In this section we concern ourselves with finding inverses of the circular (trigonometric) functions. Our immediate problem is that, owing to their periodic nature, none of the six circular functions is one-to-one. To remedy this, we restrict the domains of the circular functions in the same way we restricted the domain of the quadratic function in Example inverserestrictionex in Section InverseFunctions to obtain a one-to-one function.

We start with \(f(t) = \sin (t)\) and restrict our domain to \(\left [ -\frac {\pi }{2}, \frac {\pi }{2}\right ]\) as seen below in order to keep the range as \([-1,1]\) as well as the properties of being smooth and continuous.

Recall from Section InverseFunctions that the inverse of a function \(f\) is typically denoted \(f^{-1}\). For this reason, some textbooks use the notation \(f^{-1}(t) = \sin ^{-1}(t)\) for the inverse of \(f(t) = \sin (t)\). The obvious pitfall here is our convention of writing \((\sin (t))^2\) as \(\sin ^{2}(t)\), \((\sin (t))^3\) as \(\sin ^{3}(t)\) and so on. It is far too easy to confuse \(\sin ^{-1}(t)\) with \(\frac {1}{\sin (t)} = \csc (t)\) so we will not use this notation in our text.

Instead, we use the notation \(f^{-1}(t) = \arcsin (t)\), read ‘arc-sine of \(t\)’. We’ll explain the ‘arc’ in ‘arcsine’ shortly. For now, we graph \(f(t) = \sin (t)\) and \(f^{-1}(t) = \arcsin (t)\) below, where we obtain the latter from the former by reflecting it across the line \(y=t\), in accordance with Theorem inversefunctionprops.

Next, we consider \(g(t) = \cos (t)\). Here, we select the interval \([0,\pi ]\) for our restriction, as seen below.

Reflecting the across the line \(y = t\) produces the graph \(y = g^{-1}(t) = \arccos (t)\).

We list some important facts about the arcsine and arccosine functions in the following theorem. Everything in Theorem arccosinesinefunctionprops is a direct consequence of Theorem inversefunctionprops as applied to the (restricted) sine and cosine functions, and as such, its proof is left to the reader.

Before moving to an example, we take a moment to understand the ‘arc’ in ‘arcsine.’ Consider the figure below which illustrates the specific case of \(\arcsin \left (\frac {\sqrt {3}}{2} \right )\).

By definition, the real number \(t = \arcsin \left (\frac {\sqrt {3}}{2} \right )\) satisfies \(\sin (t) = \frac {\sqrt {3}}{2}\) with \(-\frac {\pi }{2} \leq t \leq \frac {\pi }{2}\). In other words, we are looking for angle measuring \(t\) radians between \(-\frac {\pi }{2}\) and \(\frac {\pi }{2}\) with a sine of \(\frac {\sqrt {3}}{2}\). Hence, \(\arcsin \left (\frac {\sqrt {3}}{2} \right ) = \frac {\pi }{3}\).

In terms of oriented arcs , if we start at \((1,0)\) and travel along the Unit Circle in the positive (counterclockwise) direction for \(\frac {\pi }{3}\) units, we will arrive at the point whose \(y\)-coordinate is \(\frac {\sqrt {3}}{2}\). Hence, the real number \(\frac {\pi }{3}\) also corresponds to ‘arc’ corresponding to the ‘sine’ that is \(\frac {\sqrt {3}}{2}\).

In general, the function \(f(t) = \sin (t)\) takes a real number input \(t\), associates it with the angle \(\theta = t\) radians, and returns the value \(\sin (\theta )\). The value \(\sin (\theta ) = \sin (t)\) is the \(y\)-coordinate of the terminal point on the Unit Circle of an oriented arc of length \(|t|\) whose initial point is \((1, 0)\).

Hence, we may view the inputs to \(f(t) = \sin (t)\) as oriented arcs and the outputs as \(y\)-coordinates on the Unit Circle. Therefore, the function \(f^{-1}\) reverses this process and takes \(y\)-coordinates on the Unit Circle and return oriented arcs, hence the ‘arc’ in arcsine.

It is high time for an example.

The next pair of functions we wish to discuss are the inverses of tangent and cotangent. First, we restrict \(f(t) = \tan (t)\) to its fundamental cycle on \(\left (-\frac {\pi }{2}, \frac {\pi }{2}\right )\) to obtain the arctangent function, \(f^{-1}(t) = \arctan (t)\). Among other things, note that the vertical asymptotes \(t = -\frac {\pi }{2}\) and \(t = \frac {\pi }{2}\) of the graph of \(f(t) = \tan (t)\) become the horizontal asymptotes \(y = -\frac {\pi }{2}\) and \(y = \frac {\pi }{2}\) of the graph of \(f^{-1}(t) = \arctan (t)\).

