NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Domain and Range of Inverse Trigonometric Functions

Lesson

This chapter should be read in conjunction with the chapter on Key Features of Inverse Trigonometric Functions.

In order to define an inverse for any function, we need to identify a part of its domain over which the function is either decreasing or increasing and hence, one-to-one.

The sine function is increasing on the domain $-\frac{\pi}{2}\le x\le\frac{\pi}{2}$−π2≤`x`≤π2 and over this domain the function takes its full range of values, from $-1$−1 to $1$1.

We can, therefore, define an inverse sine function, $\arcsin y$`a``r``c``s``i``n``y`, that maps the domain$-1\le y\le1$−1≤`y`≤1 to the range $-\frac{\pi}{2}\le x\le\frac{\pi}{2}$−π2≤`x`≤π2.

In practice, we might take a number between $-1$−1 and $1$1 and, using a calculator, apply the $\sin^{-1}$`s``i``n`−1 function to it. The calculator will return a number between $-\frac{\pi}{2}$−π2 and $\frac{\pi}{2}$π2 if it is set in radians, or an angle between $-90^\circ$−90° and $90^\circ$90° if set in degrees.

The cosine function is one-to-one if its domain is restricted to the interval $0\le x\le\pi$0≤`x`≤π, over which it is decreasing. Over this domain, the cosine function has the full range of values from $-1$−1 to $1$1.

We can define an inverse cosine function, $\arccos y$`a``r``c``c``o``s``y`, with domain $-1\le y\le1$−1≤`y`≤1 and range $0\le x\le\pi$0≤`x`≤π.

If a calculator is used to find the inverse cosine of a number between $-1$−1 and $1$1, it returns a number in the range $[0,\pi]$[0,π] if the calculator is set to radians, or an angle in the range $[0,180]^\circ$[0,180]° if set to degrees.

The tangent function is increasing on the restricted domain $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2) and its range on this domain is the set of real numbers.

The inverse tangent function, $\arctan y$`a``r``c``t``a``n``y` is defined for every real number, and its range is the domain of the tangent function.

A calculator will return a valid answer for $\tan^{-1}(x)$`t``a``n`−1(`x`) for every number $x$`x`. The result will be strictly between $-\frac{\pi}{2}$−π2 and $\frac{\pi}{2}$π2. in radians or between $-90^\circ$−90° and $90^\circ$90° if the calculator is set in degrees.

Find the smallest positive angle (in degrees) whose tangent is $-23.5$−23.5.

Using the $\tan^{-1}$`t``a``n`−1 function on a calculator set in degrees, we obtain the answer $-87.56^\circ$−87.56°.

We know that the tangent function, on its unrestricted domain, is periodic with period $180^\circ$180°. Therefore, we will obtain another solution if one period is added to the solution given by the calculator. This is $92.44^\circ$92.44° and it must be the smallest positive angle whose tangent is $-23.5$−23.5.

The equation $\frac{\sqrt{5}}{3}=\sin y$√53=`s``i``n``y` has infinitely many solutions. Find the three of them that are closest to zero. Express the results in both radians and degrees.

It must be that $y=\arcsin\frac{\sqrt{5}}{3}$`y`=`a``r``c``s``i``n`√53. Setting the calculator to radians, we find that $y=0.841$`y`=0.841. This is less that $\frac{\pi}{2}$π2 and is, therefore, a 'first quadrant' solution. another solution must be $\pi-0.841=2.301$π−0.841=2.301. We can find further solutions by adding or subtracting multiples of the period of the sine function. The third nearest to zero must be $2.301-2\pi=-3.983$2.301−2π=−3.983.

If we do the same calculator steps but with the calculator set in degrees, we obtain $48.2^\circ$48.2°, then $180-48.2=131.8^\circ$180−48.2=131.8°, and finally, $131.8-360=-228^\circ$131.8−360=−228°.

Consider the inverse sine, inverse cosine, and inverse tangent functions.

Which of the following is the domain of $\sin^{-1}x$

`s``i``n`−1`x`?$\left(-1,1\right)$(−1,1)

A$\left[-1,1\right]$[−1,1]

B$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2)

C$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[−π2,π2]

D$\left(-1,1\right)$(−1,1)

A$\left[-1,1\right]$[−1,1]

B$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2)

C$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[−π2,π2]

DWhich of the following is the domain of $\cos^{-1}x$

`c``o``s`−1`x`?$\left(0,\pi\right)$(0,π)

A$\left[0,\pi\right]$[0,π]

B$\left[-1,1\right]$[−1,1]

C$\left(-1,1\right)$(−1,1)

D$\left(0,\pi\right)$(0,π)

A$\left[0,\pi\right]$[0,π]

B$\left[-1,1\right]$[−1,1]

C$\left(-1,1\right)$(−1,1)

DWhich of the following is the domain of $\tan^{-1}x$

`t``a``n`−1`x`?$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[−π2,π2]

A$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2)

B$\left[-\infty,\infty\right]$[−∞,∞]

C$\left(-\infty,\infty\right)$(−∞,∞)

D$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$[−π2,π2]

A$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2)

B$\left[-\infty,\infty\right]$[−∞,∞]

C$\left(-\infty,\infty\right)$(−∞,∞)

D

The graph of $y=\sin^{-1}x$`y`=`s``i``n`−1`x` has been plotted below.

Loading Graph...

How would the graph of $y=\sin^{-1}\left(\frac{x}{5}\right)$

`y`=`s``i``n`−1(`x`5) compare to that of $y=\sin^{-1}x$`y`=`s``i``n`−1`x`?Its domain will span a larger interval than the range of $y=\sin^{-1}x$

`y`=`s``i``n`−1`x`.AIts domain will span a smaller interval than the domain of $y=\sin^{-1}x$

`y`=`s``i``n`−1`x`.BIts range will span a larger interval than the range of $y=\sin^{-1}x$

`y`=`s``i``n`−1`x`.CIts range will span a smaller interval than the range of $y=\sin^{-1}x$

`y`=`s``i``n`−1`x`.DIts domain will span a larger interval than the range of $y=\sin^{-1}x$

`y`=`s``i``n`−1`x`.AIts domain will span a smaller interval than the domain of $y=\sin^{-1}x$

`y`=`s``i``n`−1`x`.BIts range will span a larger interval than the range of $y=\sin^{-1}x$

`y`=`s``i``n`−1`x`.CIts range will span a smaller interval than the range of $y=\sin^{-1}x$

`y`=`s``i``n`−1`x`.DHence, write the domain and range of $f(x)=\sin^{-1}\left(\frac{x}{5}\right)$

`f`(`x`)=`s``i``n`−1(`x`5).Domain: $\left[\editable{},\editable{}\right]$[,]

Range: $\left[\editable{},\editable{}\right]$[,]

Consider the function $f(x)=4\cos^{-1}x$`f`(`x`)=4`c``o``s`−1`x`.

State the range of $f(x)$

`f`(`x`). Give your answer using interval notation.State the domain of $f(x)$

`f`(`x`). Give your answer using interval notation.

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