Lesson

Recall that 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 |

The sine and cosine functions vary continuously between $-1$−1 and $1$1, passing through zero twice in every period. When $\sin\left(x\right)=0$`s``i``n`(`x`)=0 we should have $\csc\left(x\right)=\frac{1}{0}$`c``s``c`(`x`)=10 and $\cot x=\frac{1}{0}$`c``o``t``x`=10 which are undefined. Similarly, when $\cos\left(x\right)=0$`c``o``s`(`x`)=0, the we would get that $\sec\left(x\right)=\frac{1}{0}$`s``e``c`(`x`)=10.

We say that the secant function has vertical asymptotes at the points where the cosine function is zero. That is, $\sec\left(x\right)$`s``e``c`(`x`) has asymptotes at $x=\frac{\pi}{2}+n\pi$`x`=π2+`n`π, or $x=90^\circ+180^\circ n$`x`=90°+180°`n`, where $n$`n` is an integer.

Similarly, $\csc\left(x\right)$`c``s``c`(`x`) and $\cot x$`c``o``t``x` have vertical asymptotes wherever $\sin\left(x\right)=0$`s``i``n`(`x`)=0. That is, at $x=n\pi$`x`=`n`π, or $x=180^\circ n$`x`=180°`n`, where $n$`n` is an integer.

*Comparing the location of the asymptotes of each reciprocal trigonometric function.*

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 $x=0^\circ,180^\circ,360^\circ,...$`x`=0°,180°,360°,... and, to be complete, when $x=180^\circ\times n$`x`=180°×`n`, for all integer values of $n$`n`.

As seen above, both tangent and cotangent have the range $-\infty`y`<∞ so that neither function has an absolute maximum or minimum.

It is also true that secant and cosecant can attain any large value by taking points close to the asymptotes. Therefore, for both of these functions there is also no maximum or minimum.

We can, however, talk about **local*** *maxima and minima as distinct from a global or absolute maximum or minimum.

Both sine and cosine have the range $-1\le y\le1$−1≤`y`≤1. The reciprocals of numbers in this range must be greater than or equal to $1$1 or less than or equal to $-1$−1. You could confirm by looking at the graphs that as $\sin\left(x\right)$`s``i``n`(`x`) approaches its maximum value of $1$1, its reciprocal $\csc\left(x\right)$`c``s``c`(`x`) must approach a **local*** *minimum of $1$1. The same fact is also true for the cosine and secant functions.

Similarly, as sine and cosine approach their minimum values, the reciprocal functions, cosecant and secant, must approach local maxima.

Shown below are the graphs of $\cos\left(x\right)$`c``o``s`(`x`) and $\sec\left(x\right)$`s``e``c`(`x`) illustrating this pattern.

*The graph of *$y=\cos\left(x\right)$`y`=`c``o``s`(`x`)* (in green) and *$y=\sec\left(x\right)$`y`=`s``e``c`(`x`)* (in blue)*.

As we have already noticed in terms of local maxima and minima, taking the reciprocal of a function inverts the relative size of the function values. That is, when we take the reciprocal of two numbers, the bigger number becomes the smaller and the smaller number becomes the bigger one. We could write the fact like this:

If $a>b$`a`>`b`, then $\frac{1}{a}<\frac{1}{b}$1`a`<1`b`.

This means that if, going left to right on the graph of a function, we move from a higher $y$`y`-value to a lower $y$`y`-value, the same movement on the graph of the reciprocal will be from a lower $y$`y`-value to a higher $y$`y`-value. The increasing trend on the first function's graph turns into a decreasing trend on the graph of the reciprocal. Likewise, a decreasing trend turns into an increasing one when you take the reciprocal of a function.

We can see this fact in the graphs of $y=\tan\left(x\right)$`y`=`t``a``n`(`x`) and $y=\cot\left(x\right)$`y`=`c``o``t`(`x`) below.

*Where *$y=\tan\left(x\right)$`y`=`t``a``n`(`x`)* (in green) is increasing, its reciprocal *$y=\cot\left(x\right)$`y`=`c``o``t`(`x`)* (in blue) is decreasing.*

The period of a function is the distance on the $x$`x`-axis between repeated parts of its graph. Since the cosecant, secant and cotangent functions are the reciprocals of functions that do repeat, then these reciprocal functions must also repeat. In fact, they will repeat at the same rate as the function to which they are the reciprocal.

Consider a function, $f(x)$`f`(`x`) that has a period of $\alpha$`α`. This means that $f(x)=f(x+\alpha)$`f`(`x`)=`f`(`x`+`α`) for all values of $x$`x`. and that $\alpha$`α` is the smallest positive value for which this fact is true.

So, it must be the case that $\frac{1}{f(x)}=\frac{1}{f(x+\alpha)}$1`f`(`x`)=1`f`(`x`+`α`). If we say that $g(x)$`g`(`x`) is the reciprocal function $\frac{1}{f(x)}$1`f`(`x`), then we have that $g(x)=g(x+\alpha)$`g`(`x`)=`g`(`x`+`α`). So, $g(x)$`g`(`x`) has the same period as $f(x)$`f`(`x`).

The period of reciprocal trigonometric functions

Trigonometric functions have the same period as their reciprocal functions.

Function | Reciprocal | Period (radians) | Period (degrees) |
---|---|---|---|

$\sin\left(x\right)$sin(x) |
$\csc\left(x\right)$csc(x) |
$2\pi$2π | $360^\circ$360° |

$\cos\left(x\right)$cos(x) |
$\sec\left(x\right)$sec(x) |
$2\pi$2π | $360^\circ$360° |

$\tan\left(x\right)$tan(x) |
$\cot\left(x\right)$cot(x) |
$\pi$π | $180^\circ$180° |

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^\circ$0° | $45^\circ$45° | $90^\circ$90° | $135^\circ$135° | $180^\circ$180° | $225^\circ$225° | $270^\circ$270° | $315^\circ$315° | $360^\circ$360° |
---|---|---|---|---|---|---|---|---|---|

$\cos x$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 $\left[0^\circ,360^\circ\right]$[0°,360°] 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^\circ$0° $45^\circ$45° $90^\circ$90° $135^\circ$135° $180^\circ$180° $225^\circ$225° $270^\circ$270° $315^\circ$315° $360^\circ$360° $\sec x$ `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 graph of $y=\operatorname{cosec}x$`y`=`c``o``s``e``c``x` below.

Loading Graph...

When $x=30^\circ$

`x`=30°, $y=2$`y`=2.What is the next positive $x$

`x`-value for which $y=2$`y`=2?What is the period of the graph?

What is the smallest value of $x$

`x`greater than $360^\circ$360° for which $y=2$`y`=2?What is the first $x$

`x`-value less than $0^\circ$0° for which $y=2$`y`=2?

What is the first negative value of $x$`x` for which $\cot x$`c``o``t``x` has an asymptote?

Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline.