Trigonometric Graphs

NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Amplitude of sine and cosine

Lesson

The graphs of trigonometric functions like $y=\sin x$`y`=`s``i``n``x` and $y=\cos x$`y`=`c``o``s``x` have sections that are repeated in the $x$`x`-direction. The length of these repeated sections is called the wavelength, and we can understand a lot about the behaviour of the whole function by looking only at what happens within a single wavelength.

Typically, each trigonometric graph will have a minimum value, a maximum value, and average value along the wavelength. The distance between the maximum (or minimum) value and the average value is called the amplitude.

If the range of the function is given by $\text{maximum value - minimum value }$maximum value - minimum value , the amplitude is then $\frac{\text{max - min }}{2}$max - min 2. Notice that the amplitude is always a positive number.

An equation of the form $y=a\sin x$`y`=`a``s``i``n``x` has an amplitude of $a$`a` units. If we start with the equation $y=\sin x$`y`=`s``i``n``x`, where $a=1$`a`=1, we can transform this equation by changing the amplitude.

Graphically this transformation corresponds to **stretching** the graph of $\sin x$`s``i``n``x` in the vertical direction when $a>1$`a`>1, and **compressing** the graph of $\sin x$`s``i``n``x` in the vertical direction when $00<`a`<1. We can also **reflect** the graph across the $x$`x`-axis by multiplying the whole function by $-1$−1.

This action of stretching and compressing the graph of a function is known as dilation. For trigonometric graphs like $\sin x$`s``i``n``x` and $\cos x$`c``o``s``x`, a vertical dilation by a factor of $a$`a` is equivalent to increasing the amplitude by a factor of $a$`a`.

State the amplitude of the function $f\left(x\right)=5\sin x$`f`(`x`)=5`s``i``n``x`.

**Think**: When we compare $f\left(x\right)$`f`(`x`) to the standard function $\sin x$`s``i``n``x` we can see that all the function values of $5\sin x$5`s``i``n``x` will be five times larger than all the function values of $\sin x$`s``i``n``x`. This means that the amplitude of $5\sin x$5`s``i``n``x` is also five times larger than the amplitude of $\sin x$`s``i``n``x`.

**Do**: The amplitude of $\sin x$`s``i``n``x` is $1$1, so the amplitude of $f\left(x\right)=5\sin x$`f`(`x`)=5`s``i``n``x` is $5\times1=5$5×1=5.

**Reflect**: We can obtain the graph of $f\left(x\right)=5\sin x$`f`(`x`)=5`s``i``n``x` by starting with the graph of $y=\sin x$`y`=`s``i``n``x` and applying a vertical dilation by a factor of $5$5.

The graph of $y=\cos x$`y`=`c``o``s``x` is reflected across the $x$`x`-axis, then compressed in the vertical direction so that its minimum value is $\frac{3}{4}$34. What is the equation of the resulting function. What is the amplitude?

**Think**: Recall that a reflection across the $x$`x`-axis corresponds to multiplying the function by $-1$−1. This will "switch" the location of the maximum and minimum values of the graph, but the amplitude will still be a positive value.

**Do**: Let's keep track of how the equation of the graph changes at each stage of the transformation.

$y=\cos x$y=cosx |
$\rightarrow$→ | $y=-\cos x$y=−cosx |
(Reflection across $x$x-axis) |

$y=-\cos x$y=−cosx |
$\rightarrow$→ | $y=-\frac{3}{4}\cos x$y=−34cosx |
(Vertical compression) |

The final equation of the resulting graph is $y=-\frac{3}{4}\cos x$`y`=−34`c``o``s``x`. The amplitude of this equation is $\frac{3}{4}$34.

**Reflect**: Compare the resulting equation with the original equation. The only difference is the constant multiple of $-\frac{3}{4}$−34. Notice that although this number is negative, the amplitude of the resulting equation is **positive**. In general, the amplitude of the equation $y=a\cos x$`y`=`a``c``o``s``x` is $\left|a\right|$|`a`|.

Consider the graph of the equation $y=\sin x$`y`=`s``i``n``x`. If this graph was vertically dilated by a factor of $9$9 from the $x$`x`-axis, what would be the equation of the resulting graph?

$y=9\sin x$

`y`=9`s``i``n``x`A$y=\sin\frac{1}{9}x$

`y`=`s``i``n`19`x`B$y=\sin9x$

`y`=`s``i``n`9`x`C$y=\frac{1}{9}\sin x$

`y`=19`s``i``n``x`D$y=9\sin x$

`y`=9`s``i``n``x`A$y=\sin\frac{1}{9}x$

`y`=`s``i``n`19`x`B$y=\sin9x$

`y`=`s``i``n`9`x`C$y=\frac{1}{9}\sin x$

`y`=19`s``i``n``x`D

Determine the equation of the graphed function given that it is of the form $y=a\sin x$`y`=`a``s``i``n``x` or $y=a\cos x$`y`=`a``c``o``s``x`.

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Consider the function $y=-3\cos x$`y`=−3`c``o``s``x`.

What is the maximum value of the function?

What is the minimum value of the function?

What is the amplitude of the function?

Select the two transformations that are required to turn the graph of $y=\cos x$

`y`=`c``o``s``x`into the graph of $y=-3\cos x$`y`=−3`c``o``s``x`.Vertical dilation.

AHorizontal translation.

BReflection about the $x$

`x`-axis.CVertical translation.

DVertical dilation.

AHorizontal translation.

BReflection about the $x$

`x`-axis.CVertical translation.

D

Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions