Reciprocal Trigonometric functions

When we take 1 divided by each of the original trigonometric functions we obtain the reciprocal trigonometric functions:  

\text{cosec}(x)=\frac{1}{\sin(x)},\hspace{30pt}\text{sec}(x)=\frac{1}{\cos(x)},\hspace{30pt}\text{cot}(x)=\frac{1}{\tan(x)}.

An easy way to remember which one is which is to look at the 3rd letter in each reciprocal function. For example, the third letter of cosec is s and so this is one over sin. We can solve equations involving reciprocal trigonometric functions in the same way we solve trigonometric equations. See Example 1. You might also notice that \text{cot}(x)=\frac{1}{\tan(x)}=\frac{\cos(x)}{\sin(x)}.

These reciprocal functions can be evaluated for angles given in degrees and radians. For example, \text{cosec}(270^\circ)=\frac{1}{\sin(270^\circ)}=\frac{1}{-1}=-1 or \text{sec}(2\pi)=\frac{1}{\cos(2\pi)}=\frac{1}{1}=1. However, like the graph of tan(x), all of the reciprocal trigonometric functions have periodic asymptotes. We cannot evaluate them when the denominator is 0. See graphs below.

Graphs of the Reciprocal Trigonometric Functions

Here are the graphs of the \text{cosec}(x), \text{sec}(x) and \text{cotan}(x) respectively. As you can see the angles are given in radians:

Consider the graph of \text{cosec}(x), for example. It is defined by \frac{1}{\sin(x)} and so has asymptotes whenever \sin(x)=0. As you can see from the graph, this occurs for multiples of \pi i.e, when x=n\pi, n\in{\mathbb Z}. As x tends towards these values, the curve grows infinitely positive if \sin(x) is positive and infinitely negative if sin(x) is negative. This asymptotic behaviour can be seen in the other two curves as well. 

Now consider the graph of \text{sec}(x). This graph has periodic maxima and minima at multiples of \pi. For example, there is a maximum at (\pi,-1) since \text{sec}(\pi)=\frac{1}{\cos(\pi)}=\frac{1}{-1}=-1. Notice that \cos(x) takes values between -1 and 1 and so \frac{1}{\cos(x)} will take values less than -1 or greater than 1. This follows from the fact that dividing by a number less than 1 gives a result greater than 1, etc. Hence, the domain and range for \text{sec}(x) are:

DOMAIN of \text{sec}(x):\hspace{30pt} \left\lbrace x\in {\mathbb R}: x\neq \frac{(2n+1)\pi}{2},n\in{\mathbb Z}\right\rbrace

RANGE of \text{sec}(x): \hspace{30pt}\left\lbrace y\in {\mathbb R}:y\leq-1,y\geq 1\right\rbrace

where we note that we cannot evaluate \cos(x) nor \text{sec}(x) at odd multiples of \frac{\pi}{2}. See more on domain and range. The domain of both \text{cosec(x)} and \text{cotan}(x) is \left\lbrace x\in {\mathbb R}: x\neq n\pi,n\in{\mathbb Z}\right\rbrace. The range of \text{cosec(x)} is the same as that of \text{sec(x)}. Since the range of \tan(x) is y\in{\mathbb R} with no exceptions, the range of \text{cotan}(x) is also y\in{\mathbb R}. See Example 2 for a transformation example.

Reciprocal Trigonometric Function Identities

There are two more trigonometric identities you should learn. See other trigonometric identities.  These two new identities are useful when solving certain trigonometric equations:

1+\tan^2(x)=\text{sec}^2(x)

1+\text{cot}^2(x)=\text{cosec}^2(x)

See Example 3. There is a quirky way to remember these identities. ‘The one with the tan is sexy’ and ‘the one with the coat is cosy’. They both come from the trigonometric identity \sin^2(x)+\cos^2(x)=1. Dividing both sides by \cos^2(x) gives the first identity and by \sin^2(x) gives the second. 

Examples

Solve 5\text{cosec}(2x)=10 for -2\pi<x<2\pi.

Solution:

By dividing both sides by 5 and writing  \text{cosec}(2x) as \frac{1}{\sin(2x)} we get the equation:

\frac{1}{\sin(2x)}=2

which can be rearranged to \sin(2x)=\frac{1}{2}. Let \theta=2x so that we are looking for solutions to \sin(\theta)=\frac{1}{2} for -4\pi<\theta<4\pi. See more on solving trigonometric equations. The \theta solutions are:

\theta=\frac{-23}{6}\pi,\frac{-19}{6}\pi,\frac{-11}{6}\pi, \frac{-7}{6}\pi,\frac{1}{6}\pi, \frac{5}{6}\pi, \frac{13}{6}\pi, \frac{17}{6}\pi

and so the x solutions are

x=\frac{-23}{12}\pi,\frac{-19}{12}\pi,\frac{-11}{12}\pi, \frac{-7}{12}\pi,\frac{1}{12}\pi, \frac{5}{12}\pi, \frac{13}{12}\pi, \frac{17}{12}\pi.

Sketch and state the domain and range of the graph of y=\text{cosec}(2x)+1.

Solution:

Recall from the transformations page that replacing x with 2x shrinks the original graph by half in the x-direction. Adding 1 translates the graph upwards by 1. Hence the graph of  y=\text{cosec}(2x)+1 is as follows:

reciprocal trigonometric functions

The domain of y=\text{cosec}(2x)+1 is \left\lbrace x\in {\mathbb R}: x\neq \frac{n\pi}{2},n\in{\mathbb Z}\right\rbrace. The range is \left\lbrace y\in {\mathbb R}:y\leq 0,y\geq 2\right\rbrace.

  1. Given that \tan(\theta)=-\frac{3}{4} and \theta is obtuse, find the value of \text{sec}(\theta).
  2. Show that \text{sec}^2(\theta)+\text{cosec}^2(\theta)\equiv\text{cosec}^2(\theta)\text{sec}^2(\theta)

Solution:

1. Using the identity 1+\tan^2(\theta)=\text{sec}^2(\theta) we have \text{sec}^2(\theta)=1+\left(-\frac{3}{4}\right)^2=\frac{25}{16}. Hence, \text{sec}(\theta)=\pm \frac{5}{4}. \cos(\theta) is negative when \theta is obtuse and so \text{sec}(\theta)=-\frac{5}{4}.

2. Using the definitions:

\begin{array}{lll}\text{sec}^2(\theta)+\text{cosec}^2(\theta)&\equiv&\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)}\\&\equiv&\frac{\sin^2(\theta)}{\sin^2(\theta)\cos^2(\theta)}+\frac{\cos^2(\theta)}{\sin^2(\theta)\cos^2(\theta)}\\&\equiv&\frac{\sin^2(\theta)+\cos^2(\theta)}{\sin^2(\theta)\cos^2(\theta)}\\&\equiv&\frac{1}{\sin^2(\theta)\cos^2(\theta)}\\&\equiv&\frac{1}{\sin^2(\theta)}\times\frac{1}{\cos^2(\theta)}\\&\equiv&\text{cosec}^2(\theta)\text{sec}^2(\theta)\end{array}

Extra Resources

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A2ReciprocalTrigExamQuestions

Open Reciprocal Trigonometric Functions in a New Window

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