12.3 The Inverse Circular Functions

In this section we concern ourselves with finding inverses of the circular (trigonometric) functions.¹ 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 5.6.3 in Section 5.6 to obtain a one-to-one function.

We start with $f (t) = \sin (t)$ and restrict our domain to $[- \frac{π}{2}, \frac{π}{2}]$ in order to keep the range as $open bracket minus 1, 1 close bracket$ as well as the properties of being smooth and continuous.

Figure: Restricting the domain of to . — **Figure 12.10** Restricting the domain of $f (t) = \sin (t)$ to $[- \frac{π}{2}, \frac{π}{2}]$ .

Recall from Section 5.6 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.²

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 $f^{- 1} (t) = \arcsin (t)$ , where we obtain the latter from the former by reflecting it across the line $y = t$ , in accordance with Theorem 5.13.

Figure: , . — **Figure 12.11** $f (t) = \sin (t)$ , $- \frac{π}{2} \leq t \leq \frac{π}{2}$ .

$\overset{\overset{reflect across y = t}{\overset{}{\to}}}{switch t and y coordinates}$

Next, we consider $g (t) = \cos (t)$ . Here, we select the interval $open bracket 0, π close bracket$ for our restriction.

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

$\overset{\overset{reflect across y = t}{\overset{}{\to}}}{switch t and y coordinates}$

We list some important facts about the arcsine and arccosine functions in the following theorem.³ Everything in Theorem 12.14 is a direct consequence of Theorem 5.13 as applied to the (restricted) sine and cosine functions, and as such, its proof is left to the reader.

Theorem

Properties of the Arccosine and Arcsine Functions

Properties of $F (x) = \arcsin (x)$
- Domain: $open bracket minus 1, 1 close bracket$
- Range: $[- \frac{π}{2}, \frac{π}{2}]$
- $\arcsin (x) = t$ if and only if $\sin (t) = x$ and $- \frac{π}{2} \leq t \leq \frac{π}{2}$
- $\sin (\arcsin (x)) = x$ provided $minus 1 less than or equal to x less than or equal to 1$
- $\arcsin (\sin (t)) = t$ provided $- \frac{π}{2} \leq t \leq \frac{π}{2}$
- $F (x) = \arcsin (x)$ is odd
Properties of $G (x) = \arccos (x)$
- Domain: $open bracket minus 1, 1 close bracket$
- Range: $open bracket 0, π close bracket$
- $\arccos (x) = t$ if and only if $\cos (t) = x$ and $0 less than or equal to t less than or equal to π$
- $\cos (\arccos (x)) = x$ provided $minus 1 less than or equal to x less than or equal to 1$
- $\arccos (\cos (t)) = t$ provided $0 less than or equal to t less than or equal to π$

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 $\arcsin (\frac{\sqrt{3}}{2})$ .

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

In terms of oriented arcs⁴, if we start at $open parenthesis 1, 0 close parenthesis$ and travel along the Unit Circle in the positive (counterclockwise) direction for $\frac{π}{3}$ units, we will arrive at the point whose $y$ -coordinate is $\frac{\sqrt{3}}{2}$ . Hence, the real number $\frac{π}{3}$ also corresponds to `arc' corresponding to the `sine' that is $\frac{\sqrt{3}}{2}$ .

Coordinate-plane figure. — **Figure 12.16**

$y$ -value on Unit Circle is $\frac{\sqrt{3}}{2}$ .

corresponding oriented arc: $t = \frac{π}{3}$

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

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.

Example 1

Find the exact values of the following.
1. $\arcsin (\frac{\sqrt{2}}{2})$
2. $\arccos (\frac{1}{2})$
1. $\arcsin (- \frac{1}{2})$
2. $\arccos (- \frac{\sqrt{2}}{2})$
1. $\arccos (\cos (\frac{π}{6}))$
2. $\arccos (\cos (\frac{11 π}{6}))$
1. $\cos (\arccos (- \frac{3}{5}))$
2. $\sin (\arccos (- \frac{3}{5}))$
Rewrite each of the following composite functions as algebraic functions of $x$ and state the domain.
1. $f (x) = \tan (\arccos (x))$
2. $g ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis 2 ⁢ arcsin ⁡ open parenthesis x close parenthesis close parenthesis$

The best way to approach these problems is to remember that $\arcsin (x)$ and $\arccos (x)$ are real numbers which correspond to the radian measure of angles that fall within a certain prescribed range.

1. To find $\arcsin (\frac{\sqrt{2}}{2})$ , we need the angle measuring $t$ radians which lies between $- \frac{π}{2}$ and $\frac{π}{2}$ with $\sin (t) = \frac{\sqrt{2}}{2}$ . Hence, $\arcsin (\frac{\sqrt{2}}{2}) = \frac{π}{4}$ .
2. To find $\arccos (\frac{1}{2})$ , we are looking for the angle measuring $t$ radians which lies between $0$ and $π$ that has $\cos (t) = \frac{1}{2}$ . Our answer is $\arccos (\frac{1}{2}) = \frac{π}{3}$ .
3. For $\arcsin (- \frac{1}{2})$ , we are looking for an angle measuring $t$ radians which lies between $- \frac{π}{2}$ and $\frac{π}{2}$ with $\sin (t) = - \frac{1}{2}$ . Hence, $\arcsin (- \frac{1}{2}) = - \frac{π}{6}$ .
  Alternatively, we could use the fact that the arcsine function is odd, so $\arcsin (- \frac{1}{2}) = - \arcsin (\frac{1}{2})$ .
  We find $\arcsin (\frac{1}{2}) = \frac{π}{6}$ , so $\arcsin (- \frac{1}{2}) = - \arcsin (\frac{1}{2}) = - \frac{π}{6}$ .
4. For $\arccos (- \frac{\sqrt{2}}{2})$ , we need the angle measuring $t$ radians which lies between $0$ and $π$ with $\cos (t) = - \frac{\sqrt{2}}{2}$ . Hence, $\arccos (- \frac{\sqrt{2}}{2}) = \frac{3 π}{4}$ .
5. Since $0 \leq \frac{π}{6} \leq π$ , we could simply invoke Theorem 12.14 to get $\arccos (\cos (\frac{π}{6})) = \frac{π}{6}$ .
  However, in order to make sure we understand why this is the case, we choose to work the example through using the definition of arccosine.
  Working from the inside out, $\arccos (\cos (\frac{π}{6})) = \arccos (\frac{\sqrt{3}}{2})$ . To find $\arccos (\frac{\sqrt{3}}{2})$ , we need an angle measuring $t$ radians which lies between $0$ and $π$ that has $\cos (t) = \frac{\sqrt{3}}{2}$ . We get $t = \frac{π}{6}$ , so that $\arccos (\cos (\frac{π}{6})) = \arccos (\frac{\sqrt{3}}{2}) = \frac{π}{6}$ .
6. Since $\frac{11 π}{6}$ does not fall between $0$ and $π$ , Theorem 12.14 does not apply. We are forced to work through from the inside out starting with $\arccos (\cos (\frac{11 π}{6})) = \arccos (\frac{\sqrt{3}}{2})$ . From the previous problem, we know $\arccos (\frac{\sqrt{3}}{2}) = \frac{π}{6}$ . Hence, $\arccos (\cos (\frac{11 π}{6})) = \frac{π}{6}$ .
7. One way to simplify $\cos (\arccos (- \frac{3}{5}))$ is to use Theorem 12.14 directly. Since $- \frac{3}{5}$ is between $minus 1$ and $1$ , we have that $\cos (\arccos (- \frac{3}{5})) = - \frac{3}{5}$ and we are done.
  However, as before, to really understand why this cancellation occurs, we let $t = \arccos (- \frac{3}{5})$ . By definition, $\cos (t) = - \frac{3}{5}$ . Hence, $\cos (\arccos (- \frac{3}{5})) = \cos (t) = - \frac{3}{5}$ , and we are finished in (nearly) the same amount of time.
8. As in the previous example, we let $t = \arccos (- \frac{3}{5})$ so that $\cos (t) = - \frac{3}{5}$ for some angle measuring $t$ radians between $0$ and $π$ .
  Since $cos ⁡ open parenthesis t close parenthesis less than 0$ , we can narrow this down a bit and conclude that $\frac{π}{2} < t < π$ , so that $t$ corresponds to an angle in Quadrant II.
  In terms of $t$ , then, we need to find $\sin (\arccos (- \frac{3}{5})) = \sin (t)$ , and since we know $\cos (t)$ , the fastest route is through the Pythagorean Identity.
  We get $\sin^{2} (t) = 1 - \cos^{2} (t) = 1 - {(- \frac{3}{5})}^{2} = \frac{16}{25}$ . Since $t$ corresponds to a Quadrant II angle, we choose the positive root, $\sin (t) = \frac{4}{5}$ , so $\sin (\arccos (- \frac{3}{5})) = \frac{4}{5}$ .
1. We begin this problem in the same manner we began the previous two problems. We let $t = \arccos (x)$ , so our goal is to find a way to express $\tan (\arccos (x)) = \tan (t)$ in terms of $x$ .
  Since $t = \arccos (x)$ , we know $\cos (t) = x$ where $0 less than or equal to t less than or equal to π$ . One approach⁵ to finding $\tan (t)$ is to use the quotient identity $\tan (t) = \frac{\sin (t)}{\cos (t)}$ . Since we know $\cos (t)$ , we just need to find $\sin (t)$ .
  Using the Pythagorean Identity, we get $\sin^{2} (t) = 1 - \cos^{2} (t) = 1 - x^{2}$ so that $\sin (t) = \pm \sqrt{1 - x^{2}}$ . Since $0 less than or equal to t less than or equal to π$ , $sin ⁡ open parenthesis t close parenthesis greater than or equal to 0$ , so we choose $\sin (t) = \sqrt{1 - x^{2}}$ .
  Thus, $\tan (t) = \frac{\sin (t)}{\cos (t)} = \frac{\sqrt{1 - x^{2}}}{x}$ , so $f (x) = \tan (\arccos (x)) = \frac{\sqrt{1 - x^{2}}}{x}$ .
  To determine the domain, we harken back to Section 5.3. The function $f (x) = \tan (\arccos (x))$ can be thought of as a two step process: first, take the arccosine of a number, and second, take the tangent of whatever comes out of the arccosine.
  Since the domain of $\arccos (x)$ is $minus 1 less than or equal to x less than or equal to 1$ , the domain of $f$ will be some subset of $open bracket minus 1, 1 close bracket$ . The range of $\arccos (x)$ is $open bracket 0, π close bracket$ , and of these values, only $\frac{π}{2}$ will cause a problem for the tangent function. Since $\arccos (x) = \frac{π}{2}$ happens when $x = \cos (\frac{π}{2}) = 0$ , we exclude $x equals 0$ from our domain. Hence, the domain of $f (x) = \tan (\arccos (x)) = \frac{\sqrt{1 - x^{2}}}{x}$ is $open bracket minus 1, 0 close parenthesis union open parenthesis 0, 1 close bracket$ .
  Note that in this particular case, we could have obtained the correct domain of $f$ using its algebraic description: $f (x) = \tan (\arccos (x)) = \frac{\sqrt{1 - x^{2}}}{x}$ . This is not always true, however, as we'll see in the next problem.
2. We proceed as in the previous problem by writing $t = \arcsin (x)$ so that $t$ lies in the interval $[- \frac{π}{2}, \frac{π}{2}]$ with $\sin (t) = x$ . We aim to express $cos ⁡ open parenthesis 2 ⁢ arcsin ⁡ open parenthesis x close parenthesis close parenthesis equals cos ⁡ open parenthesis 2 ⁢ t close parenthesis$ in terms of $x$ .
  Thanks to Theorem 12.9, we have three choices for rewriting $cos ⁡ open parenthesis 2 ⁢ t close parenthesis$ : $\cos (2 t) = \cos^{2} (t) - \sin^{2} (t)$ , $\cos (2 t) = 2 \cos^{2} (t) - 1$ and $\cos (2 t) = 1 - 2 \sin^{2} (t)$ .
  Since we know $x = \sin (t)$ , we choose: $\cos (2 \arcsin (x)) = \cos (2 t) = 1 - 2 \sin^{2} (t) = 1 - 2 x^{2}$ . Hence, $g (x) = \cos (2 \arcsin (x)) = 1 - 2 x^{2}$ .
  To find the domain of $g ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis 2 ⁢ arcsin ⁡ open parenthesis x close parenthesis close parenthesis$ , we once again appeal to what we learned in Section 5.3. The domain of $\arcsin (x)$ is $open bracket minus 1, 1 close bracket$ , and since there are no domain restrictions on cosine, the domain of $g$ is $open bracket minus 1, 1 close bracket$ .
  It is important to note that in this case, even though the algebraic expression $1 minus 2 ⁢ x squared$ is defined for all real numbers, the domain of $g$ is limited to that of $\arcsin (x)$ , namely $open bracket minus 1, 1 close bracket$ . The adage `find the domain before you simplify' rings as true here as it did in Chapter 5.

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 $(- \frac{π}{2}, \frac{π}{2})$ to obtain the arctangent function, $f^{- 1} (t) = \arctan (t)$ . Among other things, note that the vertical asymptotes $t = - \frac{π}{2}$ and $t = \frac{π}{2}$ of the graph of $f (t) = \tan (t)$ become the horizontal asymptotes $y = - \frac{π}{2}$ and $y = \frac{π}{2}$ of the graph of $f^{- 1} (t) = \arctan (t)$ .

$\overset{\overset{reflect across y = t}{\overset{}{\to}}}{switch t and y coordinates}$

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

$\overset{\overset{reflect across y = t}{\overset{}{\to}}}{switch t and y coordinates}$

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

Theorem

Properties of the Arctangent and Arccotangent Functions

Properties of $F (x) = \arctan (x)$
- Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- Range: $(- \frac{π}{2}, \frac{π}{2})$
- $lim x → minus ∞ arctan ⁡ open parenthesis x close parenthesis$ $= - \frac{π}{2}$ ; $lim x → ∞ arctan ⁡ open parenthesis x close parenthesis$ $= \frac{π}{2}$
- $\arctan (x) = t$ if and only if $\tan (t) = x$ and $- \frac{π}{2} < t < \frac{π}{2}$
- $\arctan (x) = arccot (\frac{1}{x})$ for $x greater than 0$
- $\tan (\arctan (x)) = x$ for all real numbers $x$
- $\arctan (\tan (t)) = t$ provided $- \frac{π}{2} < t < \frac{π}{2}$
- $F (x) = \arctan (x)$ is odd
Properties of $G (x) = arccot (x)$
- Domain: $open parenthesis minus ∞, ∞ close parenthesis$
- Range: $open parenthesis 0, π close parenthesis$
- $lim_{x \to - \infty} arccot (x) = π$ ; $lim x → ∞ arccot ⁢ open parenthesis x close parenthesis equals 0$
- $arccot (x) = t$ if and only if $\cot (t) = x$ and $0 less than t less than π$
- $arccot (x) = \arctan (\frac{1}{x})$ for $x greater than 0$
- $\cot (arccot (x)) = x$ for all real numbers $x$
- $arccot (\cot (t)) = t$ provided $0 less than t less than π$

The properties listed in Theorem 12.15 are consequences of the definitions of the arctangent and arccotangent functions along with Theorem 5.13, and its proof is left to the reader.

