. A bigger b makes the wave repeat more often; a smaller b stretches it out.

How do you find the equation of a sine or cosine graph?

Read the amplitude off the graph (half the distance between the highest and lowest points) to get a. Measure the length of one full cycle to get the period, then find b from period equals 2 pi over b. Choose sine or cosine to match the starting point, and add any vertical shift d for the centre line.

What is the period of tan x?

The period of y equals tan x is pi, not 2 pi. For y equals tan(bx) the period is pi over b. Unlike sine and cosine, the tangent function has no amplitude because it increases without bound.

What is the range of y = a sin x + d?

The graph oscillates about the line y equals d, reaching up to d plus the absolute value of a and down to d minus the absolute value of a. So the range is from d minus mod a to d plus mod a. For example y equals 2 cos x minus 1 has range from -3 to 1.