Trigonometric Graphs

Lesson

The cosecant function at a point $x$`x` is written as $\csc x$`c``s``c``x` and it is defined by $\csc x=\frac{1}{\sin x}$`c``s``c``x`=1`s``i``n``x`. Similarly, the secant function is defined by $\sec x=\frac{1}{\cos x}$`s``e``c``x`=1`c``o``s``x`. And, the cotangent function is defined by $\cot x=\frac{\cos x}{\sin x}$`c``o``t``x`=`c``o``s``x``s``i``n``x`. The graph of each function is drawn below.

Graph of $y=\csc x$y=cscx |

Graph of $y=\sec x$y=secx |

Graph of $y=\cot x$y=cotx |

All three of these reciprocal trigonometric functions have asymptotes. These occur at points where the relevant parent function $(\sin x$(`s``i``n``x` or $\cos x)$`c``o``s``x`) has value zero. For example, $\sec x=\frac{1}{\cos x}$`s``e``c``x`=1`c``o``s``x` is undefined at $x=\frac{\pi}{2}$`x`=π2 or at $x=\frac{3\pi}{2}$`x`=3π2, and so on, because at these points $\cos x=0$`c``o``s``x`=0. In addition, all three functions share the same periodicity as their parent functions.

Since $\csc x$`c``s``c``x` and $\sec x$`s``e``c``x` are reciprocals of the functions $\sin x$`s``i``n``x` and $\cos x$`c``o``s``x`, the reciprocal functions never attain values strictly between $y=-1$`y`=−1 and $y=1$`y`=1. So equations like $\csc x=\frac{1}{2}$`c``s``c``x`=12 have no solutions. This is not true for $\cot x$`c``o``t``x` which can attain any value.

At what values of $x$`x` is the function $y=\cot x$`y`=`c``o``t``x` undefined?

**Think:** The function is defined by $\cot x=\frac{\cos x}{\sin x}$`c``o``t``x`=`c``o``s``x``s``i``n``x`. It is undefined whenever the denominator is zero.

**Do:** The denominator is zero when $\sin x=0$`s``i``n``x`=0.

This occurs at $x=0,\pi,2\pi,...$`x`=0,π,2π,... and, to be complete, when $x=\pi n$`x`=π`n`, for all integer values of $n$`n`.

Consider the identity $\sec x=\frac{1}{\cos x}$`s``e``c``x`=1`c``o``s``x` and the table of values below.

$x$x |
$0$0 | $\frac{\pi}{4}$π4 | $\frac{\pi}{2}$π2 | $\frac{3\pi}{4}$3π4 | $\pi$π | $\frac{5\pi}{4}$5π4 | $\frac{3\pi}{2}$3π2 | $\frac{7\pi}{4}$7π4 | $2\pi$2π |
---|---|---|---|---|---|---|---|---|---|

$\cos x$cosx |
$1$1 | $\frac{1}{\sqrt{2}}$1√2 | $0$0 | $-\frac{1}{\sqrt{2}}$−1√2 | $-1$−1 | $-\frac{1}{\sqrt{2}}$−1√2 | $0$0 | $\frac{1}{\sqrt{2}}$1√2 | $1$1 |

For which values of $x$

`x`in the interval $\left[0,2\pi\right]$[0,2π] is $\sec x$`s``e``c``x`not defined?Write all $x$

`x`-values on the same line separated by commas.Complete the table of values:

$x$ `x`$0$0 $\frac{\pi}{4}$π4 $\frac{\pi}{2}$π2 $\frac{3\pi}{4}$3π4 $\pi$π $\frac{5\pi}{4}$5π4 $\frac{3\pi}{2}$3π2 $\frac{7\pi}{4}$7π4 $2\pi$2π $\sec x$ `s``e``c``x`$\editable{}$ $\editable{}$ undefined $\editable{}$ $\editable{}$ $\editable{}$ undefined $\editable{}$ $\editable{}$ What is the minimum positive value of $\sec x$

`s``e``c``x`?What is the maximum negative value of $\sec x$

`s``e``c``x`?Plot the graph of $y=\sec x$

`y`=`s``e``c``x`on the same set of axes as $y=\cos x$`y`=`c``o``s``x`.Loading Graph...

Consider the following polynomial, which has a root of multiplicity $3$3.

$P\left(x\right)=x^4-9x^3+30x^2-44x+24$`P`(`x`)=`x`4−9`x`3+30`x`2−44`x`+24.

Find the second derivative of $P\left(x\right)$

`P`(`x`).Solve $P''\left(x\right)=0$

`P`′′(`x`)=0.Find the root of multiplicity $3$3 of $P\left(x\right)=0$

`P`(`x`)=0, ensuring working is shown to justify your answer.Let $x=a$

`x`=`a`be the other root of $P\left(x\right)=0$`P`(`x`)=0. Find the value of $a$`a`.

Consider the graph of $y=\operatorname{cosec}x$`y`=`c``o``s``e``c``x` below.

Loading Graph...

When $x=\frac{\pi}{4}$

`x`=π4, $y=\sqrt{2}$`y`=√2.What is the next positive $x$

`x`-value for which $y=\sqrt{2}$`y`=√2?What is the period of the graph?

What is the smallest value of $x$

`x`greater than $2\pi$2π for which $y=\sqrt{2}$`y`=√2?What is the first $x$

`x`-value less than $0$0 for which $y=\sqrt{2}$`y`=√2?

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