Example 2

Find the exact values of the following.
1. $arctan ⁡ open parenthesis the square root of 3 close parenthesis$
2. $arccot (- \sqrt{3})$
1. $cot ⁡ open parenthesis arccot ⁢ open parenthesis minus 5 close parenthesis close parenthesis$
2. $\sin (\arctan (- \frac{4}{3}))$
Rewrite each of the following composite functions as algebraic functions of $x$ and state the domain.
1. $tan ⁡ open parenthesis 2 ⁢ arctan ⁡ open parenthesis x close parenthesis close parenthesis$
2. $cos ⁡ open parenthesis arccot ⁢ open parenthesis 2 ⁢ x close parenthesis close parenthesis$

1. To find $arctan ⁡ open parenthesis the square root of 3 close parenthesis$ , we need the angle measuring $t$ radians which lies between $- \frac{π}{2}$ and $\frac{π}{2}$ with $\tan (t) = \sqrt{3}$ . We find $\arctan (\sqrt{3}) = \frac{π}{3}$ .
2. To find $arccot (- \sqrt{3})$ , we need the angle measuring $t$ radians which lies between $0$ and $π$ with $\cot (t) = - \sqrt{3}$ . Hence, $arccot (- \sqrt{3}) = \frac{5 π}{6}$ .
3. We can apply Theorem 12.15 directly and obtain $\cot (arccot (- 5)) = - 5$ . However, working it through provides us with yet another opportunity to understand why this is the case.
  Letting $t = arccot (- 5)$ , by definition, $\cot (t) = - 5$ . Hence, $\cot (arccot (- 5)) = \cot (t) = - 5$ .
4. We start simplifying $\sin (\arctan (- \frac{4}{3}))$ by letting $t = \arctan (- \frac{4}{3})$ . By definition, $\tan (t) = - \frac{4}{3}$ for some angle measuring $t$ radians which lies between $- \frac{π}{2}$ and $\frac{π}{2}$ . Since $tan ⁡ open parenthesis t close parenthesis less than 0$ , we know, in fact, $t$ corresponds to a Quadrant IV angle.
  We are given $\tan (t)$ but wish to know $\sin (t)$ . Since there is no direct identity to marry the two, we make a quick sketch of the situation below. Since $\tan (t) = - \frac{4}{3} = \frac{- 4}{3}$ , we take $P ⁢ open parenthesis 3, minus 4 close parenthesis$ as a point on the terminal side of $θ = t = \arctan (- \frac{4}{3})$ radians.
  Figure 12.22 $P ⁢ open parenthesis 3, minus 4 close parenthesis$ is on the terminal side of $θ$ .
  We find $r = \sqrt{x^{2} + y^{2}} = \sqrt{{(3)}^{2} + {(- 4)}^{2}} = 5$ , so $\sin (t) = - \frac{4}{5}$ . Hence, $\sin (\arctan (- \frac{4}{3})) = - \frac{4}{5}$ .
1. We proceed as above and let $t = \arctan (x)$ . We have $\tan (t) = x$ where $- \frac{π}{2} < t < \frac{π}{2}$ . Our goal is to express $tan ⁡ open parenthesis 2 ⁢ arctan ⁡ open parenthesis x close parenthesis close parenthesis equals tan ⁡ open parenthesis 2 ⁢ t close parenthesis$ in terms of $x$ .
  From Theorem 12.9, we know $\tan (2 t) = \frac{2 \tan (t)}{1 - \tan^{2} (t)} = \frac{2 x}{1 - x^{2}}$ . Hence $f (x) = \tan (2 \arctan (x)) = \frac{2 x}{1 - x^{2}}$ .
  To find the domain, we once again think of $f ⁢ open parenthesis x close parenthesis equals tan ⁡ open parenthesis 2 ⁢ arctan ⁡ open parenthesis x close parenthesis close parenthesis$ as a sequence of steps and work from the inside out.
  The first step is to find the arctangent of a real number. Since the domain of $\arctan (x)$ is all real numbers, we have no restrictions here and we get out all values $(- \frac{π}{2}, \frac{π}{2})$ .
  The next step is to multiply $\arctan (x)$ by $2$ . There are no restrictions here, either. Since the range of $\arctan (x)$ is $(- \frac{π}{2}, \frac{π}{2})$ , the range of $2 ⁢ arctan ⁡ open parenthesis x close parenthesis$ is $(- π, π)$ .
  The last step is to take the tangent of $2 ⁢ arctan ⁡ open parenthesis x close parenthesis$ . Since we are taking the tangent of values in the interval $(- π, π)$ , we will run into trouble if $2 \arctan (x) = \pm \frac{π}{2}$ , that is, if $\arctan (x) = \pm \frac{π}{4}$ . Since this happens exactly when $x = \tan (\pm \frac{π}{4}) = \pm 1$ , we must exclude $x = \pm 1$ from the domain of $f$ .
  Hence, the domain of $f ⁢ open parenthesis x close parenthesis equals tan ⁡ open parenthesis 2 ⁢ arctan ⁡ open parenthesis x close parenthesis close parenthesis$ is $(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$ . In this example, we could have obtained the correct answer by looking at the algebraic equivalence, $f (x) = \frac{2 x}{1 - x^{2}}$ . However, as we saw in Example 12.3.1, number, this is not always the case.
2. To get started, we let $t equals arccot ⁢ open parenthesis 2 ⁢ x close parenthesis$ so that $cot ⁡ open parenthesis t close parenthesis equals 2 ⁢ x$ where $0 less than t less than π$ . In terms of $t$ , $cos ⁡ open parenthesis arccot ⁢ open parenthesis 2 ⁢ x close parenthesis close parenthesis equals cos ⁡ open parenthesis t close parenthesis$ , and our goal is to express the latter in terms of $x$ .
  One way to proceed is to rewrite the identity $\cot (t) = \frac{\cos (t)}{\sin (t)}$ as $\cos (t) = \cot (t) \sin (t)$ and use the fact that $cot ⁡ open parenthesis t close parenthesis equals 2 ⁢ x$ to find $\sin (t)$ in terms of $x$ . This isn't as hopeless as it might seem, since the Pythagorean Identity $\csc^{2} (t) = 1 + \cot^{2} (t)$ relates cotangent to cosecant, and $\sin (t) = \frac{1}{\csc (t)}$ .
  Following this strategy, we get $\csc^{2} (t) = 1 + \cot^{2} (t) = 1 + {(2 x)}^{2} = 1 + 4 x^{2}$ so $\csc (t) = \pm \sqrt{4 x^{2} + 1}$ . Since $t$ is between $0$ and $π$ , $csc ⁡ open parenthesis t close parenthesis greater than 0$ . Hence, $\csc (t) = \sqrt{4 x^{2} + 1}$ , so $\sin (t) = \frac{1}{\csc (t)} = \frac{1}{\sqrt{4 x^{2} + 1}}$ .
  We find $\cos (t) = \cot (t) \sin (t) = \frac{2 x}{\sqrt{4 x^{2} + 1}}$ . Hence, $g (x) = \cos (arccot (2 x)) = \frac{2 x}{\sqrt{4 x^{2} + 1}}$ .
  Viewing $g ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis arccot ⁢ open parenthesis 2 ⁢ x close parenthesis close parenthesis$ as a sequence of steps, we see we first double the input $x$ , then take the arccotangent, and, finally, take the cosine. Since each of these processes are valid for all real numbers, the domain of $g$ is $open parenthesis minus ∞, ∞ close parenthesis$ .

The reader may well wonder if there isn't a more direct way to handle Example 12.3.2 number. Indeed, we can take some inspiration from Section 11.4 and imagine an angle $θ$ measuring $t$ radians so that $\cot (θ) = \cot (t) = 2 x$ where $0 less than θ less than π$ .

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

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

$0 < θ < \frac{π}{2}$ , $x greater than 0$ .

$θ = \frac{π}{2}$ , $x equals 0$ .

$\frac{π}{2} < θ < π$ , $x less than 0$ .

If $0 < θ < \frac{π}{2}$ , then $cot ⁡ open parenthesis θ close parenthesis greater than 0$ . Hence, $x greater than 0$ so the point $P ⁢ open parenthesis 2 ⁢ x, 1 close parenthesis$ is in Quadrant I, as required. If $θ = \frac{π}{2}$ , then $x equals 0$ , and our point $P ⁢ open parenthesis 2 ⁢ x, 1 close parenthesis equals open parenthesis 0, 1 close parenthesis$ , as required. If $\frac{π}{2} < θ < π$ , then $cot ⁡ open parenthesis θ close parenthesis less than 0$ . Hence, $x less than 0$ , so $P ⁢ open parenthesis 2 ⁢ x, 1 close parenthesis$ is in Quadrant II, as required.

Hence, in all three cases, our formula for the point $P ⁢ open parenthesis 2 ⁢ x, 1 close parenthesis$ determines a point in the same quadrant as the terminal side of $θ$ , as illustrated above.

This allows us to use Theorem 11.9 from Section 11.4. We find $r = \sqrt{{(2 x)}^{2} + 1^{2}} = \sqrt{4 x^{2} + 1}$ , and hence, $\cos (θ) = \frac{2 x}{\sqrt{4 x^{2} + 1}}$ , which agrees with our answer from Example 12.3.2.

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 12.3.3.)

The last two functions to invert are secant and cosecant. A portion of each of their graphs, which were first discussed in Subsection 11.5.1, are given below with the fundamental cycles highlighted.

Figure: The graph of . — **Figure 12.26** The graph of $y = \sec (x)$ .

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 $open bracket 1, ∞ close parenthesis$ , 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.

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 $[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$

Figure: on — **Figure 12.28** $f (t) = \sec (t)$ on $[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$

$\overset{\overset{reflect across y = t}{\overset{}{\to}}}{switch t and y coordinates}$

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

$\overset{\overset{reflect across y = t}{\overset{}{\to}}}{switch t and y coordinates}$

Note that for both arcsecant and arccosecant, the domain is $(- \infty, - 1] \cup [1, \infty)$ . Taking a page from Section A.7, we can rewrite this as $open brace x vertical bar vertical bar x vertical bar greater than or equal to 1 close brace$ . (This is often done in Calculus textbooks, so we include it here for completeness.)

Using these definitions along with Theorem 5.13, we get the following properties of the arcsecant and arccosecant functions.

Theorem

Properties of the Arcsecant and Arccosecant Functions ⁶

Properties of $F (x) = arcsec (x)$
- Domain: ${x | | x | \geq 1} = (- \infty, - 1] \cup [1, \infty)$
- Range: $[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$
- $lim x → minus ∞ arcsec ⁢ open parenthesis x close parenthesis$ $= \frac{π}{2}$ ; $lim x → ∞ arcsec ⁢ open parenthesis x close parenthesis$ $= \frac{π}{2}$
- $arcsec (x) = t$ if and only if $\sec (t) = x$ and $0 \leq t < \frac{π}{2}$ or $\frac{π}{2} < t \leq π$
- $arcsec (x) = \arccos (\frac{1}{x})$ provided $vertical bar x vertical bar greater than or equal to 1$
- $\sec (arcsec (x)) = x$ provided $vertical bar x vertical bar greater than or equal to 1$
- $arcsec (\sec (t)) = t$ provided $0 \leq t < \frac{π}{2}$ or $\frac{π}{2} < t \leq π$
Properties of $G (x) = arccsc (x)$
- Domain: ${x | | x | \geq 1} = (- \infty, - 1] \cup [1, \infty)$
- Range: $[- \frac{π}{2}, 0) \cup (0, \frac{π}{2}]$
- $lim_{x \to - \infty} arccsc (x) = 0$ ; $lim x → ∞ arccsc ⁢ open parenthesis x close parenthesis equals 0$
- $arccsc (x) = t$ if and only if $\csc (t) = x$ and $- \frac{π}{2} \leq t < 0$ or $0 < t \leq \frac{π}{2}$
- $arccsc (x) = \arcsin (\frac{1}{x})$ provided $vertical bar x vertical bar greater than or equal to 1$
- $\csc (arccsc (x)) = x$ provided $vertical bar x vertical bar greater than or equal to 1$
- $arccsc (\csc (t)) = t$ provided $- \frac{π}{2} \leq t < 0$ or $0 < t \leq \frac{π}{2}$
- $G (x) = arccsc (x)$ is odd

The reason the ranges here are called `Trigonometry Friendly' is specifically because of two properties listed in Theorem 12.16: $arcsec (x) = \arccos (\frac{1}{x})$ and $arccsc (x) = \arcsin (\frac{1}{x})$ .

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.

