Trig Functions and Graphs (radians)

Hong Kong

Stage 4 - Stage 5

Lesson

Consider the graphs of $y=\sin x$`y`=`s``i``n``x` and $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$`y`=−2`s``i``n`(3`x`+π4)+2 which are drawn below.

The graphs of $y=\sin x$y=sinx and $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4)+2 |

Starting with the graph of $y=\sin x$`y`=`s``i``n``x`, we can work through a series of transformations so that it coincides with the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$`y`=−2`s``i``n`(3`x`+π4)+2.

We can first reflect the graph of $y=\sin x$`y`=`s``i``n``x` about the $x$`x`-axis. This is represented by applying a negative sign to the function (multiplying the function by $-1$−1).

The graph of $y=-\sin x$y=−sinx |

Then we can increase the amplitude of the function to match. This is represented by multiplying the $y$`y`-value of every point on $y=-\sin x$`y`=−`s``i``n``x` by $2$2.

The graph of $y=-2\sin x$y=−2sinx |

Next we can apply the period change that is the result of multiplying the $x$`x`-value inside the function by $3$3. This means that to get a particular $y$`y`-value, we can put in an $x$`x`-value that is $3$3 times smaller than before. Notice that the points on the graph of $y=-2\sin x$`y`=−2`s``i``n``x` move towards the vertical axis by a factor of $3$3 as a result.

The graph of $y=-2\sin3x$y=−2sin3x |

Our next step will be to obtain the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)$`y`=−2`s``i``n`(3`x`+π4), and we can do so by applying a horizontal translation. In order to see what translation to apply, however, we first factorise the function into the form $y=-2\sin\left(3\left(x+\frac{\pi}{12}\right)\right)$`y`=−2`s``i``n`(3(`x`+π12)).

In this form, we can see that the $x$`x`-values are increased by $\frac{\pi}{12}$π12 inside the function. This means that to get a particular $y$`y`-value, we can put in an $x$`x`-value that is $\frac{\pi}{12}$π12 smaller than before. Graphically, this corresponds to shifting the entire function to the **left** by $\frac{\pi}{12}$π12 units.

The graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)$y=−2sin(3x+π4) |

Lastly, we translate the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)$`y`=−2`s``i``n`(3`x`+π4) upwards by $2$2 units, to obtain the final graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$`y`=−2`s``i``n`(3`x`+π4)+2.

The graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4)+2 |

Careful!

When we geometrically apply each transformation to the graph of $y=\sin x$`y`=`s``i``n``x`, it's important to consider the order of operations. If we had wanted to vertically translate the graph before reflecting about the $x$`x`-axis, we would have needed to translate the graph downwards first.

In the example above we were transforming the graph of $y=\sin x$`y`=`s``i``n``x`. The particular function $y=\sin x$`y`=`s``i``n``x` was not important, however. We could have just as easily transformed the graph of $y=\cos x$`y`=`c``o``s``x`, or even a non-trigonometric function, using the same method!

Consider a function $y=f\left(x\right)$`y`=`f`(`x`). Then we can obtain the graph of $y=af\left(b\left(x-c\right)\right)+d$`y`=`a``f`(`b`(`x`−`c`))+`d`, where $a,b,c,d$`a`,`b`,`c`,`d` are constants, by applying a series of transformations to the graph of $y=f\left(x\right)$`y`=`f`(`x`). These transformations are summarised below.

Summary

To obtain the graph of $y=af\left(b\left(x-c\right)\right)+d$`y`=`a``f`(`b`(`x`−`c`))+`d` from the graph of $y=f\left(x\right)$`y`=`f`(`x`):

- $a$
`a`vertically dilates the graph of $y=f\left(x\right)$`y`=`f`(`x`). - $b$
`b`horizontally dilates the graph of $y=f\left(x\right)$`y`=`f`(`x`). - $c$
`c`horizontally translates the graph of $y=f\left(x\right)$`y`=`f`(`x`). - $d$
`d`vertically translates the graph of $y=f\left(x\right)$`y`=`f`(`x`).

In the case that $a$`a` is negative, it has the additional property of reflecting the graph of $y=f\left(x\right)$`y`=`f`(`x`) about the horizontal axis.

If $y=f\left(x\right)$`y`=`f`(`x`) is the equation of a trigonometric function, then a vertical dilation corresponds to an amplitude change, a horizontal dilation corresponds to a period change and a horizontal translation corresponds to a phase shift.

