Section 5.5: Inverses of the Circular Functions

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.

To find $\arcsin\left(\frac{\sqrt{2}}{2}\right)$, we need the angle measuring $t$ radians which lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ with $\sin(t) = \frac{\sqrt{2}}{2}$. Hence, $\arcsin\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$.

To find $\arccos\left(\frac{1}{2}\right)$, we are looking for the angle measuring $t$ radians which lies between $0$ and $\pi$ that has $\cos(t) = \frac{1}{2}$. Our answer is $\arccos\left(\frac{1}{2}\right)= \frac{\pi}{3}$.

For $\arcsin\left(-\frac{1}{2}\right)$, we are looking for an angle measuring $t$ radians which lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ with $\sin(t) = -\frac{1}{2}$. We want to look in QIV of the unit circle, but we need to be sure to give our answer a negative angle name, so that our final answer is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ as required. In QIV, we see that the angle $\frac{11\pi}{6}$ has a sine of $-\frac{1}{2}$. The negative angle coterminal with $\frac{11\pi}{6}$ is $-\frac{\pi}{6}$. Hence, $\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$.

Alternatively, we could use the fact that the arcsine function is odd, so $\arcsin\left(-\frac{1}{2}\right) = - \arcsin\left(\frac{1}{2}\right)$. We find $\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$, so $\arcsin\left(-\frac{1}{2}\right) = - \arcsin\left(\frac{1}{2}\right) = -\frac{\pi}{6}$.

For $\arccos\left(-\frac{\sqrt{2}}{2}\right)$, we need the angle measuring $t$ radians which lies between $0$ and $\pi$ with $\cos(t) = -\frac{\sqrt{2}}{2}$ . Hence, $\arccos\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}$.

Since $0 \leq \frac{\pi}{6} \leq \pi$, we could simply invoke the properties of inverse trig functions to get $\arccos\left( \cos\left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6}$.

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, $\arccos\left( \cos\left(\frac{\pi}{6}\right)\right) = \arccos\left( \frac{\sqrt{3}}{2}\right)$. To find $\arccos\left( \frac{\sqrt{3}}{2}\right)$, we need an angle measuring $t$ radians which lies between $0$ and $\pi$ that has $\cos(t) = \frac{\sqrt{3}}{2}$. We get $t = \frac{\pi}{6}$, so that $\arccos\left( \cos\left(\frac{\pi}{6}\right)\right) = \arccos\left( \frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$.

Since $\frac{11\pi}{6}$ does not fall between $0$ and $\pi$, we can't look at the outer arccosine as simply "undoing" the inner cosine. We are forced to work through from the inside out starting with $\arccos\left( \cos\left(\frac{11\pi}{6}\right)\right) = \arccos\left(\frac{\sqrt{3}}{2}\right)$. From the previous problem, we know $\arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$. Hence, $\arccos\left( \cos\left(\frac{11\pi}{6}\right)\right) = \frac{\pi}{6}$.

First, let's think about what this jumble of information means. When we write $\arccos\left(-\frac{3}{5}\right)$, we mean that there is an angle between $0$ and $\pi$ that has an $x$-coordinate of $-\frac{3}{5}$ on the unit circle. Let's temporarily name that angle $t$, so that $t = \arccos\left(-\frac{3}{5}\right)$. Well then, by definition of arccosine, $\cos(t) = -\frac{3}{5}$. We also know that the terminal side of $t$ must fall in QII, since $t$ must be between $0$ and $\pi$ and its cosine is negative.

Now, we can use our friend the Pythagorean Identity to figure out the sine of $t$.

Because we know that $t$ is a QII angle, its sine must be positive, so $\sin(t) = \frac{4}{5}$. Therefore, $\sin\left(\arccos\left(-\frac{3}{5}\right)\right) = \frac{4}{5}$.

Note that if you prefer to draw a reference triangle in Quadrant II, you can arrive at the same final answer.

Again, we let $t =\arcsin\left(\frac{5}{7}\right)$. Then we can start by saying that there is an angle, $t$, between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ that has a $y$-coordinate of $\frac{5}{7}$ on the Unit Circle. That places our $t$ in QI, and we know that $\sin(t) = \frac{5}{7}$. Using the Pythagorean Identity, then, we have $$\cos^2(t) + \left(\frac{5}{7}\right)^2 = 1$$ $$\cos^2(t) + \frac{25}{49} = 1$$ $$\cos^2(t) = \frac{24}{49}$$ $$\cos(t) = \frac{\sqrt{24}}{7}$$ We have chosen the positive solution, since cosine is positive in QI.