Example 3

Find the exact values of the following.
1. $arcsec ⁢ open parenthesis 2 close parenthesis$
2. $arccsc ⁢ open parenthesis minus 2 close parenthesis$
3. $arcsec (\sec (\frac{5 π}{4}))$
4. $cot ⁡ open parenthesis arccsc ⁢ open parenthesis minus 3 close parenthesis close parenthesis$
Rewrite each of the following composite functions as algebraic functions of $x$ and state the domain.
1. $f (x) = \tan (arcsec (x))$
2. $g ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis close parenthesis$

1. Using Theorem 12.16, we have $arcsec (2) = \arccos (\frac{1}{2}) = \frac{π}{3}$ .
2. Once again, Theorem 12.16 comes to our aid giving $arccsc (- 2) = \arcsin (- \frac{1}{2}) = - \frac{π}{6}$ .
3. Since $\frac{5 π}{4}$ doesn't fall between $0$ and $\frac{π}{2}$ or $\frac{π}{2}$ and $π$ , we cannot use the inverse property stated in Theorem 12.16. Hence, we work from the `inside out.'
  We get: $arcsec (\sec (\frac{5 π}{4})) = arcsec (- \sqrt{2}) = \arccos (- \frac{1}{\sqrt{2}}) = \arccos (- \frac{\sqrt{2}}{2}) = \frac{3 π}{4}$ .
4. We begin simplifying $cot ⁡ open parenthesis arccsc ⁢ open parenthesis minus 3 close parenthesis close parenthesis$ by letting $t = arccsc (- 3)$ . Then, $\csc (t) = - 3$ . Since $csc ⁡ open parenthesis t close parenthesis less than 0$ , $t$ lies in the interval $[- \frac{π}{2}, 0)$ , so $t$ corresponds to a Quadrant IV angle.
  To find $\cot (arccsc (- 3)) = \cot (t)$ , we use the Pythagorean Identity: $\cot^{2} (t) = \csc^{2} (t) - 1$ . We get $\csc^{2} (t) = {(- 3)}^{2} - 1 = 8$ , or $\cot (t) = \pm \sqrt{8} = \pm 2 \sqrt{2}$ .
  Since $t$ corresponds to a Quadrant IV angle, $cot ⁡ open parenthesis t close parenthesis less than 0$ . Hence, $\cot (arccsc (- 3)) = - 2 \sqrt{2}$ .
1. Proceeding as above, we let $t = arcsec (x)$ . Then, $\sec (t) = x$ for $t$ in $[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$ . We seek a formula for $\tan (arcsec (x)) = \tan (t)$ in terms of $x$ .
  To relate $\sec (t)$ to $\tan (t)$ , we use the Pythagorean Identity: $\tan^{2} (t) = \sec^{2} (t) - 1$ . Substituting $\sec (t) = x$ , we get $\tan^{2} (t) = \sec^{2} (t) - 1 = x^{2} - 1$ , so $\tan (t) = \pm \sqrt{x^{2} - 1}$ .
  If $t$ belongs to $[0, \frac{π}{2})$ then $tan ⁡ open parenthesis t close parenthesis greater than or equal to 0$ . On the the other hand, if $t$ belongs to $(\frac{π}{2}, π]$ then $tan ⁡ open parenthesis t close parenthesis less than or equal to 0$ . As a result, we get a piecewise defined function for $\tan (t)$ :
  $\tan (t) = {\begin{matrix} \sqrt{x^{2} - 1}, & if 0 \leq t < \frac{π}{2} \\ - \sqrt{x^{2} - 1}, & if \frac{π}{2} < t \leq π \end{matrix}$
  Now we need to determine what these conditions on $t$ mean for $x$ . Since $x = \sec (t)$ , when $0 \leq t < \frac{π}{2}$ , $x greater than or equal to 1$ , and when $\frac{π}{2} < t \leq π$ , $x less than or equal to minus 1$ . Hence,
  $f (x) = \tan (arcsec (x)) = {\begin{matrix} \sqrt{x^{2} - 1}, & if x \geq 1 \\ - \sqrt{x^{2} - 1}, & if x \leq - 1 \end{matrix}$
  To find the domain of $f$ , we consider $f (x) = \tan (arcsec (x))$ as a two step process. First, we have the arcsecant function, whose domain is $open parenthesis ∞, minus 1 close bracket union open bracket 1, ∞ close parenthesis$ .
  Since the range of $arcsec (x)$ is $[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$ , taking the tangent of any output from $arcsec (x)$ is defined. Hence, the domain of $f$ is $(- \infty, - 1] \cup [1, \infty)$ .
2. Taking a cue from the previous problem, we start by letting $t equals arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis$ . Then $csc ⁡ open parenthesis t close parenthesis equals 4 ⁢ x$ for $t$ in $[- \frac{π}{2}, 0) \cup (0, \frac{π}{2}]$ . Our goal is to rewrite $cos ⁡ open parenthesis arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis close parenthesis equals cos ⁡ open parenthesis t close parenthesis$ in terms of $x$ .
  From $csc ⁡ open parenthesis t close parenthesis equals 4 ⁢ x$ , we get $\sin (t) = \frac{1}{4 x}$ , so to find $\cos (t)$ , we can make use of the Pythagorean Identity: $\cos^{2} (t) = 1 - \sin^{2} (t)$ . Substituting $\sin (t) = \frac{1}{4 x}$ gives $\cos^{2} (t) = 1 - {(\frac{1}{4 x})}^{2} = 1 - \frac{1}{16 x^{2}}$ . Getting a common denominator and extracting square roots, we obtain:
  $\cos (t) = \pm \sqrt{\frac{16 x^{2} - 1}{16 x^{2}}} = \pm \frac{\sqrt{16 x^{2} - 1}}{4 | x |} .$
  Since $t$ belongs to $[- \frac{π}{2}, 0) \cup (0, \frac{π}{2}]$ , we know $cos ⁡ open parenthesis t close parenthesis greater than or equal to 0$ , so we choose $\cos (t) = \frac{\sqrt{16 - x^{2}}}{4 | x |}$ . (The absolute values here are necessary, since $x$ could be negative.) Hence,
  $g (x) = \cos (arccsc (4 x)) = \frac{\sqrt{16 - x^{2}}}{4 | x |} .$
  To find the domain of $g ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis close parenthesis$ , as usual, we think of $g$ as a series of processes. First, we take the input, $x$ , and multiply it by $4$ . Since this can be done to any real number, we have no restrictions here.
  Next, we take the arccosecant of $4 ⁢ x$ . Using interval notation, the domain of the arccosecant function is written as: $(- \infty, - 1] \cup [1, \infty)$ . Hence to take the arccosecant of $4 ⁢ x$ , the quantity $4 ⁢ x$ must lie in one of these two intervals.⁷ That is, $4 ⁢ x less than or equal to minus 1$ or $4 ⁢ x greater than or equal to 1$ , so $x \leq - \frac{1}{4}$ or $x \geq \frac{1}{4}$ .
  The third and final process coded in $g ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis close parenthesis$ is to take the cosine of $arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis$ . Since the cosine accepts any real number, we have no additional restrictions. Hence, the domain of $g$ is $(- \infty, - \frac{1}{4}] \cup [\frac{1}{4}, \infty)$ .

As promised in the discussion following Example 12.3.2, in which we used the methods from Section 11.4 to circumvent some onerous identity work, we take some time here to revisit number to see what issues arise when we take a Section 11.4 approach here.

As above, we start rewriting $f (x) = \tan (arcsec (x))$ by letting $t = arcsec (x)$ so that $\sec (t) = x$ where $0 \leq t < \frac{π}{2}$ or $\frac{π}{2} < t \leq π$ . We let $θ = t$ radians and wish to view $\sec (θ) = \sec (t) = x$ as described in Theorem 11.9: the ratio of the radius of a circle, $r$ centered at the origin, divided by the abscissa⁸ of a point on the terminal side of $θ$ which intersects said circle.

If we make the usual identification $\sec (θ) = x = \frac{x}{1}$ , we see that if $0 \leq θ < \frac{π}{2}$ , then $x equals sec ⁡ θ greater than or equal to 1$ , 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 $θ$ 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, $\frac{π}{2} < t \leq π$ , then $x = \sec (t) \leq - 1$ , so we need to rewrite $\sec (θ) = x = \frac{x}{1} = \frac{- x}{- 1}$ in order to keep the radius of the circle, $r = - x > 0$ and the abscissa, $minus 1 less than 0$ . From ${(- 1)}^{2} + y^{2} = {(- x)}^{2}$ , we still get $y = \sqrt{x^{2} - 1}$ , as shown below on the right.

$0 < θ < \frac{π}{2}$ , $x greater than or equal to 1$ , $r = x$ .

$\frac{π}{2} < θ < π$ , $x less than or equal to minus 1$ , $r = - x$ .

In the Quadrant I case, when $x greater than or equal to 1$ , we get $\tan (θ) = \frac{\sqrt{x^{2} - 1}}{1} = \sqrt{x^{2} - 1}$ . In Quadrant II, when $x less than or equal to minus 1$ , we obtain $\tan (θ) = \frac{\sqrt{x^{2} - 1}}{- 1} = - \sqrt{x^{2} - 1}$ . Hence, we get the piecewise definition for $f (x)$ as we did in number above: $f (x) = \tan (arcsec (x)) = \sqrt{x^{2} - 1}$ if $x greater than or equal to 1$ and $f (x) = \tan (arcsec (x)) = - \sqrt{x^{2} - 1}$ if $x less than or equal to minus 1$ .

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 12.3.3 number. 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.

Inverses of Secant and Cosecant: Calculus Friendly Approach

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

$\overset{\overset{reflect across y = t}{\overset{}{\to}}}{switch t and y coordinates}$

$\overset{\overset{reflect across y = t}{\overset{}{\to}}}{switch x and y coordinates}$

Theorem

Properties of the Arcsecant and Arccosecant Functions ⁹

Properties of $F (x) = arcsec (x)$
- Domain: ${x | | x | \geq 1} = (- \infty, - 1] \cup [1, \infty)$
- Range: $[0, \frac{π}{2}) \cup [π, \frac{3 π}{2})$
- $lim x → minus ∞ arcsec ⁢ open parenthesis x close parenthesis$ $= \frac{3 π}{2}$ ; $lim x → ∞ arcsec ⁢ open parenthesis x close parenthesis$ $= \frac{π}{2}$
- $arcsec (x) = t$ if and only if $\sec (t) = x$ and $0 \leq t < \frac{π}{2}$ or $π \leq t < \frac{3 π}{2}$ .
- $arcsec (x) = \arccos (\frac{1}{x})$ for $x greater than or equal to 1$ only
- $\sec (arcsec (x)) = x$ provided $vertical bar x vertical bar greater than or equal to 1$
- $arcsec (\sec (t)) = t$ provided $0 \leq t < \frac{π}{2}$ or $π \leq t < \frac{3 π}{2}$
Properties of $G (x) = arccsc (x)$
- Domain: ${x | | x | \geq 1} = (- \infty, - 1] \cup [1, \infty)$
- Range: $(0, \frac{π}{2}] \cup (π, \frac{3 π}{2}]$
- $lim_{x \to - \infty} arccsc (x) = π$ ; $lim x → ∞ arccsc ⁢ open parenthesis x close parenthesis equals 0$
- $arccsc (x) = t$ if and only if $\csc (t) = x$ and $0 < t \leq \frac{π}{2}$ or $π < t \leq \frac{3 π}{2}$
- $arccsc (x) = \arcsin (\frac{1}{x})$ for $x greater than or equal to 1$ only
- $\csc (arccsc (x)) = x$ provided $vertical bar x vertical bar greater than or equal to 1$
- $arccsc (\csc (t)) = t$ provided $0 < t \leq \frac{π}{2}$ or $π < t \leq \frac{3 π}{2}$

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 $arcsec (x) = \arccos (\frac{1}{x})$ and $arccsc (x) = \arcsin (\frac{1}{x})$ hold for $x greater than or equal to 1$ 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 12.3.3. The interested reader is invited to see what differences are to be had as a consequence of the change in ranges.

Example 4

Find the exact values of the following.
1. $arcsec ⁢ open parenthesis 2 close parenthesis$
2. $arccsc ⁢ open parenthesis minus 2 close parenthesis$
3. $arcsec (\sec (\frac{5 π}{4}))$
4. $cot ⁡ open parenthesis arccsc ⁢ open parenthesis minus 3 close parenthesis close parenthesis$
Rewrite each of the following composite functions as algebraic functions of $x$ and state the domain.
1. $\tan (arcsec (x))$
2. $cos ⁡ open parenthesis arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis close parenthesis$