Next, we restrict \(g(t) = \cot (t)\) to its fundamental cycle on \((0,\pi )\) to obtain \(g^{-1}(t) = \text {arccot}(t)\), the arccotangent function. Once again, the vertical asymptotes \(t=0\) and \(t=\pi \) of the graph of \(g(t) = \cot (t)\) become the horizontal asymptotes \(y = 0\) and \(y = \pi \) of the graph of \(g^{-1}(t) = \text {arccot}(t)\).

Below we summarize the important properties of the arctangent and arccotangent functions.

The properties listed in Theorem arctangentcotangentfunctionprops are consequences of the definitions of the arctangent and arccotangent functions along with Theorem inversefunctionprops, and its proof is left to the reader.

The reader may well wonder if there isn’t a more direct way to handle Example arctanarccotexample number cosineofarccotex. Indeed, we can take some inspiration from Section TheOtherCircularFunctions and imagine an angle \(\theta \) measuring \(t\) radians so that \(\cot (\theta ) = \cot (t) = 2x\) where \(0< \theta < \pi \).

Thinking of \(\cot (\theta )\) as a ratio of coordinates on a circle, we may rewrite \(\cot (\theta ) = 2x = \frac {2x}{1}\) and we would like to identify a point \(P(2x,1)\) on the terminal side of \(\theta \).

We need to be careful here. Since \(\cot (\theta ) = 2x\), \(x = \frac {1}{2} \cot (\theta )\), so as \(\theta \) ranges between \(0\) and \(\pi \), \(x\) can take on positive or negative values or \(0\). We need to argue that the point \(P(2x,1)\) lies in the quadrant we expect (as depicted below) in all cases before we delve too far into our analysis.

If \(0 < \theta < \frac {\pi }{2}\), then \(\cot (\theta ) > 0\). Hence, \(x>0\) so the point \(P(2x,1)\) is in Quadrant I, as required. If \(\theta = \frac {\pi }{2}\), then \(x = 0\), and our point \(P(2x,1) = (0,1)\), as required. If \(\frac {\pi }{2} < \theta < \pi \), then \(\cot (\theta )<0\). Hence, \(x<0\), so \(P(2x,1)\) is in Quadrant II, as required.

Hence, in all three cases, our formula for the point \(P(2x,1)\) determines a point in the same quadrant as the terminal side of \(\theta \), as illustrated above.

This allows us to use Theorem circularfunctionscircle from Section TheOtherCircularFunctions. We find \(r = \sqrt {(2x)^2+1^2} = \sqrt {4x^2+1}\), and hence, \(\cos (\theta ) = \frac {2x}{\sqrt {4x^2+1}}\), which agrees with our answer from Example arctanarccotexample.

It shouldn’t surprise the reader that there are some cases where the approach outlined above doesn’t go as smoothly (as we’ll see in the discussion following Example arcsecantcosecantex1.)

The last two functions to invert are secant and cosecant. First we graph secant with its fundamental cycle highlighted.

Next, we graph cosecant with its fundamental cycle highlighted.

It is clear from the graph of secant that we cannot find one single continuous piece of its graph which covers its entire range of \((-\infty , -1] \cup [1, \infty )\) and restricts the domain of the function so that it is one-to-one. The same is true for cosecant.

Thus in order to define the arcsecant and arccosecant functions, we must settle for a piecewise approach wherein we choose one piece to cover the top of the range, namely \([1, \infty )\), and another piece to cover the bottom, namely \((-\infty , -1]\).

There are two generally accepted ways make these choices which restrict the domains of these functions so that they are one-to-one. One approach simplifies the Trigonometry associated with the inverse functions, but complicates the Calculus; the other makes the Calculus easier, but the Trigonometry less so.

For completeness, we present both points of view, each in its own subsection.

1 Inverses of Secant and Cosecant: Trigonometry Friendly Approach

In this subsection, we restrict the secant and cosecant functions to coincide with the restrictions on cosine and sine, respectively. For \(f(t) = \sec (t)\), we restrict the domain to \(\left [0, \frac {\pi }{2}\right ) \cup \left ( \frac {\pi }{2}, \pi \right ]\)

and we restrict \(g(t) = \csc (t)\) to \(\left [-\frac {\pi }{2}, 0\right ) \cup \left (0, \frac {\pi }{2}\right ]\).

Note that for both arcsecant and arccosecant, the domain is \((-\infty , -1] \cup [1, \infty )\). Taking a page from Section ??, we can rewrite this as \(\left \{ x \, | \, |x| \geq 1\right \}\). (This is often done in Calculus textbooks, so we include it here for completeness.)