The signs of $c$`c` and $d$`d` determine the direction of the horizontal and vertical translations respectively. If $c$`c` is positive the transformation describes a translation to the right, and if $c$`c` is negative the transformation describes a translation to the left. If $d$`d` is positive the transformation describes a translation upwards, and if $d$`d` is negative the transformation describes a translation downwards.

Careful!

If $c$`c` is negative, it may be convenient to represent the equation in the form $y=af\left(b\left(x+c\right)\right)+d$`y`=`a``f`(`b`(`x`+`c`))+`d` instead, where we've redefined $c$`c` using its absolute value. In this case, the value of $c$`c` represents translation to the left.

Similarly, if $d$`d` is negative, it may be convenient to represent the equation in the form $y=af\left(b\left(x-c\right)\right)-d$`y`=`a``f`(`b`(`x`−`c`))−`d`, where we've redefined $d$`d` using its absolute value. In this case, the value of $d$`d` represents translation downwards.

Lastly, the magnitude of $a$`a` and $b$`b` determine whether the vertical and horizontal dilations each describe a compression or an expansion.

For a value of $a$`a` where $\left|a\right|>1$|`a`|>1, the graph of $y=f\left(x\right)$`y`=`f`(`x`) vertically expands or stretches. For a trigonometric function, we say that the amplitude increases. If $\left|a\right|<1$|`a`|<1, the graph of $y=f\left(x\right)$`y`=`f`(`x`) vertically compresses. For a trigonometric function, we say that the amplitude decreases.

For a value of $b$`b` where $\left|b\right|>1$|`b`|>1, the graph of $y=f\left(x\right)$`y`=`f`(`x`) horizontally compresses. If $\left|b\right|<1$|`b`|<1, then the graph horizontally expands or stretches. In the case that the graph describes a trigonometric function, a horizontal compression means the period decreases and a horizontal expansion means the period increases.

Try experimenting with the value of each of these variables in the applet below!

Consider the function $y=\sin x+4$`y`=`s``i``n``x`+4.

Complete the table of values.

$x$ `x`$0$0 $\frac{\pi}{2}$π2 $\pi$π $\frac{3\pi}{2}$3π2 $2\pi$2π $y$ `y`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Graph the function.

Loading Graph...What transformation of the graph of $y=\sin x$

`y`=`s``i``n``x`results in the graph of $y=\sin x+4$`y`=`s``i``n``x`+4?Horizontal translation $4$4 units to the right.

AHorizontal translation $4$4 units to the left.

BVertical translation $4$4 units down.

CVertical translation $4$4 units up.

DWhat is the maximum value of $y=\sin x+4$

`y`=`s``i``n``x`+4?What is the minimum value of $y=\sin x+4$

`y`=`s``i``n``x`+4?

Which combinations of transformations could be used to turn the graph of $y=\cos x$`y`=`c``o``s``x` into the graph of $y=-\cos x+3$`y`=−`c``o``s``x`+3?

Select the two correct options.

Reflection about the $x$

`x`-axis, then translation $3$3 units down.AReflection about the $x$

`x`-axis, then translation $3$3 units up.BTranslation $3$3 units up, then reflection about the $x$

`x`-axis.CTranslation $3$3 units down, then reflection about the $x$

`x`-axis.D

Consider the given graph of $y=\sin x$`y`=`s``i``n``x`.

Loading Graph...

How can we transform the graph of $y=\sin x$

`y`=`s``i``n``x`to create the graph of $y=\sin\left(x-\frac{\pi}{2}\right)+3$`y`=`s``i``n`(`x`−π2)+3?Move the graph to the left by $\frac{\pi}{2}$π2 radians and up by $3$3 units.

AMove the graph to the right by $\frac{\pi}{2}$π2 radians and up by $3$3 units.

BMove the graph to the right by $\frac{\pi}{2}$π2 radians and down by $3$3 units.

CMove the graph to the left by $\frac{\pi}{2}$π2 radians and down by $3$3 units.

DHence graph $y=\sin\left(x-\frac{\pi}{2}\right)+3$

`y`=`s``i``n`(`x`−π2)+3 on the same set of axes.Loading Graph...What is the period of the curve $y=\sin\left(x-\frac{\pi}{2}\right)+3$

`y`=`s``i``n`(`x`−π2)+3 in radians?