Now, since $t=\arcsin\left(\frac{5}{7}\right)$, our expression $\tan\left(\arcsin\left(\frac{5}{7}\right)\right)$ is equivalent to $\tan(t)$. Now, we know the sine and cosine of our angle, but in the end, we want to know its tangent. Recall that $\tan(t) = \frac{\sin(t)}{\cos(t)}$. So to calculate our tangent, we use the sine and cosine values from above: $$\tan(t) = \frac{5/7}{\sqrt{24}/7}$$ $$=\frac{5}{\sqrt{24}}$$ $$=\frac{5\sqrt{24}}{24}$$ $$=\frac{5\cdot 2\sqrt{6}}{24}$$ $$\tan(t)=\frac{5\sqrt{6}}{12}$$

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 $\tan\left(\arccos\left(x \right)\right) = \tan(t)$ in terms of $x$.

Since $t = \arccos(x)$, we know $\cos(t) = x$ where $0 \leq t \leq \pi$. One approach 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 $0 \leq t \leq \pi$, $\sin(t) \geq 0$, 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 consider that the function $f(x) = \tan\left(\arccos\left(x \right)\right)$ 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 $-1 \leq x \leq 1$, the domain of $f$ will be some subset of $[-1,1]$. The range of $\arccos(x)$ is $[0,\pi]$, and of these values, only $\frac{\pi}{2}$ will cause a problem for the tangent function. Since $\arccos(x) = \frac{\pi}{2}$ happens when $x = \cos\left(\frac{\pi}{2} \right) = 0$, we exclude $x=0$ from our domain. Hence, the domain of $f(x) = \tan\left(\arccos\left(x \right)\right) =\frac{\sqrt{1-x^2}}{x}$ is $[-1,0)\cup(0,1]$.

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, so proceed with caution!

We proceed as in the previous problem by writing $t = \arcsin(x)$ so that $t$ lies in the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2}\right]$ with $\sin(t) = x$. We aim to express $\cos\left( \arcsin(x)\right) = \cos(t)$ in terms of $x$.

Since we know $x = \sin(t)$, then by the Pythagorean Identity we have $$\cos^2(t) + x^2 = 1$$ $$\cos^2(t) = 1- x^2$$ $$\cos(t) = \pm \sqrt{1-x^2}$$

Again, because $t$ lies in the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2}\right]$, the cosine of $t$ must be positive, so $\cos(t) = \sqrt{1-x^2}$.

To find the domain of $g(x) = \cos\left(\arcsin(x)\right)$, we once again consider the inner function first. The domain of $\arcsin(x)$ is $[-1,1]$, and since there are no domain restrictions on cosine, the domain of $g$ is $[-1,1]$.

To find $\arctan(\sqrt{3})$, we need the angle measuring $t$ radians which lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ with $\tan(t) = \sqrt{3}$. We find $\arctan(\sqrt{3}) = \frac{\pi}{3}$.

To find $\mbox{arccot}(-\sqrt{3})$, we need the angle measuring $t$ radians which lies between $0$ and $\pi$ with $\cot(t) = -\sqrt{3}$. Hence, $\mbox{arccot}(-\sqrt{3}) = \frac{5\pi}{6}$.

This is another case where we are better off taking things a step at a time, so that we don't make assumptions that result in the wrong answer. In other words, the arccotangent does not directly undo the cotangent here, because the $\frac{7\pi}{6}$ input to the cotangent is not a valid output for the arccotangent function; arccotangent only gives answers between 0 and $\pi$!

First, $\cot\left(\frac{7\pi}{6}\right) = \frac{\cos(7\pi/6)}{\sin(7\pi/6)} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$. That means we can write $\mbox{arccot}\left(\cot\left(\frac{7\pi}{6}\right)\right)$ as $\mbox{arccot}\left(\sqrt{3}\right)$. The angle between 0 and $\pi$ with a cotangent of $\sqrt{3}$ is $\frac{\pi}{6}$, so our final answer is $\mbox{arccot}\left(\cot\left(\frac{7\pi}{6}\right)\right) = \frac{\pi}{6}$.

