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# Definition of Inverse Reciprocal Trig Functions

## Inverse secant, cosecant, and cotangent functions.

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Definition of Inverse Reciprocal Trig Functions

So far you've had to deal with trig functions, reciprocal functions, and inverse functions. Now you'll start to see inverse reciprocal functions. For example, can you compute

sec123\begin{align*}\sec^{-1} \frac{2}{\sqrt{3}}\end{align*}

As it turns out, this can be readily computed.

### Inverse Reciprocal Trigonometric Functions

We already know that the cosecant function is the reciprocal of the sine function. This will be used to derive the reciprocal of the inverse sine function.

yx1xcsc11xcsc11x=sin1x=siny=cscy=y=sin1x\begin{align*}y & = \sin^{-1} x\\ x & = \sin y\\ \frac{1}{x} & = \csc y\\ \csc^{-1} \frac{1}{x} & = y\\ \csc^{-1} \frac{1}{x} & = \sin^{-1} x\end{align*}

Because cosecant and secant are inverses, sin11x=csc1x\begin{align*}\sin^{-1} \frac{1}{x} = \csc^{-1} x\end{align*} is also true.

The inverse reciprocal identity for cosine and secant can be proven by using the same process as above. However, remember that these inverse functions are defined by using restricted domains and the reciprocals of these inverses must be defined with the intervals of domain and range on which the definitions are valid.

sec11x=cos1xcos11x=sec1x\begin{align*}\sec^{-1} \frac{1}{x} = \cos^{-1} x \leftrightarrow \cos^{-1} \frac{1}{x} = \sec^{-1} x\end{align*}

Tangent and cotangent have a slightly different relationship. Recall that the graph of cotangent differs from tangent by a reflection over the y\begin{align*}y-\end{align*}axis and a shift of π2\begin{align*}\frac{\pi}{2}\end{align*}. As an equation, this can be written as cotx=tan(xπ2)\begin{align*}\cot x = -\tan \left ( x-\frac{\pi}{2} \right )\end{align*}. Taking the inverse of this function will show the inverse reciprocal relationship between arccotangent and arctangent.

yyxxtan1(x)π2+tan1(x)π2tan1x=cot1x=tan1(xπ2)=tan(yπ2)=tan(yπ2)=yπ2=y=y\begin{align*}y & = \cot^{-1} x\\ y & = -\tan^{-1} \left ( x-\frac{\pi}{2} \right )\\ x & = -\tan \left( y-\frac{\pi}{2} \right )\\ -x & = \tan \left ( y-\frac{\pi}{2} \right )\\ \tan^{-1} (-x) & = y-\frac{\pi}{2}\\ \frac{\pi}{2} + \tan^{-1} (-x) & = y\\ \frac{\pi}{2} - \tan^{-1} x & = y\end{align*}

Remember that tangent is an odd function, so that tan(x)=tan(x)\begin{align*}\tan(-x) = -\tan(x)\end{align*}. Because tangent is odd, its inverse is also odd. So, this tells us that cot1x=π2tan1x\begin{align*}\cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x\end{align*} and tan1x=π2cot1x\begin{align*}\tan^{-1} x =\frac{\pi}{2} - \cot^{-1} x\end{align*}. To graph arcsecant, arccosecant, and arccotangent in your calculator you will use these conversion identities: sec1x=cos11x,csc1x=sin11x,cot1x=π2tan1x\begin{align*}\sec^{-1} x = \cos^{-1} \frac{1}{x} , \csc^{-1} x = \sin^{-1} \frac{1}{x} , \cot^{-1} x = \frac{\pi}{2} -\tan^{-1} x\end{align*}. Note: It is also true that cot1x=tan11x\begin{align*}\cot^{-1} x = \tan^{-1} \frac{1}{x}\end{align*}.

#### sec−12√\begin{align*}\sec^{-1}\sqrt{2}\end{align*}

Use the inverse reciprocal property. sec1x=cos11xsec12=cos112\begin{align*}\sec^{-1} x = \cos^{-1} \frac{1}{x} \rightarrow \sec^{-1} \sqrt{2} = \cos^{-1} \frac{1}{\sqrt{2}}\end{align*}. Recall that 12=1222=22\begin{align*}\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}\end{align*}. So, sec12=cos122\begin{align*}\sec^{-1} \sqrt{2} = \cos^{-1} \frac{\sqrt{2}}{2}\end{align*}, and we know that cos122=π4\begin{align*}\cos^{-1} \frac{\sqrt{2}}{2} = \frac{\pi}{4}\end{align*}. Therefore, sec12=π4\begin{align*}\sec^{-1} \sqrt{2} = \frac{\pi}{4}\end{align*}.