1. Since $2 greater than or equal to 1$ , we may invoke Theorem 12.17 to get $arcsec (2) = \arccos (\frac{1}{2}) = \frac{π}{3}$ .
2. Unfortunately, $minus 2$ is not greater to or equal to $1$ , so we cannot apply Theorem 12.17 to $arccsc ⁢ open parenthesis minus 2 close parenthesis$ and convert this into an arcsine problem. Instead, we appeal to the definition.
  To find $t = arccsc (- 2)$ , we need the angle measuring $t$ radians with $\csc (t) = - 2$ and is either between $0$ and $\frac{π}{2}$ or between $π$ and $\frac{3 π}{2}$ .
  Since $csc ⁡ open parenthesis t close parenthesis less than 0$ , we know $t$ corresponds to an angle between $π$ and $\frac{3 π}{2}$ , and since $\csc (t) = - 2$ , we know $\sin (t) = - \frac{1}{2}$ . Hence, $t = arccsc (- 2) = \frac{7 π}{6}$ .
3. Since $\frac{5 π}{4}$ lies between $π$ and $\frac{3 π}{2}$ , Theorem 12.17 applies: $arcsec (\sec (\frac{5 π}{4})) = \frac{5 π}{4}$ .
  We encourage the reader to work this through using the definition as we have done in the previous examples to see how it goes.
4. To simplify $cot ⁡ open parenthesis arccsc ⁢ open parenthesis minus 3 close parenthesis close parenthesis$ we let $t = arccsc (- 3)$ so that $\cot (arccsc (- 3)) = \cot (t)$ .
  We know $\csc (t) = - 3$ , and since this is negative, $t$ lies in $(π, \frac{3 π}{2}]$ . To get from $\csc (t)$ to $\cot (t)$ , we use the Pythagorean Identity: $\cot^{2} (t) = \csc^{2} (t) - 1$ . We find $\cot^{2} (t) = {(- 3)}^{2} - 1 = 8$ so that $\cot (t) = \pm \sqrt{8} = \pm 2 \sqrt{2}$ .
  Since $t$ is in the interval $(π, \frac{3 π}{2}]$ , $t$ corresponds to a Quadrant III angle, so we know $cot ⁡ open parenthesis t close parenthesis greater than 0$ . Hence, our answer is $\cot (arccsc (- 3)) = 2 \sqrt{2}$ .
1. We begin rewriting $f (x) = \tan (arcsec (x))$ by letting $t = arcsec (x)$ . Hence, $\sec (t) = x$ where either $0 \leq t < \frac{π}{2}$ or $π \leq t < \frac{3 π}{2}$ . Our goal is to find an expression for $\tan (t)$ in terms of $x$ .
  To relate $\sec (t)$ to $\tan (t)$ , we use the Pythagorean Identity: $\tan^{2} (t) = \sec^{2} (t) - 1 = x^{2} - 1$ so that $\tan (t) = \pm \sqrt{x^{2} - 1}$ . Since $t$ lies in $[0, \frac{π}{2}) \cup [π, \frac{3 π}{2})$ , $tan ⁡ open parenthesis t close parenthesis greater than or equal to 0$ , so we choose $\tan (t) = \sqrt{x^{2} - 1}$ . Hence, $f (x) = \tan (arcsec (x)) = \sqrt{x^{2} - 1}$ .
  For the domain of $f$ , we note that the domain of $arcsec (x)$ is $(- \infty, - 1] \cup [1, \infty)$ . Since all values in the range of arcsecant, $[0, \frac{π}{2}) \cup [π, \frac{3 π}{2})$ , are in the domain of the tangent function, we find the domain of $f$ is $(- \infty, - 1] \cup [1, \infty)$ .
2. To rewrite $g ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis close parenthesis$ , we start by letting $t equals arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis$ . Then $csc ⁡ open parenthesis t close parenthesis equals 4 ⁢ x$ where either $0 < t \leq \frac{π}{2}$ or $π < t \leq \frac{3 π}{2}$ . Our goal is to find an expression for $\cos (t)$ in terms of $x$ .
  From $csc ⁡ open parenthesis t close parenthesis equals 4 ⁢ x$ , we get $\sin (t) = \frac{1}{4 x}$ , so to find $\cos (t)$ , we can make use of the Pythagorean Identity: $\cos^{2} (t) = 1 - \sin^{2} (t)$ . Substituting $\sin (t) = \frac{1}{4 x}$ gives $\cos^{2} (t) = 1 - {(\frac{1}{4 x})}^{2} = 1 - \frac{1}{16 x^{2}}$ . Getting a common denominator and extracting square roots, we obtain:
  $\cos (t) = \pm \sqrt{\frac{16 x^{2} - 1}{16 x^{2}}} = \pm \frac{\sqrt{16 x^{2} - 1}}{4 | x |} .$
  If $t$ lies in $(0, \frac{π}{2}]$ , then $cos ⁡ open parenthesis t close parenthesis greater than or equal to 0$ , and we choose $\cos (t) = \frac{\sqrt{16 x^{2} - 1}}{4 | x |}$ . Here, $x = \frac{1}{4} \csc (t) > 0$ as well, so we can disregard the absolute values here and write $\cos (t) = \frac{\sqrt{16 x^{2} - 1}}{4 x}$ .
  If $t$ belongs to $(π, \frac{3 π}{2}]$ , then $cos ⁡ open parenthesis t close parenthesis less than or equal to 0$ , so, we choose $\cos (t) = - \frac{\sqrt{16 x^{2} - 1}}{4 | x |}$ . In this case, $x = \frac{1}{4} \csc (t) < 0$ , so $| x | = - x$ (see Section 1.3 for a refresher, if needs be!) and so,
  $\cos (t) = - \frac{\sqrt{16 x^{2} - 1}}{4 | x |} = - \frac{\sqrt{16 x^{2} - 1}}{4 (- x)} = \frac{\sqrt{16 x^{2} - 1}}{4 x} .$
  Hence, in both cases, we get
  $g (x) = \cos (arccsc (4 x)) = \frac{\sqrt{16 x^{2} - 1}}{4 x} .$
  To find the domain of $g ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis close parenthesis$ , as usual, we think of $g$ as a series of processes. First, we take the input, $x$ , and multiply it by $4$ . Since this can be done to any real number, we have no restrictions here.
  Next, we take the arccosecant of $4 ⁢ x$ . Using interval notation, the domain of the arccosecant function is written as: $(- \infty, - 1] \cup [1, \infty)$ . Hence to take the arccosecant of $4 ⁢ x$ , the quantity $4 ⁢ x$ must lie in one of these two intervals.¹⁰ That is, $4 ⁢ x less than or equal to minus 1$ or $4 ⁢ x greater than or equal to 1$ , so $x \leq - \frac{1}{4}$ or $x \geq \frac{1}{4}$ .
  The third and final process coded in $g ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis close parenthesis$ is to take the cosine of $arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis$ . Since the cosine accepts any real number, we have no additional restrictions. Hence, the domain of $g$ is $(- \infty, - \frac{1}{4}] \cup [\frac{1}{4}, \infty)$ .

For completeness, we embark here on a discussion of how the techniques from Section 11.4, in particular Theorem 11.9 can be used to circumvent some of the identity work in number above.¹¹

As above, we start rewriting $f (x) = \tan (arcsec (x))$ by letting $t = arcsec (x)$ so that $\sec (t) = x$ where $0 \leq t < \frac{π}{2}$ or $π \leq t < \frac{3 π}{2}$ . We let $θ = t$ radians and wish to view $\sec (θ) = \sec (t) = x$ as described in Theorem 11.9: the ratio of the radius of a circle, $r$ centered at the origin, divided by the abscissa¹² of a point on the terminal side of $θ$ which intersects said circle.

If, however, $π \leq t < \frac{3 π}{2}$ , then $x = \sec (t) \leq - 1$ , so we need to rewrite $\sec (θ) = x = \frac{x}{1} = \frac{- x}{- 1}$ in order to keep the radius of the circle, $r = - x > 0$ and the abscissa, $minus 1 less than 0$ . 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.

$0 < θ < \frac{π}{2}$ , $x greater than or equal to 1$ , $r = x$ .

$π < θ < \frac{3 π}{2}$ , $x less than or equal to minus 1$ , $r = - x$ .

In the Quadrant I case, when $x greater than or equal to 1$ , we get $\tan (θ) = \frac{\sqrt{x^{2} - 1}}{1} = \sqrt{x^{2} - 1}$ . In Quadrant III, when $x less than or equal to minus 1$ , we obtain $\tan (θ) = \frac{- \sqrt{x^{2} - 1}}{- 1} = \sqrt{x^{2} - 1}$ . Hence, in both cases, we obtain the same answer as we did in number above: $f (x) = \tan (arcsec (x)) = \sqrt{x^{2} - 1}$ for $x$ in $(- \infty, - 1] \cup [1, \infty)$ .

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.

Example 5

Use a calculator to approximate the following values to four decimal places.
1. $arccot ⁢ open parenthesis 2 close parenthesis$
2. $arcsec ⁢ open parenthesis 5 close parenthesis$
3. $arccot ⁢ open parenthesis minus 2 close parenthesis$
4. $arccsc (- \frac{3}{2})$
Find the domain and range of the following functions. Check your answers using a graphing utility.
1. $f (x) = \frac{π}{2} - \arccos (\frac{x}{5})$
2. $g ⁢ open parenthesis x close parenthesis equals 3 ⁢ arctan ⁡ open parenthesis 4 ⁢ x close parenthesis$ .

1. Since $2 greater than 0$ , we can use the property listed in Theorem 12.15 to get: $arccot (2) = \arctan (\frac{1}{2})$ . In `radian' mode, we find $arccot (2) = \arctan (\frac{1}{2}) \approx 0.4636$ .
2. Since $5 greater than or equal to 1$ , we can use the property from either Theorem 12.16 or Theorem 12.17 to write $arcsec (5) = \arccos (\frac{1}{5}) \approx 1.3694$ .
3. Since the argument $minus 2$ is negative, we cannot directly apply Theorem 12.15 to help us find $arccot ⁢ open parenthesis minus 2 close parenthesis$ , so we appeal to the definition.
  The number $t = arccot (- 2)$ corresponds to an angle $θ = t$ radians with $\cot (θ) = - 2$ which lies between $0$ and $π$ . Moreover, since $cot ⁡ open parenthesis θ close parenthesis less than 0$ , we know $θ$ is a Quadrant II angle.
  Let $α$ be the reference angle for $θ$ , as pictured below on the left. By definition, $α$ is an acute angle which means $0 < α < \frac{π}{2}$ . By The Reference Angle Theorem, Theorem 11.1, we also know $cot ⁡ open parenthesis α close parenthesis equals 2$ . Hence, by definition, $α equals arccot ⁢ open parenthesis 2 close parenthesis$ radians.
  Since the argument of arccotangent is now a positive $2$ , we can use Theorem 12.15 to get $α = arccot (2) = \arctan (\frac{1}{2})$ radians. Since $θ = π - α = π - \arctan (\frac{1}{2}) \approx 2.6779$ radians, we get $arccot (- 2) \approx 2.6779$ .
  Figure 12.40 $α$ is a reference angle of $θ$
  Figure 12.41 $β = \arctan (- \frac{1}{2})$ radians
  Another way to attack the problem is to use $\arctan (- \frac{1}{2})$ . By definition, we have the real number $t = \arctan (- \frac{1}{2})$ satisfies $\tan (t) = - \frac{1}{2}$ with $- \frac{π}{2} < t < \frac{π}{2}$ . Since $tan ⁡ open parenthesis t close parenthesis less than 0$ , we know more specifically that $- \frac{π}{2} < t < 0$ , so $t$ corresponds to an angle $β$ in Quadrant IV. We sketch $β$ along with $θ = arccot (- 2)$ radians above on the right.
  To find the value of $arccot ⁢ open parenthesis minus 2 close parenthesis$ , we once again visualize the angle $θ = arccot (- 2)$ radians and note that it is a Quadrant II angle with $\tan (θ) = - \frac{1}{2}$ . This means it is exactly $π$ units away from $β$ , and we get $θ = π + β = π + \arctan (- \frac{1}{2}) \approx 2.6779$ radians. Hence, as before, $arccot (- 2) \approx 2.6779$ .
4. If the range of arccosecant is taken to be $[- \frac{π}{2}, 0) \cup (0, \frac{π}{2}]$ , we can use Theorem 12.16 to get $arccsc (- \frac{3}{2}) = \arcsin (- \frac{2}{3}) \approx - 0.7297$ .
  If, on the other hand, the range of arccosecant is taken to be $(0, \frac{π}{2}] \cup (π, \frac{3 π}{2}]$ , then we proceed as in the previous problem by letting $t = arccsc (- \frac{3}{2})$ .
  Then $t$ is a real number with $\csc (t) = - \frac{3}{2}$ and since $csc ⁡ open parenthesis t close parenthesis less than 0$ , we know $π < θ \leq \frac{3 π}{2}$ , Hence, $t$ corresponds to a Quadrant III angle, $θ$ , as depicted below.
  As above, we let $α$ be the reference angle for $θ$ . Then $0 < α < \frac{π}{2}$ and $\csc (α) = \frac{3}{2}$ , which means $α = arccsc (\frac{3}{2})$ radians.
  Since the argument of arccosecant is now positive, Theorem 12.17 applies so we can rewrite $α = arccsc (\frac{3}{2}) = \arcsin (\frac{2}{3})$ radians. Since $θ = π + α = π + \arcsin (\frac{2}{3}) \approx 3.8713$ radians, we have that in this case, $arccsc (- \frac{3}{2}) \approx 3.8713$ .
  Figure 12.42
1. To find the domain of $f$ , we can think of the function as a sequence of steps and track our inputs through each step and track the restrictions that arise. To that end, we rewrite $f (x)$ as we did in Section 5.4: $f (x) = \frac{π}{2} - \arccos (\frac{x}{5}) = - \arccos (\frac{x}{5}) + \frac{π}{2}$ .
  Starting with a real number $x$ , we divide by $5$ to obtain $\frac{x}{5}$ . So far, we have no restrictions.
  Next, we take the arccosine of $\frac{x}{5}$ . Since the arccosine function only admits inputs between $minus 1$ and $1$ inclusive, we require that $- 1 \leq \frac{x}{5} \leq 1$ . Solving, we get $minus 5 less than or equal to x less than or equal to 5$ .
  Moving outside the arccosine, we multiply the outputs from the arccosine by $minus 1$ and then add $\frac{π}{2}$ . Since these are defined for all real numbers, we have our domain restricted only by the arccosine itself. Hence, the domain of $f$ is $open bracket minus 5, 5 close bracket$ .
  To determine the range of $f$ , we work through the steps above, this time paying attention to the outputs from each step. We know our domain is restricted to $open bracket minus 5, 5 close bracket$ due to the arccosine, so we start with the range of arccosine: $open bracket 0, π close bracket$ .
  Let $y$ represent a typical output from $\arccos (x)$ . Then $0 less than or equal to y less than or equal to π$ . From our work above, we know the arccosine values are first multiplied by $minus 1$ and then added to $\frac{π}{2}$ , so we apply these same operations to the inequality $0 less than or equal to y less than or equal to π$ .
  Multiplying this inequality through by $minus 1$ gives $- π \leq - y \leq 0$ . Adding through by $\frac{π}{2}$ gives $- \frac{π}{2} \leq - y + \frac{π}{2} \leq \frac{π}{2}$ . Hence the range of $f (x) = - \arccos (\frac{x}{5}) + \frac{π}{2}$ is $[- \frac{π}{2}, \frac{π}{2}]$ . Our graph below on the left confirms our results.¹³
2. As with the previous example, we think of $g ⁢ open parenthesis x close parenthesis equals 3 ⁢ arctan ⁡ open parenthesis 4 ⁢ x close parenthesis$ as a series of steps in order to find the domain and track the range.
  Starting with an input $x$ , we multiply it by $4$ , take the arctangent, and then multiply that result by $3$ . Since all of these operations are defined for all real numbers, we conclude the domain of $g$ is also all real numbers, or $open parenthesis minus ∞, ∞ close parenthesis$ .
  Looking at outputs, we find that the range first becomes limited when taking the arctangent. If $y$ represent a typical output from $\arctan (x)$ , then $- \frac{π}{2} < y < \frac{π}{2}$ .
  In our formula for $g (x)$ , the outputs from the arctangent are multiplied by $3$ . Multiplying the inequality $- \frac{π}{2} < y < \frac{π}{2}$ through by $3$ gives $- \frac{3 π}{2} < 3 y < \frac{3 π}{2}$ .
  Hence the range of $g$ is $(- \frac{3 π}{2}, \frac{3 π}{2})$ . Our answers are confirmed by examining the graph of $g$ below on the right.
  Figure 12.43
  Figure 12.44
  $y = f (x) = \frac{π}{2} - \arccos (\frac{x}{5})$
  $y = g (x) = 3 \arctan (4 x)$

Solving Equations Using the Inverse Trigonometric Functions.

In Sections 11.2 and 11.4, we learned how to solve equations like $\sin (θ) = \frac{1}{2}$ 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 the related section.

If, on the other hand, we had been asked to find all angles with $\sin (θ) = \frac{1}{3}$ 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 $x squared equals 4$ is a lot like $\sin (θ) = \frac{1}{2}$ in that it has friendly, `common value' answers $x = \pm 2$ . The equation $x squared equals 7$ , on the other hand, is a lot like $\sin (θ) = \frac{1}{3}$ . We know there are answers, but we can't express them using `friendly' numbers.

To solve $x squared equals 7$ , 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 $the square root of 7$ 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 $\sin (θ) = \frac{1}{3}$ . 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 $\arcsin (\frac{1}{3})$ , as our next example illustrates.