We start simplifying $\sin\left(\arctan\left(-\frac{4}{3}\right)\right)$ by letting $t = \arctan\left(-\frac{4}{3}\right)$. By definition, $\tan(t) = -\frac{4}{3}$ for some angle measuring $t$ radians which lies between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Since $\tan(t) \lt 0$, 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 $P(3,-4)$ as a point on the terminal side of $\theta = t = \arctan\left(-\frac{4}{3}\right)$ radians.

We find $r = \sqrt{x^2+y^2} = \sqrt{(3)^2+(-4)^2} = 5$, so $\sin(t) = -\frac{4}{5}$. Hence, $\sin\left(\arctan\left(-\frac{4}{3}\right)\right) = -\frac{4}{5}$.

To get started, we let $t = \mbox{arccot}(2x)$ so that $\cot(t) = 2x$ where $0 \lt t \lt \pi$. In terms of $t$, $\cos(\mbox{arccot}(2x)) = \cos(t)$, and our goal is to express the latter in terms of $x$.

Note that if $t = \mbox{arccot}(2x)$, then $\mbox{cot}(t) = 2x$. If we think way back to right triangles, this means we could draw a right triangle with reference angle $t$, where the ratio of the adjacent side to the opposite side is $2x$. The easiest way to make such a triangle is to let the adjacent side be $2x$ and the opposite side be 1.

Since we want to find $\cos(t)$, we will need to find an expression for the hypotenuse of this triangle. We can once more lean on the Pythagorean Theorem: $$(2x)^2+1^2 = c^2$$ $$4x^2 + 1 = c^2$$ $$c = \sqrt{4x^2+1}$$

Note: It may be tempting to try to simplify this expression, but the addition inside of the root makes that impossible!

Because $\cos(t)$ is adjacent over hypotenuse, our answer is $$\cos(t) = \frac{2x}{\sqrt{4x^2+1}}$$

To find the domain, viewing $g(x) = \cos(\mbox{arccot}(2x))$ 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 $(-\infty, \infty)$.

Using the property highlighted above, we have $\mbox{arcsec}(2) = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}$.

Once again, we can rewrite in terms of sine, giving $\mbox{arccsc}(-2) = \arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$.

Since $\frac{5\pi}{4}$ doesn't fall between $0$ and $\frac{\pi}{2}$ or $\frac{\pi}{2}$ and $\pi$, we cannot use the inverse property directly. Hence, we work from the 'inside out.'

We get: $\mbox{arcsec}\left( \sec\left( \frac{5\pi}{4} \right) \right) = \mbox{arcsec}(-\sqrt{2}) = \arccos\left(-\frac{1}{\sqrt{2}}\right) =\arccos\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}$.

We begin simplifying $\cot\left(\mbox{arccsc}\left(-3\right)\right)$ by letting $t = \mbox{arccsc}(-3)$. Then, $\csc(t) = -3$. Since $\csc(t) \lt 0$, $t$ lies in the interval $\left[ -\frac{\pi}{2},0\right)$, so $t$ corresponds to a Quadrant IV angle.

All right, so we know that our angle lies in QIV, with $\csc(t) = -3$. Therefore, $\sin(t) = -\frac{1}{3}$. Again, let's draw a simple triangle with these properties:

In the end we want to know $\cot(t)$, which is the ratio of the adjacent side over the opposite side, so we'll need to solve for the length of the missing leg using the Pythagorean Theorem.

Since our missing side lies on the positive x-axis, we have $x = \sqrt{8} = 2\sqrt{2}$ for this side. Therefore $$\cot(t) = \frac{2\sqrt{2}}{-1}$$ $$\cot(t) = -2\sqrt{2}$$

Proceeding as above, we let $t = \mbox{arcsec}(x)$. Then, $\sec(t) = x$ for $t$ in $\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi \right]$. We seek a formula for $\tan(\mbox{arcsec}(x)) = \tan(t)$ in terms of $x$.

Since $\sec(t) = x$, then we know that $\cos(t) = \frac{1}{x}$. Here we have to be careful; $t$ can be in $\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi \right]$, so $\frac{1}{x}$ might be negative or positive depending on the quadrant we find ourselves in. The pictures here can get unneccessarily confusing, so we're going to try a different strategy: rewriting the Pythagorean Identity. If $$\cos^2(t) + \sin^2(t) = 1$$ then we can divide every term by $\cos^2(t)$ to get $$\frac{\cos^2(t)}{\cos^2(t)} + \frac{\sin^2(t)}{\cos^2(t)} = \frac{1}{\cos^2(t)}$$ $$\Rightarrow 1 + \tan^2(t) = \sec^2(t)$$