#### Find the exact value of each expression within the restricted domain, without a calculator.

For each of these problems, first find the reciprocal and then determine the angle from that.

sec12\begin{align*}\sec^{-1} \sqrt{2} \end{align*}

\begin{align*}\sec^{-1} \sqrt{2} = \cos^{-1} \frac{\sqrt{2}}{2} \end{align*} From the unit circle, we know that the answer is \begin{align*}\frac{\pi}{4}\end{align*}.

\begin{align*}\cot^{-1} \left ( -\sqrt{3} \right )\end{align*}

\begin{align*}\cot^{-1} \left ( -\sqrt{3} \right ) = \frac{\pi}{2} - \tan^{-1} \left ( -\sqrt{3} \right )\end{align*} From the unit circle, the answer is \begin{align*}\frac{5\pi}{6}\end{align*}.

\begin{align*}\csc^{-1} \frac{2\sqrt{3}}{3} \end{align*}

\begin{align*}\csc^{-1} \frac{2\sqrt{3}}{3} = \sin^{-1} \frac{\sqrt{3}}{2} \end{align*} Within our interval, there are is one answer, \begin{align*}\frac{\pi}{3}\end{align*}.

#### Using technology, find the value in radian measure, of each of the following:

\begin{align*}\arcsin 0.6384\end{align*}

\begin{align*}\arccos (-0.8126)\end{align*}

\begin{align*}\arctan (-1.9249)\end{align*}

### Examples

#### Example 1

Earlier, you were asked to evaluate \begin{align*}\sec^{-1} \frac{2}{\sqrt{3}}\end{align*}

\begin{align*}\sec^{-1} x = \cos^{-1} \frac{1}{x}\end{align*}

Substituting in values for "x" gives:

\begin{align*}\sec^{-1} \frac{2}{\sqrt{3}} = \cos^{-1} \frac{1}{\frac{2}{\sqrt{3}}}\end{align*}

This can be rewritten as:

\begin{align*}\cos^{-1}\frac{\sqrt{3}}{2}\end{align*}

And

\begin{align*}\cos^{-1}\frac{\sqrt{3}}{2} = \frac{\pi}{6}\end{align*}

Therefore,

\begin{align*}\sec^{-1} \frac{2}{\sqrt{3}} = \frac{\pi}{6}\end{align*}

#### Example 2

Evaluate \begin{align*}\sec^{-1} (-2)\end{align*}

\begin{align*}\frac{2\pi}{3}\end{align*}

#### Example 3

Evaluate \begin{align*}\cot^{-1} (-1)\end{align*}

\begin{align*}-\frac{\pi}{4}\end{align*}

#### Example 4

Evaluate \begin{align*}\csc^{-1} \left ( \sqrt{2} \right )\end{align*}

\begin{align*}\frac{\pi}{4}\end{align*}

### Review

Using technology, find the value in radian measure, of each of the following.

1. \begin{align*}\sin^{-1}(.345)\end{align*}
2. \begin{align*}\cos^{-1}(.87)\end{align*}
3. \begin{align*}\csc^{-1}(4)\end{align*}
4. \begin{align*}\sec^{-1}(2.32)\end{align*}
5. \begin{align*}\cot^{-1}(5.2)\end{align*}

Find the exact value of each expression within the restricted domain, without a calculator.

1. \begin{align*}\sec^{-1}(\frac{2\sqrt{3}}{3})\end{align*}
2. \begin{align*}\csc^{-1}(1)\end{align*}
3. \begin{align*}\cot^{-1}(\sqrt{3})\end{align*}
4. \begin{align*}\csc^{-1}(2)\end{align*}
5. \begin{align*}\sec^{-1}(\sqrt{2})\end{align*}
6. \begin{align*}\cot^{-1}(1)\end{align*}
7. \begin{align*}\cos^{-1}(\frac{1}{2})\end{align*}
8. \begin{align*}\sec^{-1}(2)\end{align*}
9. \begin{align*}\cot^{-1}(\frac{\sqrt{3}}{3})\end{align*}
10. \begin{align*}\sin^{-1}(\frac{\sqrt{3}}{2})\end{align*}

To see the Review answers, open this PDF file and look for section 4.7.

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### Vocabulary Language: English

inverse function

Inverse functions are functions that 'undo' each other. Formally $f(x)$ and $g(x)$ are inverse functions if $f(g(x)) = g(f(x)) = x$.

Reciprocal Function

A reciprocal function is a function with the parent function $y=\frac{1}{x}$.

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