Example 6

Solve the following equations.

Find all angles $θ$ for which $\sin (θ) = \frac{1}{3}$ .
Find all real numbers $t$ for which $\tan (t) = - 2$
Solve $\sec (x) = - \frac{5}{3}$ for $x$ .

If $\sin (θ) = \frac{1}{3}$ , then the terminal side of $θ$ , when plotted in standard position, intersects the Unit Circle at $y = \frac{1}{3}$ . Geometrically, we see that this happens at two places: in Quadrant I and Quadrant II.
If we let $α$ denote the acute solution to the equation, then all the solutions to this equation in Quadrant I are coterminal with $α$ , and $α$ serves as the reference angle for all of the solutions to this equation in Quadrant II as seen below.
Since $\frac{1}{3}$ isn't the sine of any of the `common angles' we've encountered, we use the arcsine functions to express our answers. By definition, real number $t = \arcsin (\frac{1}{3})$ $\sin (t) = \frac{1}{3}$ with $0 < t < \frac{π}{2}$ .
Hence, $α = \arcsin (\frac{1}{3})$ radians is an acute angle with $\sin (α) = \frac{1}{3}$ . Since all of the Quadrant I solutions $θ$ are all coterminal with $α$ , we get $θ = α + 2 π k = \arcsin (\frac{1}{3}) + 2 π k$ for integers $k$ .
Turning our attention to Quadrant II, we get one solution to be $π - α$ . Hence, the Quadrant II solutions are $θ = π - α + 2 π k = π - \arcsin (\frac{1}{3}) + 2 π k$ , for integers $k$ .
Figure 12.45
Figure 12.46
The real number solutions to $\tan (t) = - 2$ correspond to angles $θ$ with $\tan (θ) = - 2$ . Since tangent is negative only in Quadrants II and IV, we focus our efforts there.
The real number $t = \arctan (- 2)$ satisfies $\tan (t) = - 2$ and $- \frac{π}{2} < t < 0$ . If we let $β = \arctan (- 2)$ radians, then all of the Quadrant IV solutions to $\tan (θ) = - 2$ are coterminal with $β$ .
Moreover, as seen below on the right, the solutions from Quadrant II differ by exactly $π$ units from the solutions in Quadrant IV (recall, the period of the tangent function is $π$ .) Hence, all of the solutions to $\tan (θ) = - 2$ are of the form $θ = β + π k = \arctan (- 2) + π k$ for some integer $k$ . Switching back to the variable $t$ , we record our final answer to $\tan (t) = - 2$ as $t = \arctan (- 2) + π k$ for integers $k$ .
Figure 12.47
Figure 12.48
Another tact we could have taken to solve this problem is to use reference angles. Consider the (angle) equation: $\tan (θ) = - 2$ . If we let $α$ be the reference angle for the solutions $θ$ , we know $α$ is an acute angle with $tan ⁡ open parenthesis α close parenthesis equals 2$ .
By definition, the real number $t equals arctan ⁡ open parenthesis 2 close parenthesis$ satisfies $0 < t < \frac{π}{2}$ with $tan ⁡ open parenthesis t close parenthesis equals 2$ . Hence, the angle $α equals arctan ⁡ open parenthesis 2 close parenthesis$ radians is the reference angle for our solutions to $\tan (θ) = - 2$ .
Adjusting for quadrants, we get our answers to $\tan (θ) = - 2$ are $θ = - α + π k = - \arctan (2) + π k$ for integers $k$ . Again, we cosmetically change the variable from $θ$ back to $t$ so our answer to $\tan (t) = - 2$ is $t = - \arctan (2) + π k$ . Thanks to the odd property of arctangent, $\arctan (- 2) = - \arctan (2)$ and we see this family of solutions is identical to what we obtained earlier.
Figure 12.49
Figure 12.50
In the last equation, $\sec (x) = - \frac{5}{3}$ , we are not told whether or not $x$ represents an angle or a real number. This isn't really much of an issue, since we attack both problems the same way.
Taking a cue from our work in Section 11.4 and use a Reciprocal Identity to convert the equation $\sec (x) = - \frac{5}{3}$ to $\cos (x) = - \frac{3}{5}$ . Thinking geometrically, we are looking for angles $θ$ with $\cos (θ) = - \frac{3}{5}$ . Since $cos ⁡ open parenthesis θ close parenthesis less than 0$ , we are looking for solutions in Quadrants II and III. Since $- \frac{3}{5}$ isn't the cosine of any of the `common angles', we'll need to express our solutions in terms of the arccosine function.
Figure 12.51
Figure 12.52
The real number $t = \arccos (- \frac{3}{5})$ is defined so that $\frac{π}{2} < t < π$ with $\cos (t) = - \frac{3}{5}$ . Hence, the angle $β = \arccos (- \frac{3}{5})$ radians is a Quadrant II angle which satisfies $\cos (β) = - \frac{3}{5}$ . To obtain a Quadrant III angle solution, we may simply use $- β = - \arccos (- \frac{3}{5})$ as seen above on the right.
Since all angle solutions are coterminal with $β$ or $- β$ , we get our solutions to $\cos (θ) = - \frac{3}{5}$ to be $θ = β + 2 π k = \arccos (- \frac{3}{5}) + 2 π k$ or $θ = - β + 2 π k = - \arccos (- \frac{3}{5}) + 2 π k$ for integers $k$ .
Switching back to the variable $x$ , we record our final answer to $\sec (x) = - \frac{5}{3}$ as $x = \arccos (- \frac{3}{5}) + 2 π k$ or $x = - \arccos (- \frac{3}{5}) + 2 π k$ for integers $k$ .
As with the previous problem, we can approach solving $\cos (θ) = - \frac{3}{5}$ using reference angles. Letting $α$ represent the reference angle for the solutions $θ$ , we know $α$ is an acute angle with $\cos (α) = \frac{3}{5}$ .
We know the real number $t = \arccos (\frac{3}{5})$ satisfies $\cos (t) = \frac{3}{5}$ and $0 < t < \frac{π}{2}$ , hence $α = \arccos (\frac{3}{5})$ radians is the reference angle for the solutions to $\cos (θ) = - \frac{3}{5}$ .
Hence, the Quadrant II solutions to $\cos (θ) = - \frac{3}{5}$ are $θ = π - α + 2 π k = π - \arccos (\frac{3}{5}) + 2 π k$ while the Quadrant IV solutions to $\cos (θ) = - \frac{3}{5}$ are $θ = π + α + 2 π k = π + \arccos (\frac{3}{5}) + 2 π k$ for integers $k$ .
Figure 12.53
Figure 12.54
Shifting back to the variable $x$ , we get our solution to $\sec (x) = - \frac{5}{3}$ are $x = π - \arccos (\frac{3}{5}) + 2 π k$ or $x = π + \arccos (\frac{3}{5}) + 2 π k$ for integers $k$ . While these certainly look quite different than what we obtained before, $x = \arccos (- \frac{3}{5}) + 2 π k$ or $x = - \arccos (- \frac{3}{5}) + 2 π k$ for integers $k$ , they are, in fact, equivalent. To show this, we start with $\arccos (- \frac{3}{5}) = π - \arccos (\frac{3}{5})$ and begin writing out specific solutions from each family by choosing specific values of $k$ . We leave these details to the reader.

We close this section with one last sinusoid example.

Example 7

Consider the function $f (t) = 3 \cos (6 t) - 4 \sin (6 t)$ . Find a formula for $f (t)$ :

in the form $C (t) = A \cos (ω t + ϕ) + B$ for $ω greater than 0$
in the form $S (t) = A \sin (ω t + ϕ) + B$ for $ω greater than 0$

As in Example 12.2.7, we compare the expanded form of $C (t) = A \cos (ω t) \cos (ϕ) - A \sin (ω t) \sin (ϕ) + B$ with $f (t) = 3 \cos (6 t) - 4 \sin (6 t)$ . We identify $ω equals 6$ and $B equals 0$ and by equating coefficients of $cos ⁡ open parenthesis 6 ⁢ t close parenthesis$ and $sin ⁡ open parenthesis 6 ⁢ t close parenthesis$ get the two equations: $A ⁢ cos ⁡ open parenthesis ϕ close parenthesis equals 3$ and $A ⁢ sin ⁡ open parenthesis ϕ close parenthesis equals 4$ .
Using the Pythagorean Identity to eliminate $ϕ$ , we get $A^{2} = {(A \cos (ϕ))}^{2} + {(A \sin (ϕ))}^{2} = 3^{2} + 4^{2} = 25$ . We choose $A equals 5$ and work to find the phase angle $ϕ$ .
Substituting $A equals 5$ into our two equations relating $A$ and $ϕ$ , we get $5 ⁢ cos ⁡ open parenthesis ϕ close parenthesis equals 3$ , or $\cos (ϕ) = \frac{3}{5}$ and $5 ⁢ sin ⁡ open parenthesis ϕ close parenthesis equals 4$ , so $\sin (ϕ) = \frac{4}{5}$ . Since both $\sin (ϕ)$ and $\cos (ϕ)$ are positive, we know $ϕ$ is a Quadrant I angle. However, since neither the sine nor cosine value of $ϕ$ corresponds to a common angle, we need to express $ϕ$ in terms of either an arcsine or arccosine.
Since the real number $t = \arccos (\frac{3}{5})$ satisfies $\cos (t) = \frac{3}{5}$ and $0 < t < \frac{π}{2}$ , we know the angle $ϕ = \arccos (\frac{3}{5})$ radians is an acute (Quadrant I) angle which satisfies $\cos (ϕ) = \frac{3}{5}$ . Hence, we can take $ϕ = \arccos (\frac{3}{5})$ and write $f (t) = 5 \cos (6 t + \arccos (\frac{3}{5}))$ .
In addition, the real number $t = \arcsin (\frac{4}{5})$ satisfies $\sin (t) = \frac{4}{5}$ and $0 < t < \frac{π}{2}$ . Hence $ϕ = \arcsin (\frac{4}{5})$ radians is Quadrant I angle with $\sin (ϕ) = \frac{4}{5}$ . This means we could also take $ϕ = \arcsin (\frac{4}{5})$ and write $f (t) = 5 \cos (6 t + \arcsin (\frac{4}{5}))$ . (We could also express $ϕ$ in terms of arctangents, if we wanted!)
We leave it to the reader to verify (both) solutions analytically and graphically.
Once again, we equate the expanded form of $S (t) = A \sin (ω t) \cos (ϕ) + A \cos (ω t) \sin (ϕ) + B$ with $f (t) = 3 \cos (6 t) - 4 \sin (6 t)$ . Once again, we get $ω equals 6$ and $B equals 0$ . Here, our two equations for $A$ and $ϕ$ are $A \cos (ϕ) = - 4$ and $A ⁢ sin ⁡ open parenthesis ϕ close parenthesis equals 3$ .
As before, we get $A^{2} = {(A \cos (ϕ))}^{2} + {(A \sin (ϕ))}^{2} = {(- 4)}^{2} + 3^{2} = 25$ , and we choose $A equals 5$ . Our equations for $ϕ$ become: $\cos (ϕ) = - \frac{4}{5}$ and $\sin (ϕ) = \frac{3}{5}$ . Since $cos ⁡ open parenthesis ϕ close parenthesis less than 0$ but $sin ⁡ open parenthesis ϕ close parenthesis greater than 0$ , we know $ϕ$ is a Quadrant II angle. As before, since neither the sine nor cosine value of $ϕ$ corresponds to a common angle, we need to express $ϕ$ in terms of either an arcsine or arccosine.
Here, we opt to use the arccosine function, since the range of arccosine, $open bracket 0, π close bracket$ covers Quadrant II. From $\cos (ϕ) = - \frac{4}{5}$ , we get $ϕ = \arccos (- \frac{4}{5})$ , so $f (t) = 5 \sin (6 t + \arccos (- \frac{4}{5}))$ .
Had we chosen to work with arcsines, we would need a Quadrant II solution to $\sin (ϕ) = \frac{3}{5}$ . Going through the usual machinations, we arrive at $ϕ = π - \arcsin (\frac{3}{5})$ . Hence, an alternative form of our answer is $f (t) = 5 \sin (6 t + π - \arcsin (\frac{3}{5}))$ . We leave the checks to the reader.

Exercises

In Exercises -, find the exact value.

$arcsin ⁡ open parenthesis minus 1 close parenthesis$
$\arcsin (- \frac{\sqrt{3}}{2})$
$\arcsin (- \frac{\sqrt{2}}{2})$
$\arcsin (- \frac{1}{2})$
$arcsin ⁡ open parenthesis 0 close parenthesis$
$\arcsin (\frac{1}{2})$
$\arcsin (\frac{\sqrt{2}}{2})$
$\arcsin (\frac{\sqrt{3}}{2})$
$arcsin ⁡ open parenthesis 1 close parenthesis$
$arccos ⁡ open parenthesis minus 1 close parenthesis$
$\arccos (- \frac{\sqrt{3}}{2})$
$\arccos (- \frac{\sqrt{2}}{2})$
$\arccos (- \frac{1}{2})$
$arccos ⁡ open parenthesis 0 close parenthesis$
$\arccos (\frac{1}{2})$
$\arccos (\frac{\sqrt{2}}{2})$
$\arccos (\frac{\sqrt{3}}{2})$
$arccos ⁡ open parenthesis 1 close parenthesis$
$\arctan (- \sqrt{3})$
$arctan ⁡ open parenthesis minus 1 close parenthesis$
$\arctan (- \frac{\sqrt{3}}{3})$
$arctan ⁡ open parenthesis 0 close parenthesis$
$\arctan (\frac{\sqrt{3}}{3})$
$arctan ⁡ open parenthesis 1 close parenthesis$
$arctan ⁡ open parenthesis the square root of 3 close parenthesis$
$arccot (- \sqrt{3})$
$arccot ⁢ open parenthesis minus 1 close parenthesis$
$arccot (- \frac{\sqrt{3}}{3})$
$arccot ⁢ open parenthesis 0 close parenthesis$
$arccot (\frac{\sqrt{3}}{3})$
$arccot ⁢ open parenthesis 1 close parenthesis$
$arccot ⁢ open parenthesis the square root of 3 close parenthesis$
$arcsec ⁢ open parenthesis 2 close parenthesis$
$arccsc ⁢ open parenthesis 2 close parenthesis$
$arcsec ⁢ open parenthesis the square root of 2 close parenthesis$
$arccsc ⁢ open parenthesis the square root of 2 close parenthesis$
$arcsec (\frac{2 \sqrt{3}}{3})$
$arccsc (\frac{2 \sqrt{3}}{3})$
$arcsec ⁢ open parenthesis 1 close parenthesis$
$arccsc ⁢ open parenthesis 1 close parenthesis$

In Exercises -, assume that the range of arcsecant is $[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$ and that the range of arccosecant is $[- \frac{π}{2}, 0) \cup (0, \frac{π}{2}]$ when finding the exact value. (See Section 12.3.1.)

