# Reciprocal Trig Functions

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Reciprocal Trig Functions

Remember how we said that $\sin ^{-1}$, $\cos ^{-1}$ and $\tan ^{-1}$ aren’t quite the same as $\dfrac{1}{\textcolor{blue}{\sin}}$, $\dfrac{1}{\textcolor{limegreen}{\cos}}$ and $\dfrac{1}{\textcolor{red}{\tan}}$? Here’s why…

A Level

## Introducing $\mathbf{\cosec}$, $\mathbf{\sec}$ and $\mathbf{\cot}$

We have three definitions, now:

• Cosecant: $\textcolor{blue}{\cosec x} = \dfrac{1}{\textcolor{blue}{\sin x}}$
• Secant: $\textcolor{limegreen}{\sec x} = \dfrac{1}{\textcolor{limegreen}{\cos x}}$
• Cotangent: $\textcolor{orange}{\cot x} = \dfrac{1}{\textcolor{red}{\tan x}}$

We’ve given these three reciprocal functions a special name so we don’t get them confused with the inverse functions.

The easiest way to remember which is which, is to look at the third letter of each function, so, Cosecant $= \dfrac{1}{\textcolor{blue}{\sin}}$, Secant $= \dfrac{1}{\textcolor{limegreen}{\cos}}$ and Cotangent $= \dfrac{1}{\textcolor{red}{\tan}}$.

A Level

## In Graphical Form

Now, we’ll represent the graph of $\textcolor{blue}{y = \cosec x}$:

Notice how there are now asymptotes at $x = -180°, 0°, 180°$.

Well, $\textcolor{blue}{\cosec x} = \dfrac{1}{\sin x}$, so it will not have values where $\sin x = 0°$, i.e. where $x = -180°, 0°, 180°$.

Notice, also, how there are no values between $y = 1$ and $y = -1$. Think on that for a while…

Anyway, here’s the graph for $\textcolor{limegreen}{y = \sec x}$:

See how the asymptotes are now at $x = -270°, -90°, 90°, 270°$. Again, this is because we’ll find that $\cos x = 0$ at these points.

Again, there are no values between $y = 1$ and $y = -1$

To complete the set, here’s $\textcolor{orange}{y = \cot x}$:

This time, there are no undefined values for $y$, but we do have vertical asymptotes at $x = -180°, 0°, 180°$ again.

A Level

## Domains, Ranges and Key Values

You might want to reference from this table of ranges and domains:

and here’s some key points that you might need to know.

In case you were wondering, $(-1,1)$ means the value is between but not including $-1$ and $1$.

A Level

## Example Questions

$\sec x = 2$ means that $\cos x = \dfrac{1}{2}$, which has solutions at $x = 60°, 300°, 420°, 660°$.

$\cosec x = \dfrac{1}{\sin x}$ and $\sec x = \dfrac{1}{\cos x}$.

Since $\sin x$ and $\cos x$ have a minimum of $-1$ and maximum of $1$, we conclude that $\cosec x$ and $\sec x$ have a positive minimum of $\dfrac{1}{1} = 1$, and a negative maximum of $\dfrac{1}{-1} = -1$.

To have $\cosec x$ or $\sec x$ between $-1$ and $1$, we require $\sin x$ or $\cos x$ to be greater than $1$, or less than $-1$, which is not possible.

Rewrite this as

$\dfrac{\sqrt{3}}{\cos x} = \dfrac{1}{\sin x}$

We can rearrange to get

$\tan x = \dfrac{1}{\sqrt{3}}$

This gives

$x = \tan ^{-1}\dfrac{1}{\sqrt{3}} = \dfrac{-5\pi}{6}, \dfrac{\pi}{6}$

A Level

A Level

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