Using these definitions along with Theorem ??, we get the following properties of the arcsecant and arccosecant functions.

The reason the ranges here are called ‘Trigonometry Friendly’ is specifically because of two properties listed in Theorem 3: \(\text {arcsec}(x) = \arccos \left (\frac {1}{x}\right )\) and \(\text {arccsc}(x) = \arcsin \left (\frac {1}{x}\right )\).

These formulas essentially allow us to always convert arcsecants and arccosecants back to arccosines and arcsines, respectively. We see this play out in our next example.

As promised in the discussion following Example 2, in which we used the methods from Section ?? to circumvent some onerous identity work, we take some time here to revisit number 2(i) to see what issues arise when we take a Section ?? approach here.

As above, we start rewriting \(f(x) = \tan (\text {arcsec}(x))\) by letting \(t = \text {arcsec}(x)\) so that \(\sec (t) = x\) where \(0 \leq t < \frac {\pi }{2}\) or \( \frac {\pi }{2} < t \leq \pi \). We let \(\theta = t\) radians and wish to view \(\sec (\theta ) = \sec (t) = x\) as described in Theorem ?? : the ratio of the radius of a circle, \(r\) centered at the origin, divided by the abscissa of a point on the terminal side of \(\theta \) which intersects said circle.

If we make the usual identification \(\sec (\theta ) = x = \frac {x}{1}\), we see that if \(0 \leq \theta < \frac {\pi }{2}\), then \(x = \sec {\theta } \geq 1\), so it makes sense to identify the quantity \(x\) as the radius of the circle with \(1\) as the abscissa of the point where the terminal side of \(\theta \) intersects said circle. To find the associated ordinate (\(y\)-coordinate), we have \(1^2 + y^2 = x^2\) so \(y = \sqrt {x^2-1}\), where we have chosen the positive root since we are in Quadrant I. We sketch out this scenario below on the left.

If, however, \( \frac {\pi }{2} < t \leq \pi \), then \(x = \sec (t) \leq -1\), so we need to rewrite \(\sec (\theta ) = x = \frac {x}{1} = \frac {-x}{-1}\) in order to keep the radius of the circle, \(r = -x > 0\) and the abscissa, \(-1 < 0\). From \((-1)^2+y^2 = (-x)^2\), we still get \(y = \sqrt {x^2-1}\), as shown below on the right.

In the Quadrant I case, when \(x \geq 1\), we get \(\tan (\theta ) = \frac {\sqrt {x^2-1}}{1} = \sqrt {x^2-1}\). In Quadrant II, when \(x \leq -1\), we obtain \(\tan (\theta ) = \frac {\sqrt {x^2-1}}{-1} = - \sqrt {x^2-1}\). Hence, we get the piecewise definition for \(f(x)\) as we did in number 2(i) above: \(f(x) = \tan (\text {arcsec}(x)) = \sqrt {x^2-1}\) if \(x \geq 1\) and \(f(x) = \tan (\text {arcsec}(x)) = -\sqrt {x^2-1}\) if \(x \leq -1\).

The moral of the story here is that you are free to choose whichever route you like to simplify expressions like those found in Example 3 number 2(i). Whether you choose identities or a more geometric route, just be careful to keep in mind which quadrants are in play, which variables represent which quantities, and what signs (\(\pm \)) each should have.

2 Inverses of Secant and Cosecant: Calculus Friendly Approach

In this subsection, we restrict \(f(t) = \sec (t)\) to \(\left [0, \frac {\pi }{2}\right ) \cup \left [\pi , \frac {3\pi }{2}\right )\) and \(g(t) = \csc (t)\) to \(\left (0, \frac {\pi }{2}\right ] \cup \left ( \pi , \frac {3\pi }{2}\right ]\). Using these restrictions we get the graphs and properties below.

First, the graph of arcsecant:

Next, the graph of arccosecant:

While it is difficult to explain why the choices here for the ranges for the arcsecant an arccosecant are, indeed, ‘Calculus Friendly,’ we can demonstrate how they are slightly less ‘Trigonometry Friendly.’ Note the equivalences \(\text {arcsec}(x) = \arccos \left (\frac {1}{x}\right )\) and \(\text {arccsc}(x) = \arcsin \left (\frac {1}{x}\right )\) hold for \(x \geq 1\) only, and not for all \(x\) in the domain. We will need to remember this as we work through the problems in the next example.

Speaking of which, our next example is a duplicate of Example 3. The interested reader is invited to see what differences are to be had as a consequence of the change in ranges.

For completeness, we embark here on a discussion of how the techniques from Section ??, in particular Theorem ?? can be used to circumvent some of the identity work in number 2(i) above.