$arcsec ⁢ open parenthesis minus 2 close parenthesis$
$arcsec (- \sqrt{2})$
$arcsec (- \frac{2 \sqrt{3}}{3})$
$arcsec ⁢ open parenthesis minus 1 close parenthesis$
$arccsc ⁢ open parenthesis minus 2 close parenthesis$
$arccsc (- \sqrt{2})$
$arccsc (- \frac{2 \sqrt{3}}{3})$
$arccsc ⁢ open parenthesis minus 1 close parenthesis$

In Exercises -, assume that the range of arcsecant is $[0, \frac{π}{2}) \cup [π, \frac{3 π}{2})$ and that the range of arccosecant is $(0, \frac{π}{2}] \cup (π, \frac{3 π}{2}]$ when finding the exact value. (See Section 12.3.2.)

$arcsec ⁢ open parenthesis minus 2 close parenthesis$
$arcsec (- \sqrt{2})$
$arcsec (- \frac{2 \sqrt{3}}{3})$
$arcsec ⁢ open parenthesis minus 1 close parenthesis$
$arccsc ⁢ open parenthesis minus 2 close parenthesis$
$arccsc (- \sqrt{2})$
$arccsc (- \frac{2 \sqrt{3}}{3})$
$arccsc ⁢ open parenthesis minus 1 close parenthesis$

In Exercises -, find the exact value or state that it is undefined.

$\sin (\arcsin (\frac{1}{2}))$
$\sin (\arcsin (- \frac{\sqrt{2}}{2}))$
$\sin (\arcsin (\frac{3}{5}))$
$sin ⁡ open parenthesis arcsin ⁡ open parenthesis minus 0.42 close parenthesis close parenthesis$
$\sin (\arcsin (\frac{5}{4}))$
$\cos (\arccos (\frac{\sqrt{2}}{2}))$
$\cos (\arccos (- \frac{1}{2}))$
$\cos (\arccos (\frac{5}{13}))$
$cos ⁡ open parenthesis arccos ⁡ open parenthesis minus 0.998 close parenthesis close parenthesis$
$\cos (\arccos (π))$
$tan ⁡ open parenthesis arctan ⁡ open parenthesis minus 1 close parenthesis close parenthesis$
$tan ⁡ open parenthesis arctan ⁡ open parenthesis the square root of 3 close parenthesis close parenthesis$
$\tan (\arctan (\frac{5}{12}))$
$tan ⁡ open parenthesis arctan ⁡ open parenthesis 0.965 close parenthesis close parenthesis$
$tan ⁡ open parenthesis arctan ⁡ open parenthesis 3 ⁢ π close parenthesis close parenthesis$
$cot ⁡ open parenthesis arccot ⁢ open parenthesis 1 close parenthesis close parenthesis$
$\cot (arccot (- \sqrt{3}))$
$\cot (arccot (- \frac{7}{24}))$
$cot ⁡ open parenthesis arccot ⁢ open parenthesis minus 0.001 close parenthesis close parenthesis$
$\cot (arccot (\frac{17 π}{4}))$
$sec ⁡ open parenthesis arcsec ⁢ open parenthesis 2 close parenthesis close parenthesis$
$sec ⁡ open parenthesis arcsec ⁢ open parenthesis minus 1 close parenthesis close parenthesis$
$\sec (arcsec (\frac{1}{2}))$
$sec ⁡ open parenthesis arcsec ⁢ open parenthesis 0.75 close parenthesis close parenthesis$
$sec ⁡ open parenthesis arcsec ⁢ open parenthesis 117 ⁢ π close parenthesis close parenthesis$
$csc ⁡ open parenthesis arccsc ⁢ open parenthesis the square root of 2 close parenthesis close parenthesis$
$\csc (arccsc (- \frac{2 \sqrt{3}}{3}))$
$\csc (arccsc (\frac{\sqrt{2}}{2}))$
$csc ⁡ open parenthesis arccsc ⁢ open parenthesis 1.0001 close parenthesis close parenthesis$
$\csc (arccsc (\frac{π}{4}))$

In Exercises -, find the exact value or state that it is undefined.

$\arcsin (\sin (\frac{π}{6}))$
$\arcsin (\sin (- \frac{π}{3}))$
$\arcsin (\sin (\frac{3 π}{4}))$
$\arcsin (\sin (\frac{11 π}{6}))$
$\arcsin (\sin (\frac{4 π}{3}))$
$\arccos (\cos (\frac{π}{4}))$
$\arccos (\cos (\frac{2 π}{3}))$
$\arccos (\cos (\frac{3 π}{2}))$
$\arccos (\cos (- \frac{π}{6}))$
$\arccos (\cos (\frac{5 π}{4}))$
$\arctan (\tan (\frac{π}{3}))$
$\arctan (\tan (- \frac{π}{4}))$
$\arctan (\tan (π))$
$\arctan (\tan (\frac{π}{2}))$
$\arctan (\tan (\frac{2 π}{3}))$
$arccot (\cot (\frac{π}{3}))$
$arccot (\cot (- \frac{π}{4}))$
$arccot (\cot (π))$
$arccot (\cot (\frac{π}{2}))$
$arccot (\cot (\frac{2 π}{3}))$

$arcsec (\sec (\frac{π}{4}))$
$arcsec (\sec (\frac{4 π}{3}))$
$arcsec (\sec (\frac{5 π}{6}))$
$arcsec (\sec (- \frac{π}{2}))$
$arcsec (\sec (\frac{5 π}{3}))$
$arccsc (\csc (\frac{π}{6}))$
$arccsc (\csc (\frac{5 π}{4}))$
$arccsc (\csc (\frac{2 π}{3}))$
$arccsc (\csc (- \frac{π}{2}))$
$arccsc (\csc (\frac{11 π}{6}))$
$arcsec (\sec (\frac{11 π}{12}))$
$arccsc (\csc (\frac{9 π}{8}))$

$arcsec (\sec (\frac{π}{4}))$
$arcsec (\sec (\frac{4 π}{3}))$
$arcsec (\sec (\frac{5 π}{6}))$
$arcsec (\sec (- \frac{π}{2}))$
$arcsec (\sec (\frac{5 π}{3}))$
$arccsc (\csc (\frac{π}{6}))$
$arccsc (\csc (\frac{5 π}{4}))$
$arccsc (\csc (\frac{2 π}{3}))$
$arccsc (\csc (- \frac{π}{2}))$
$arccsc (\csc (\frac{11 π}{6}))$
$arcsec (\sec (\frac{11 π}{12}))$
$arccsc (\csc (\frac{9 π}{8}))$

In Exercises -, find the exact value or state that it is undefined.

$\sin (\arccos (- \frac{1}{2}))$
$\sin (\arccos (\frac{3}{5}))$
$sin ⁡ open parenthesis arctan ⁡ open parenthesis minus 2 close parenthesis close parenthesis$
$sin ⁡ open parenthesis arccot ⁢ open parenthesis the square root of 5 close parenthesis close parenthesis$
$sin ⁡ open parenthesis arccsc ⁢ open parenthesis minus 3 close parenthesis close parenthesis$
$\cos (\arcsin (- \frac{5}{13}))$
$cos ⁡ open parenthesis arctan ⁡ open parenthesis the square root of 7 close parenthesis close parenthesis$
$cos ⁡ open parenthesis arccot ⁢ open parenthesis 3 close parenthesis close parenthesis$
$cos ⁡ open parenthesis arcsec ⁢ open parenthesis 5 close parenthesis close parenthesis$
$\tan (\arcsin (- \frac{2 \sqrt{5}}{5}))$
$\tan (\arccos (- \frac{1}{2}))$
$\tan (arcsec (\frac{5}{3}))$
$tan ⁡ open parenthesis arccot ⁢ open parenthesis 12 close parenthesis close parenthesis$
$\cot (\arcsin (\frac{12}{13}))$
$\cot (\arccos (\frac{\sqrt{3}}{2}))$
$cot ⁡ open parenthesis arccsc ⁢ open parenthesis the square root of 5 close parenthesis close parenthesis$
$cot ⁡ open parenthesis arctan ⁡ open parenthesis 0.25 close parenthesis close parenthesis$
$\sec (\arccos (\frac{\sqrt{3}}{2}))$
$\sec (\arcsin (- \frac{12}{13}))$
$sec ⁡ open parenthesis arctan ⁡ open parenthesis 10 close parenthesis close parenthesis$
$\sec (arccot (- \frac{\sqrt{10}}{10}))$
$csc ⁡ open parenthesis arccot ⁢ open parenthesis 9 close parenthesis close parenthesis$
$\csc (\arcsin (\frac{3}{5}))$
$\csc (\arctan (- \frac{2}{3}))$

In Exercises -, find the exact value or state that it is undefined.

$\sin (\arcsin (\frac{5}{13}) + \frac{π}{4})$
$cos ⁡ open parenthesis arcsec ⁢ open parenthesis 3 close parenthesis plus arctan ⁡ open parenthesis 2 close parenthesis close parenthesis$
$\tan (\arctan (3) + \arccos (- \frac{3}{5}))$
$\sin (2 \arcsin (- \frac{4}{5}))$
$\sin (2 arccsc (\frac{13}{5}))$
$sin ⁡ open parenthesis 2 ⁢ arctan ⁡ open parenthesis 2 close parenthesis close parenthesis$
$\cos (2 \arcsin (\frac{3}{5}))$
$\cos (2 arcsec (\frac{25}{7}))$
$\cos (2 arccot (- \sqrt{5}))$
$\sin (\frac{\arctan (2)}{2})$

In Exercises -, rewrite each of the following composite functions as algebraic functions of $x$ and state the domain.

$f (x) = \sin (\arccos (x))$
$f (x) = \cos (\arctan (x))$
$f (x) = \tan (\arcsin (x))$
$f (x) = \sec (\arctan (x))$
$f (x) = \csc (\arccos (x))$
$f ⁢ open parenthesis x close parenthesis equals sin ⁡ open parenthesis 2 ⁢ arctan ⁡ open parenthesis x close parenthesis close parenthesis$
$f ⁢ open parenthesis x close parenthesis equals sin ⁡ open parenthesis 2 ⁢ arccos ⁡ open parenthesis x close parenthesis close parenthesis$
$f ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis 2 ⁢ arctan ⁡ open parenthesis x close parenthesis close parenthesis$
$f ⁢ open parenthesis x close parenthesis equals sin ⁡ open parenthesis arccos ⁡ open parenthesis 2 ⁢ x close parenthesis close parenthesis$
$f (x) = \sin (\arccos (\frac{x}{5}))$
$f (x) = \cos (\arcsin (\frac{x}{2}))$
$f ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis arctan ⁡ open parenthesis 3 ⁢ x close parenthesis close parenthesis$
$f ⁢ open parenthesis x close parenthesis equals sin ⁡ open parenthesis 2 ⁢ arcsin ⁡ open parenthesis 7 ⁢ x close parenthesis close parenthesis$
$f (x) = \sin (2 \arcsin (\frac{x \sqrt{3}}{3}))$
$f ⁢ open parenthesis x close parenthesis equals cos ⁡ open parenthesis 2 ⁢ arcsin ⁡ open parenthesis 4 ⁢ x close parenthesis close parenthesis$
$f ⁢ open parenthesis x close parenthesis equals sec ⁡ open parenthesis arctan ⁡ open parenthesis 2 ⁢ x close parenthesis close parenthesis ⁢ tan ⁡ open parenthesis arctan ⁡ open parenthesis 2 ⁢ x close parenthesis close parenthesis$
$f (x) = \sin (\arcsin (x) + \arccos (x))$
$f (x) = \cos (\arcsin (x) + \arctan (x))$
$f ⁢ open parenthesis x close parenthesis equals tan ⁡ open parenthesis 2 ⁢ arcsin ⁡ open parenthesis x close parenthesis close parenthesis$
$f (x) = \sin (\frac{1}{2} \arctan (x))$
If $θ = \arcsin (\frac{x}{2})$ , find an expression for $θ plus sin ⁡ open parenthesis 2 ⁢ θ close parenthesis$ in terms of $x$ .
If $θ = \arctan (\frac{x}{7})$ , find an expression for $\frac{1}{2} θ - \frac{1}{2} \sin (2 θ)$ in terms of $x$ .
If $θ = arcsec (\frac{x}{4})$ , find an expression for $4 ⁢ tan ⁡ open parenthesis θ close parenthesis minus 4 ⁢ θ$ in terms of $x$ assuming $x greater than or equal to 4$ .

In Exercises -, solve the equation using the techniques discussed in Example 12.3.6 then approximate the solutions which lie in the interval $open bracket 0, 2 ⁢ π close parenthesis$ to four decimal places.

$\sin (θ) = \frac{7}{11}$
$\cos (θ) = - \frac{2}{9}$
$\sin (θ) = - 0.569$
$cos ⁡ open parenthesis θ close parenthesis equals 0.117$
$sin ⁡ open parenthesis θ close parenthesis equals 0.008$
$\cos (θ) = \frac{359}{360}$
$tan ⁡ open parenthesis t close parenthesis equals 117$
$\cot (t) = - 12$
$\sec (t) = \frac{3}{2}$
$\csc (t) = - \frac{90}{17}$
$\tan (t) = - \sqrt{10}$
$\sin (t) = \frac{3}{8}$
$\cos (x) = - \frac{7}{16}$
$tan ⁡ open parenthesis x close parenthesis equals 0.03$
$sin ⁡ open parenthesis x close parenthesis equals 0.3502$
$\sin (x) = - 0.721$
$cos ⁡ open parenthesis x close parenthesis equals 0.9824$
$\cos (x) = - 0.5637$
$\cot (x) = \frac{1}{117}$
$\tan (x) = - 0.6109$

In Exercises -, rewrite the given function as a sinusoid of the form $C (t) = A \cos (ω t + ϕ)$ and $S (t) = A \sin (ω t + ϕ)$ (See Example 12.3.7.) Approximate the value of $ϕ$ (which is in radians, of course) to four decimal places.