Now, since $\sec(t) = x$, we have $$1 + \tan^2(t) = x^2$$ $$\tan^2(t) = x^2 - 1$$ $$\tan(t) = \pm \sqrt{x^2 - 1}$$

If $t$ belongs to $\left[0, \frac{\pi}{2}\right)$ then $\tan(t) \geq 0$. On the the other hand, if $t$ belongs to $\left(\frac{\pi}{2}, \pi \right]$ then $\tan(t) \leq 0$. As a result, we get a \textit{piecewise defined} function for $\tan(t)$: \[ \tan(t) = \left\{ \begin{array}{rr} \sqrt{x^2-1}, & \text{if $0 \leq t \lt \frac{\pi}{2}$} \\ -\sqrt{x^2-1}, & \text{if $\frac{\pi}{2} \lt t \leq \pi$} \end{array}\right. \]

Now we need to determine what these conditions on $t$ mean for $x$. Since $x = \sec(t)$, when $0 \leq t \lt \frac{\pi}{2}$, $x \geq 1$, and when $\frac{\pi}{2} \lt t \leq \pi$, $x \leq -1$. Hence, \[f(x) = \tan(\mbox{arcsec}(x)) = \left\{ \begin{array}{rr} \sqrt{x^2-1}, & \text{if $x \geq 1$} \\ -\sqrt{x^2-1}, & \text{if $x \leq -1$} \end{array}\right. \]

To find the domain of $f$, we consider $f(x) = \tan(\mbox{arcsec}(x))$ as a two step process. First, we have the arcsecant function, whose domain is $(-\infty, -1] \cup [1, \infty)$.

Since the range of $\mbox{arcsec}(x)$ is $\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi \right]$, taking the tangent of any output from $\mbox{arcsec}(x)$ is defined. Hence, the domain of $f$ is $(-\infty, -1] \cup [1, \infty)$.

Taking a cue from the previous problem, we start by letting $t = \mbox{arccsc}(4x)$. Then $\csc(t) = 4x$ for $t$ in $\left[-\frac{\pi}{2}, 0 \right) \cup \left(0, \frac{\pi}{2}\right]$. Our goal is to rewrite $\cos(\mbox{arccsc}(4x)) = \cos(t)$ in terms of $x$.

From $\csc(t) = 4x$, we get $\sin(t) = \frac{1}{4x}$, so to find $\cos(t)$, we can make use of the Pythagorean Identity: $\cos^{2}(t)= 1- \sin^{2}(t)$. Substituting $\sin(t) = \frac{1}{4x}$ gives $\cos^{2}(t) = 1 - \left(\frac{1}{4x}\right)^2 = 1 - \frac{1}{16x^2}$. Getting a common denominator and extracting square roots, we obtain: \[\cos(t) = \pm \sqrt{\frac{16x^2-1}{16x^2}} = \pm \frac{\sqrt{16x^2-1}}{4|x|}.\]

Since $t$ belongs to $\left[-\frac{\pi}{2}, 0 \right) \cup \left(0, \frac{\pi}{2}\right]$, we know $\cos(t) \geq 0$, so we choose $\cos(t) = \frac{\sqrt{16-x^2}}{4|x|}$. (The absolute values here are necessary, since $x$ could be negative.) Hence, \[g(x) = \cos(\mbox{arccsc}(4x)) = \frac{\sqrt{16-x^2}}{4|x|}.\]

To find the domain of $g(x) = \cos(\mbox{arccsc}(4x))$, 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 $4x$. Using interval notation, the domain of the arccosecant function is written as: $(-\infty, -1] \cup [1, \infty)$. Hence to take the arccosecant of $4x$, the quantity $4x$ must lie in one of these two intervals.\footnote{Alternatively, we can write the domain of $\mbox{arccsc}(x)$ as $|x| \geq 1$, so the domain of $\mbox{arccsc}(4x)$ is $|4x| \geq 1$.} That is, $4x \leq -1$ or $4x \geq 1$, so $x \leq -\frac{1}{4}$ or $x \geq \frac{1}{4}$.

The third and final process coded in $g(x) = \cos(\mbox{arccsc}(4x))$ is to take the cosine of $\mbox{arccsc}(4x)$. Since the cosine accepts any real number, we have no additional restrictions. Hence, the domain of $g$ is $\left( -\infty, -\frac{1}{4} \right] \cup \left[ \frac{1}{4}, \infty \right)$.