As above, we start rewriting \(f(x) = \tan (\text {arcsec}(x))\) by letting \(t = \text {arcsec}(x)\) so that \(\sec (t) = x\) where \(0 \leq t < \frac {\pi }{2}\) or \(\pi \leq t < \frac {3\pi }{2}\). We let \(\theta = t\) radians and wish to view \(\sec (\theta ) = \sec (t) = x\) as described in Theorem ?? : the ratio of the radius of a circle, \(r\) centered at the origin, divided by the abscissa of a point on the terminal side of \(\theta \) which intersects said circle.

If we make the usual identification \(\sec (\theta ) = x = \frac {x}{1}\), we see that if \(0 \leq \theta < \frac {\pi }{2}\), then \(x = \sec {\theta } \geq 1\), so it makes sense to identify the quantity \(x\) as the radius of the circle with \(1\) as the abscissa of the point where the terminal side of \(\theta \) intersects said circle. To find the associated ordinate (\(y\)-coordinate), we have \(1^2 + y^2 = x^2\) so \(y = \sqrt {x^2-1}\), where we have chosen the positive root since we are in Quadrant I. We sketch out this scenario below on the left.

If, however, \( \pi \leq t < \frac {3\pi }{2}\), then \(x = \sec (t) \leq -1\), so we need to rewrite \(\sec (\theta ) = x = \frac {x}{1} = \frac {-x}{-1}\) in order to keep the radius of the circle, \(r = -x > 0\) and the abscissa, \(-1 < 0\). From \((-1)^2+y^2 = (-x)^2\), we get \(y = -\sqrt {x^2-1}\), in this case choosing the negative root since we are in Quadrant III.

In the Quadrant I case, when \(x \geq 1\), we get \(\tan (\theta ) = \frac {\sqrt {x^2-1}}{1} = \sqrt {x^2-1}\). In Quadrant III, when \(x \leq -1\), we obtain \(\tan (\theta ) = \frac {-\sqrt {x^2-1}}{-1} = \sqrt {x^2-1}\). Hence, in both cases, we obtain the same answer as we did in number 2(i) above: \(f(x) = \tan (\text {arcsec}(x)) = \sqrt {x^2-1}\) for \(x\) in \((-\infty , -1] \cup [1, \infty )\).

3 Calculators and the Inverse Circular Functions.

In the sections to come, we will have need to approximate the values of the inverse circular functions. On most calculators, only the arcsine, arccosine and arctangent functions are available and they are usually labeled as \(\sin ^{-1}, \cos ^{-1}\) and \(\tan ^{-1}\), respectively. If we are asked to approximate these values, it is a simple matter to punch up the appropriate decimal on the calculator.

If we are asked for an arccotangent, arcsecant or arccosecant, however, we often need to employ some ingenuity, as our next example illustrates.

4 Solving Equations Using the Inverse Trigonometric Functions.

In Sections ?? and ??, we learned how to solve equations like \(\sin (\theta ) = \frac {1}{2}\) and \(\tan (t) = -1\). In each case, we ultimately appealed to the Unit Circle and relied on the fact that the answers corresponded to a set of ‘common angles’ listed on page ?? .

If, on the other hand, we had been asked to find all angles with \(\sin (\theta ) = \frac {1}{3}\) or solve \(\tan (t) = -2\) for real numbers \(t\), we would have been hard-pressed to do so. With the introduction of the inverse trigonometric functions, however, we are now in a position to solve these equations.

A good parallel to keep in mind is how the square root function can be used to solve certain quadratic equations. The equation \(x^2 = 4\) is a lot like \(\sin (\theta ) = \frac {1}{2}\) in that it has friendly, ‘common value’ answers \(x = \pm 2\). The equation \(x^2 = 7\), on the other hand, is a lot like \(\sin (\theta ) = \frac {1}{3}\). We know there are answers, but we can’t express them using ‘friendly’ numbers.

To solve \(x^2 = 7\), we make use of the square root function (which is an inverse to \(f(x) = x^2\) on a restricted domain) and write our answer as \(x = \pm \sqrt {7}\). We need the \(\pm \) to adjust for the fact that \(\sqrt {7}\) is defined to be positive only, but we know we have two solutions, one positive and one negative. Using a calculator, we can certainly approximate the values \(\pm \sqrt {7}\), but as far as exact answers go, we leave them as \(x = \pm \sqrt {7}\).

In the same way, we will use the arcsine function (the inverse to the sine function on a restricted domain) to solve \(\sin (\theta ) = \frac {1}{3}\). However, we will need to adjust for the fact that there is more than one answer to this equation (infinitely many, in fact!) As it turns out, we will be able to express every solution in terms of \(\arcsin \left (\frac {1}{3}\right )\), as our next example illustrates.

We close this section with one last sinusoid example.