$f (t) = 5 \sin (3 t) + 12 \cos (3 t)$
$f (t) = 3 \cos (2 t) + 4 \sin (2 t)$
$f (t) = \cos (t) - 3 \sin (t)$
$f (t) = 7 \sin (10 t) - 24 \cos (10 t)$
$f (t) = - \cos (t) - 2 \sqrt{2} \sin (t)$
$f (t) = 2 \sin (t) - \cos (t)$

In Exercises -, find the domain of the given function. Write your answers in interval notation.

$f ⁢ open parenthesis x close parenthesis equals arcsin ⁡ open parenthesis 5 ⁢ x close parenthesis$
$f (x) = \arccos (\frac{3 x - 1}{2})$
$f ⁢ open parenthesis x close parenthesis equals arcsin ⁡ open parenthesis 2 ⁢ x squared close parenthesis$
$f (x) = \arccos (\frac{1}{x^{2} - 4})$
$f ⁢ open parenthesis x close parenthesis equals arctan ⁡ open parenthesis 4 ⁢ x close parenthesis$
$f (x) = arccot (\frac{2 x}{x^{2} - 9})$
$f (x) = \arctan (\ln (2 x - 1))$
$f (x) = arccot (\sqrt{2 x - 1})$
$f ⁢ open parenthesis x close parenthesis equals arcsec ⁢ open parenthesis 12 ⁢ x close parenthesis$
$f (x) = arccsc (x + 5)$
$f (x) = arcsec (\frac{x^{3}}{8})$
$f (x) = arccsc (e^{2 x})$
Find the following limits.
1. $lim_{x \to 1^{-}} \arcsin (x)$
2. $lim x → minus ∞ arctan ⁡ open parenthesis 3 ⁢ x close parenthesis$
3. $lim x → 1 arcsec ⁢ open parenthesis 2 ⁢ x close parenthesis$
Find a nonzero number $x$ where $arccot (x) \neq \arctan (\frac{1}{x})$ .
Find an example where $arcsec (x) \neq \arccos (\frac{1}{x})$ if we use $[0, \frac{π}{2}) \cup [π, \frac{3 π}{2})$ as the range of $f (x) = arcsec (x)$ .
Show that $\arcsin (x) + \arccos (x) = \frac{π}{2}$ for $minus 1 less than or equal to x less than or equal to 1$ .
Discuss with your classmates why $\arcsin (\frac{1}{2}) \neq 30$ .
Use the diagram below along with the accompanying questions to show:
$\arctan (1) + \arctan (2) + \arctan (3) = π$
Figure 12.55
1. Clearly $△ A O B$ and $△ B C D$ are right triangles because the line through $O$ and $A$ and the line through $C$ and $D$ are perpendicular to the $x$ -axis. Use the distance formula to show that $△ B A D$ is also a right triangle (with $∠ ⁢ B ⁢ A ⁢ D$ being the right angle) by showing that the sides of the triangle satisfy the Pythagorean Theorem.
2. Use $△ A O B$ to show that $α equals arctan ⁡ open parenthesis 1 close parenthesis$
3. Use $△ B A D$ to show that $β equals arctan ⁡ open parenthesis 2 close parenthesis$
4. Use $△ B C D$ to show that $γ equals arctan ⁡ open parenthesis 3 close parenthesis$
5. Use the fact that $O$ , $B$ and $C$ all lie on the $x$ -axis to conclude that $α + β + γ = π$ . Thus $\arctan (1) + \arctan (2) + \arctan (3) = π$ .

Answers

$\arcsin (- 1) = - \frac{π}{2}$
$\arcsin (- \frac{\sqrt{3}}{2}) = - \frac{π}{3}$
$\arcsin (- \frac{\sqrt{2}}{2}) = - \frac{π}{4}$
$\arcsin (- \frac{1}{2}) = - \frac{π}{6}$
$arcsin ⁡ open parenthesis 0 close parenthesis equals 0$
$\arcsin (\frac{1}{2}) = \frac{π}{6}$
$\arcsin (\frac{\sqrt{2}}{2}) = \frac{π}{4}$
$\arcsin (\frac{\sqrt{3}}{2}) = \frac{π}{3}$
$\arcsin (1) = \frac{π}{2}$
$\arccos (- 1) = π$
$\arccos (- \frac{\sqrt{3}}{2}) = \frac{5 π}{6}$
$\arccos (- \frac{\sqrt{2}}{2}) = \frac{3 π}{4}$
$\arccos (- \frac{1}{2}) = \frac{2 π}{3}$
$\arccos (0) = \frac{π}{2}$
$\arccos (\frac{1}{2}) = \frac{π}{3}$
$\arccos (\frac{\sqrt{2}}{2}) = \frac{π}{4}$
$\arccos (\frac{\sqrt{3}}{2}) = \frac{π}{6}$
$arccos ⁡ open parenthesis 1 close parenthesis equals 0$
$\arctan (- \sqrt{3}) = - \frac{π}{3}$
$\arctan (- 1) = - \frac{π}{4}$
$\arctan (- \frac{\sqrt{3}}{3}) = - \frac{π}{6}$
$arctan ⁡ open parenthesis 0 close parenthesis equals 0$
$\arctan (\frac{\sqrt{3}}{3}) = \frac{π}{6}$
$\arctan (1) = \frac{π}{4}$
$\arctan (\sqrt{3}) = \frac{π}{3}$
$arccot (- \sqrt{3}) = \frac{5 π}{6}$
$arccot (- 1) = \frac{3 π}{4}$
$arccot (- \frac{\sqrt{3}}{3}) = \frac{2 π}{3}$
$arccot (0) = \frac{π}{2}$
$arccot (\frac{\sqrt{3}}{3}) = \frac{π}{3}$
$arccot (1) = \frac{π}{4}$
$arccot (\sqrt{3}) = \frac{π}{6}$
$arcsec (2) = \frac{π}{3}$
$arccsc (2) = \frac{π}{6}$
$arcsec (\sqrt{2}) = \frac{π}{4}$
$arccsc (\sqrt{2}) = \frac{π}{4}$
$arcsec (\frac{2 \sqrt{3}}{3}) = \frac{π}{6}$
$arccsc (\frac{2 \sqrt{3}}{3}) = \frac{π}{3}$
$arcsec ⁢ open parenthesis 1 close parenthesis equals 0$
$arccsc (1) = \frac{π}{2}$
$arcsec (- 2) = \frac{2 π}{3}$
$arcsec (- \sqrt{2}) = \frac{3 π}{4}$
$arcsec (- \frac{2 \sqrt{3}}{3}) = \frac{5 π}{6}$
$arcsec (- 1) = π$
$arccsc (- 2) = - \frac{π}{6}$
$arccsc (- \sqrt{2}) = - \frac{π}{4}$
$arccsc (- \frac{2 \sqrt{3}}{3}) = - \frac{π}{3}$
$arccsc (- 1) = - \frac{π}{2}$
$arcsec (- 2) = \frac{4 π}{3}$
$arcsec (- \sqrt{2}) = \frac{5 π}{4}$
$arcsec (- \frac{2 \sqrt{3}}{3}) = \frac{7 π}{6}$
$arcsec (- 1) = π$
$arccsc (- 2) = \frac{7 π}{6}$
$arccsc (- \sqrt{2}) = \frac{5 π}{4}$
$arccsc (- \frac{2 \sqrt{3}}{3}) = \frac{4 π}{3}$
$arccsc (- 1) = \frac{3 π}{2}$
$\sin (\arcsin (\frac{1}{2})) = \frac{1}{2}$
$\sin (\arcsin (- \frac{\sqrt{2}}{2})) = - \frac{\sqrt{2}}{2}$
$\sin (\arcsin (\frac{3}{5})) = \frac{3}{5}$
$\sin (\arcsin (- 0.42)) = - 0.42$
$\sin (\arcsin (\frac{5}{4}))$ is undefined.
$\cos (\arccos (\frac{\sqrt{2}}{2})) = \frac{\sqrt{2}}{2}$
$\cos (\arccos (- \frac{1}{2})) = - \frac{1}{2}$
$\cos (\arccos (\frac{5}{13})) = \frac{5}{13}$
$\cos (\arccos (- 0.998)) = - 0.998$
$\cos (\arccos (π))$ is undefined.
$\tan (\arctan (- 1)) = - 1$
$\tan (\arctan (\sqrt{3})) = \sqrt{3}$
$\tan (\arctan (\frac{5}{12})) = \frac{5}{12}$
$tan ⁡ open parenthesis arctan ⁡ open parenthesis 0.965 close parenthesis close parenthesis equals 0.965$
$tan ⁡ open parenthesis arctan ⁡ open parenthesis 3 ⁢ π close parenthesis close parenthesis equals 3 ⁢ π$
$cot ⁡ open parenthesis arccot ⁢ open parenthesis 1 close parenthesis close parenthesis equals 1$
$\cot (arccot (- \sqrt{3})) = - \sqrt{3}$
$\cot (arccot (- \frac{7}{24})) = - \frac{7}{24}$
$\cot (arccot (- 0.001)) = - 0.001$
$\cot (arccot (\frac{17 π}{4})) = \frac{17 π}{4}$
$sec ⁡ open parenthesis arcsec ⁢ open parenthesis 2 close parenthesis close parenthesis equals 2$
$\sec (arcsec (- 1)) = - 1$
$\sec (arcsec (\frac{1}{2}))$ is undefined.
$sec ⁡ open parenthesis arcsec ⁢ open parenthesis 0.75 close parenthesis close parenthesis$ is undefined.
$sec ⁡ open parenthesis arcsec ⁢ open parenthesis 117 ⁢ π close parenthesis close parenthesis equals 117 ⁢ π$
$\csc (arccsc (\sqrt{2})) = \sqrt{2}$
$\csc (arccsc (- \frac{2 \sqrt{3}}{3})) = - \frac{2 \sqrt{3}}{3}$
$\csc (arccsc (\frac{\sqrt{2}}{2}))$ is undefined.
$csc ⁡ open parenthesis arccsc ⁢ open parenthesis 1.0001 close parenthesis close parenthesis equals 1.0001$
$\csc (arccsc (\frac{π}{4}))$ is undefined.
$\arcsin (\sin (\frac{π}{6})) = \frac{π}{6}$
$\arcsin (\sin (- \frac{π}{3})) = - \frac{π}{3}$
$\arcsin (\sin (\frac{3 π}{4})) = \frac{π}{4}$
$\arcsin (\sin (\frac{11 π}{6})) = - \frac{π}{6}$
$\arcsin (\sin (\frac{4 π}{3})) = - \frac{π}{3}$
$\arccos (\cos (\frac{π}{4})) = \frac{π}{4}$
$\arccos (\cos (\frac{2 π}{3})) = \frac{2 π}{3}$
$\arccos (\cos (\frac{3 π}{2})) = \frac{π}{2}$
$\arccos (\cos (- \frac{π}{6})) = \frac{π}{6}$
$\arccos (\cos (\frac{5 π}{4})) = \frac{3 π}{4}$
$\arctan (\tan (\frac{π}{3})) = \frac{π}{3}$
$\arctan (\tan (- \frac{π}{4})) = - \frac{π}{4}$
$arctan ⁡ open parenthesis tan ⁡ open parenthesis π close parenthesis close parenthesis equals 0$
$\arctan (\tan (\frac{π}{2}))$ is undefined
$\arctan (\tan (\frac{2 π}{3})) = - \frac{π}{3}$
$arccot (\cot (\frac{π}{3})) = \frac{π}{3}$
$arccot (\cot (- \frac{π}{4})) = \frac{3 π}{4}$
$arccot (\cot (π))$ is undefined
$arccot (\cot (\frac{3 π}{2})) = \frac{π}{2}$
$arccot (\cot (\frac{2 π}{3})) = \frac{2 π}{3}$
$arcsec (\sec (\frac{π}{4})) = \frac{π}{4}$
$arcsec (\sec (\frac{4 π}{3})) = \frac{2 π}{3}$
$arcsec (\sec (\frac{5 π}{6})) = \frac{5 π}{6}$
$arcsec (\sec (- \frac{π}{2}))$ is undefined.
$arcsec (\sec (\frac{5 π}{3})) = \frac{π}{3}$
$arccsc (\csc (\frac{π}{6})) = \frac{π}{6}$
$arccsc (\csc (\frac{5 π}{4})) = - \frac{π}{4}$
$arccsc (\csc (\frac{2 π}{3})) = \frac{π}{3}$
$arccsc (\csc (- \frac{π}{2})) = - \frac{π}{2}$
$arccsc (\csc (\frac{11 π}{6})) = - \frac{π}{6}$
$arcsec (\sec (\frac{11 π}{12})) = \frac{11 π}{12}$
$arccsc (\csc (\frac{9 π}{8})) = - \frac{π}{8}$
$arcsec (\sec (\frac{π}{4})) = \frac{π}{4}$
$arcsec (\sec (\frac{4 π}{3})) = \frac{4 π}{3}$
$arcsec (\sec (\frac{5 π}{6})) = \frac{7 π}{6}$
$arcsec (\sec (- \frac{π}{2}))$ is undefined.
$arcsec (\sec (\frac{5 π}{3})) = \frac{π}{3}$
$arccsc (\csc (\frac{π}{6})) = \frac{π}{6}$
$arccsc (\csc (\frac{5 π}{4})) = \frac{5 π}{4}$
$arccsc (\csc (\frac{2 π}{3})) = \frac{π}{3}$
$arccsc (\csc (- \frac{π}{2})) = \frac{3 π}{2}$
$arccsc (\csc (\frac{11 π}{6})) = \frac{7 π}{6}$
$arcsec (\sec (\frac{11 π}{12})) = \frac{13 π}{12}$
$arccsc (\csc (\frac{9 π}{8})) = \frac{9 π}{8}$
$\sin (\arccos (- \frac{1}{2})) = \frac{\sqrt{3}}{2}$
$\sin (\arccos (\frac{3}{5})) = \frac{4}{5}$
$\sin (\arctan (- 2)) = - \frac{2 \sqrt{5}}{5}$
$\sin (arccot (\sqrt{5})) = \frac{\sqrt{6}}{6}$
$\sin (arccsc (- 3)) = - \frac{1}{3}$
$\cos (\arcsin (- \frac{5}{13})) = \frac{12}{13}$
$\cos (\arctan (\sqrt{7})) = \frac{\sqrt{2}}{4}$
$\cos (arccot (3)) = \frac{3 \sqrt{10}}{10}$
$\cos (arcsec (5)) = \frac{1}{5}$
$\tan (\arcsin (- \frac{2 \sqrt{5}}{5})) = - 2$
$\tan (\arccos (- \frac{1}{2})) = - \sqrt{3}$
$\tan (arcsec (\frac{5}{3})) = \frac{4}{3}$
$\tan (arccot (12)) = \frac{1}{12}$
$\cot (\arcsin (\frac{12}{13})) = \frac{5}{12}$
$\cot (\arccos (\frac{\sqrt{3}}{2})) = \sqrt{3}$
$\cot (arccsc (\sqrt{5})) = 2$
$cot ⁡ open parenthesis arctan ⁡ open parenthesis 0.25 close parenthesis close parenthesis equals 4$
$\sec (\arccos (\frac{\sqrt{3}}{2})) = \frac{2 \sqrt{3}}{3}$
$\sec (\arcsin (- \frac{12}{13})) = \frac{13}{5}$
$\sec (\arctan (10)) = \sqrt{101}$
$\sec (arccot (- \frac{\sqrt{10}}{10})) = - \sqrt{11}$
$\csc (arccot (9)) = \sqrt{82}$
$\csc (\arcsin (\frac{3}{5})) = \frac{5}{3}$
$\csc (\arctan (- \frac{2}{3})) = - \frac{\sqrt{13}}{2}$
$\sin (\arcsin (\frac{5}{13}) + \frac{π}{4}) = \frac{17 \sqrt{2}}{26}$
$\cos (arcsec (3) + \arctan (2)) = \frac{\sqrt{5} - 4 \sqrt{10}}{15}$
$\tan (\arctan (3) + \arccos (- \frac{3}{5})) = \frac{1}{3}$
$\sin (2 \arcsin (- \frac{4}{5})) = - \frac{24}{25}$
$\sin (2 arccsc (\frac{13}{5})) = \frac{120}{169}$
$\sin (2 \arctan (2)) = \frac{4}{5}$
$\cos (2 \arcsin (\frac{3}{5})) = \frac{7}{25}$
$\cos (2 arcsec (\frac{25}{7})) = - \frac{527}{625}$
$\cos (2 arccot (- \sqrt{5})) = \frac{2}{3}$
$\sin (\frac{\arctan (2)}{2}) = \sqrt{\frac{5 - \sqrt{5}}{10}}$
$f (x) = \sin (\arccos (x)) = \sqrt{1 - x^{2}}$ for $minus 1 less than or equal to x less than or equal to 1$
$f (x) = \cos (\arctan (x)) = \frac{1}{\sqrt{1 + x^{2}}}$ for all $x$
$f (x) = \tan (\arcsin (x)) = \frac{x}{\sqrt{1 - x^{2}}}$ for $minus 1 less than x less than 1$
$f (x) = \sec (\arctan (x)) = \sqrt{1 + x^{2}}$ for all $x$
$f (x) = \csc (\arccos (x)) = \frac{1}{\sqrt{1 - x^{2}}}$ for $minus 1 less than x less than 1$
$f (x) = \sin (2 \arctan (x)) = \frac{2 x}{x^{2} + 1}$ for all $x$
$f (x) = \sin (2 \arccos (x)) = 2 x \sqrt{1 - x^{2}}$ for $minus 1 less than or equal to x less than or equal to 1$
$f (x) = \cos (2 \arctan (x)) = \frac{1 - x^{2}}{1 + x^{2}}$ for all $x$
$f (x) = \sin (\arccos (2 x)) = \sqrt{1 - 4 x^{2}}$ for $- \frac{1}{2} \leq x \leq \frac{1}{2}$
$f (x) = \sin (\arccos (\frac{x}{5})) = \frac{\sqrt{25 - x^{2}}}{5}$ for $minus 5 less than or equal to x less than or equal to 5$
$f (x) = \cos (\arcsin (\frac{x}{2})) = \frac{\sqrt{4 - x^{2}}}{2}$ for $minus 2 less than or equal to x less than or equal to 2$
$f (x) = \cos (\arctan (3 x)) = \frac{1}{\sqrt{1 + 9 x^{2}}}$ for all $x$
$f (x) = \sin (2 \arcsin (7 x)) = 14 x \sqrt{1 - 49 x^{2}}$ for $- \frac{1}{7} \leq x \leq \frac{1}{7}$
$f (x) = \sin (2 \arcsin (\frac{x \sqrt{3}}{3})) = \frac{2 x \sqrt{3 - x^{2}}}{3}$ for $- \sqrt{3} \leq x \leq \sqrt{3}$
$f (x) = \cos (2 \arcsin (4 x)) = 1 - 32 x^{2}$ for $- \frac{1}{4} \leq x \leq \frac{1}{4}$
$f (x) = \sec (\arctan (2 x)) \tan (\arctan (2 x)) = 2 x \sqrt{1 + 4 x^{2}}$ for all $x$
$f (x) = \sin (\arcsin (x) + \arccos (x)) = 1$ for $minus 1 less than or equal to x less than or equal to 1$
$f (x) = \cos (\arcsin (x) + \arctan (x)) = \frac{\sqrt{1 - x^{2}} - x^{2}}{\sqrt{1 + x^{2}}}$ for $minus 1 less than or equal to x less than or equal to 1$
¹⁴ $f (x) = \tan (2 \arcsin (x)) = \frac{2 x \sqrt{1 - x^{2}}}{1 - 2 x^{2}}$ for $x$ in $(- 1, - \frac{\sqrt{2}}{2}) \cup (- \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) \cup (\frac{\sqrt{2}}{2}, 1)$
$f (x) = \sin (\frac{1}{2} \arctan (x)) = {\begin{matrix} \sqrt{\frac{\sqrt{x^{2} + 1} - 1}{2 \sqrt{x^{2} + 1}}} & for x \geq 0 \\ - \sqrt{\frac{\sqrt{x^{2} + 1} - 1}{2 \sqrt{x^{2} + 1}}} & for x < 0 \end{matrix}$
$θ + \sin (2 θ) = \arcsin (\frac{x}{2}) + \frac{x \sqrt{4 - x^{2}}}{2}$
$\frac{1}{2} θ - \frac{1}{2} \sin (2 θ) = \frac{1}{2} \arctan (\frac{x}{7}) - \frac{7 x}{x^{2} + 49}$
$4 \tan (θ) - 4 θ = \sqrt{x^{2} - 16} - 4 arcsec (\frac{x}{4})$
$θ = \arcsin (\frac{7}{11}) + 2 π k$ or $θ = π - \arcsin (\frac{7}{11}) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $θ approximately equals 0.6898, 2.4518$
$θ = \arccos (- \frac{2}{9}) + 2 π k$ or $θ = - \arccos (- \frac{2}{9}) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $θ approximately equals 1.7949, 4.4883$
$θ = π + \arcsin (0.569) + 2 π k$ or $θ = 2 π - \arcsin (0.569) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $θ approximately equals 3.7469, 5.6779$
$θ = \arccos (0.117) + 2 π k$ or $θ = 2 π - \arccos (0.117) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $θ approximately equals 1.4535, 4.8297$
$θ = \arcsin (0.008) + 2 π k$ or $θ = π - \arcsin (0.008) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $θ approximately equals 0.0080, 3.1336$
$θ = \arccos (\frac{359}{360}) + 2 π k$ or $θ = 2 π - \arccos (\frac{359}{360}) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $θ approximately equals 0.0746, 6.2086$
$t = \arctan (117) + π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $t approximately equals 1.56225, 4.70384$
$t = \arctan (- \frac{1}{12}) + π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $t approximately equals 3.0585, 6.2000$
$t = \arccos (\frac{2}{3}) + 2 π k$ or $t = 2 π - \arccos (\frac{2}{3}) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $t approximately equals 0.8411, 5.4422$
$t = π + \arcsin (\frac{17}{90}) + 2 π k$ or $t = 2 π - \arcsin (\frac{17}{90}) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $t approximately equals 3.3316, 6.0932$
$t = \arctan (- \sqrt{10}) + π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $t approximately equals 1.8771, 5.0187$
$t = \arcsin (\frac{3}{8}) + 2 π k$ or $t = π - \arcsin (\frac{3}{8}) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $t approximately equals 0.3844, 2.7572$
$x = \arccos (- \frac{7}{16}) + 2 π k$ or $x = - \arccos (- \frac{7}{16}) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $x approximately equals 2.0236, 4.2596$
$x = \arctan (0.03) + π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $x approximately equals 0.0300, 3.1716$
$x = \arcsin (0.3502) + 2 π k$ or $x = π - \arcsin (0.3502) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $x approximately equals 0.3578, 2.784$
$x = π + \arcsin (0.721) + 2 π k$ or $x = 2 π - \arcsin (0.721) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $x approximately equals 3.9468, 5.4780$
$x = \arccos (0.9824) + 2 π k$ or $x = 2 π - \arccos (0.9824) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $x approximately equals 0.1879, 6.0953$
$x = \arccos (- 0.5637) + 2 π k$ or $x = - \arccos (- 0.5637) + 2 π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $x approximately equals 2.1697, 4.1135$
$x = \arctan (117) + π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $x approximately equals 1.5622, 4.7038$
$x = \arctan (- 0.6109) + π k$ , in $open bracket 0, 2 ⁢ π close parenthesis$ , $x approximately equals 2.5932, 5.7348$
$f (t) = 5 \sin (3 t) + 12 \cos (3 t) = 13 \sin (3 t + \arcsin (\frac{12}{13})) \approx 13 \sin (3 t + 1.1760)$
$f (t) = 5 \sin (3 t) + 12 \cos (3 t) = 13 \cos (3 t + \arcsin (- \frac{5}{13})) \approx 13 \cos (3 t - 0.3948)$
$f (t) = 3 \cos (2 t) + 4 \sin (2 t) = 5 \sin (2 t + \arcsin (\frac{3}{5})) \approx 5 \sin (2 t + 0.6435)$
$f (t) = 3 \cos (2 t) + 4 \sin (2 t) = 5 \cos (2 t + \arcsin (- \frac{4}{5})) \approx 5 \cos (2 t - 0.9273)$
$f (t) = \cos (t) - 3 \sin (t) = \sqrt{10} \sin (t + \arccos (- \frac{3 \sqrt{10}}{10})) \approx \sqrt{10} \sin (t + 2.8198)$
$f (t) = \cos (t) - 3 \sin (t) = \sqrt{10} \cos (t + \arcsin (\frac{3 \sqrt{10}}{10})) \approx \sqrt{10} \cos (t + 1.2490)$
$f (t) = 7 \sin (10 t) - 24 \cos (10 t) = 25 \sin (10 t + \arcsin (- \frac{24}{25})) \approx 25 \sin (10 t - 1.2870)$
$f (t) = 7 \sin (10 t) - 24 \cos (10 t) = 25 \cos (10 t + π + \arcsin (\frac{7}{25})) \approx 25 \cos (10 t + 3.4254)$
$f (t) = - \cos (t) - 2 \sqrt{2} \sin (t) = 3 \sin (t + π + \arcsin (\frac{1}{3})) \approx 3 \sin (t + 3.4814)$
$f (t) = - \cos (t) - 2 \sqrt{2} \sin (t) = 3 \cos (t + \arccos (- \frac{1}{3})) \approx 3 \cos (t + 1.9106)$
$f (t) = 2 \sin (t) - \cos (t) = \sqrt{5} \sin (t + \arcsin (- \frac{\sqrt{5}}{5})) \approx \sqrt{5} \sin (t - 0.4636)$
$f (t) = 2 \sin (t) - \cos (t) = \sqrt{5} \cos (t + π + \arcsin (\frac{2 \sqrt{5}}{5})) \approx \sqrt{5} \cos (t + 4.2487)$
$[- \frac{1}{5}, \frac{1}{5}]$
$[- \frac{1}{3}, 1]$
$[- \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}]$
$(- \infty, - \sqrt{5}] \cup [- \sqrt{3}, \sqrt{3}] \cup [\sqrt{5}, \infty)$
$open parenthesis minus ∞, ∞ close parenthesis$
$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$
$(\frac{1}{2}, \infty)$
$[\frac{1}{2}, \infty)$
$(- \infty, - \frac{1}{12}] \cup [\frac{1}{12}, \infty)$
$(- \infty, - 6] \cup [- 4, \infty)$
$(- \infty, - 2] \cup [2, \infty)$
$open bracket 0, ∞ close parenthesis$
1. $lim_{x \to 1^{-}} \arcsin (x)$ $= \frac{π}{2}$
2. $lim_{x \to - \infty} \arctan (3 x) = - \infty$
3. $lim x → 1 arcsec ⁢ open parenthesis 2 ⁢ x close parenthesis$ $= \frac{π}{3}$

We have already discussed this concept in Section B.2 as the `angle finder' in the context of acute angles in right triangles. ↩
But be aware that many books do! As always, be sure to check the context! ↩
We switch the input variable to the arcsine and arccosine functions to ` $x$ ' to avoid confusion with the outputs we label ` $t$ .' ↩
See page the related section if you need a review of how we associate real numbers with angles in radian measure. ↩
Alternatively, we could use the identity: $1 + \tan^{2} (t) = \sec^{2} (t)$ . Since $x = \cos (t)$ , $\sec (t) = \frac{1}{\cos (t)} = \frac{1}{x}$ . The reader is invited to work through this approach to see what, if any, difficulties arise. ↩
…assuming the “Trigonometry Friendly” ranges are used. ↩
Alternatively, we can write the domain of $arccsc (x)$ as $vertical bar x vertical bar greater than or equal to 1$ , so the domain of $arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis$ is $vertical bar 4 ⁢ x vertical bar greater than or equal to 1$ . ↩
We'll avoid the label ` $x$ -coordinate' here since as we'll see, the quantity $x$ in this problem is tied to the radius as opposed to the coordinates of points on the terminal side of $θ$ . ↩
…assuming the “Calculus Friendly” ranges are used. ↩
Alternatively, we can write the domain of $arccsc (x)$ as $vertical bar x vertical bar greater than or equal to 1$ , so the domain of $arccsc ⁢ open parenthesis 4 ⁢ x close parenthesis$ is $vertical bar 4 ⁢ x vertical bar greater than or equal to 1$ . ↩
See also the remarks following Examples 12.3.2 and 12.3.3. ↩
We'll avoid the label ` $x$ -coordinate' here since as we'll see, the quantity $x$ in this problem is tied to the radius as opposed to the coordinates of points on the terminal side of $θ$ . ↩
If this sort of analysis sounds familiar, it should. We are really just tracking the effect of transformations as in Section 5.4. ↩
The equivalence for $x = \pm 1$ can be verified independently of the derivation of the formula, but Calculus is required to fully understand what is happening at those $x$ values. You'll see what we mean when you work through the details of the identity for $tan ⁡ open parenthesis 2 ⁢ t close parenthesis.$ For now, we exclude $x = \pm 1$ from our answer. ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.