# Inverse Functions: Arcsecant (Part 27)

We now turn to a little-taught and perhaps controversial inverse function: arcsecant. As we’ve seen throughout this series, the domain of this inverse function must be chosen so that the graph of $y = \sec x$ satisfies the horizontal line test. It turns out that the choice of domain has surprising consequences that are almost unforeseeable using only the tools of Precalculus.

The standard definition of $y = \sec^{-1} x$ uses the interval $[0,\pi]$, so that

$\sec^{-1} x = \cos^{-1} \left( \displaystyle \frac{1}{x} \right)$

Why would this be controversial? Let’s fast-forward a couple of semesters and use implicit differentiation (see also https://meangreenmath.com/2014/08/08/different-definitions-of-logarithm-part-8/ for how this same logic is used for other inverse functions) to find the derivative of $y = \sec^{-1} x$:

$x = \sec y$

$\displaystyle \frac{d}{dx} (x) = \displaystyle \frac{d}{dx} (\sec y)$

$1 = \sec y \tan y \displaystyle \frac{dy}{dx}$

$\displaystyle \frac{1}{\sec y \tan y} = \displaystyle \frac{dy}{dx}$

At this point, the object is to convert the left-hand side to something involving only $x$. Clearly, we can replace $\sec y$ with $x$. However, doing the same with $\tan y$ is trickier. We begin with one of the Pythagorean identities:

$1 + \tan^2 y = \sec^2 y$

$\tan^2 y = \sec^2 y - 1$

$\tan^2 y = x^2 - 1$

$\tan y = \sqrt{x^2 - 1} \qquad \hbox{or} \tan y = -\sqrt{x^2 - 1}$

• If $0 \le y < \pi/2$ (so that $x = \sec y \ge 1$), then $\tan y$ is positive, and so $\tan y = \sqrt{\sec^2 y - 1}$.
• If $\pi/2 < y \le \pi$ (so that $x = \sec y \le 1$), then $\tan y$ is negative, and so $\tan y = -\sqrt{\sec^2 y - 1}$.

We therefore have two formulas for the derivative of $y = \sec^{-1} x$:

$\displaystyle \frac{d}{dx} sec^{-1} x = \displaystyle \frac{1}{x \sqrt{x^2-1}} \qquad \hbox{if} \qquad x > 1$

$\displaystyle \frac{d}{dx} sec^{-1} x = \displaystyle -\frac{1}{x \sqrt{x^2-1}} \qquad \hbox{if} \qquad x < 1$

These may then be combined into the single formula

$\displaystyle \frac{d}{dx} sec^{-1} x = \displaystyle \frac{1}{|x| \sqrt{x^2-1}}$

It gets better. Let’s now find the integral

$\displaystyle \int \frac{dx}{x \sqrt{x^2 -1}}$

Several calculus textbooks that I’ve seen will lazily give the answer

$\displaystyle \int \frac{dx}{x \sqrt{x^2 -1}} = \sec^{-1} x + C$

This answer works as long as $x > 1$, so that $|x|$ reduces to simply $x$. For example, the red signed area in the above picture on the interval $[2\sqrt{3}/3,2]$ may be correctly computed as

$\displaystyle \int_{2\sqrt{3}/3}^2 \frac{dx}{x \sqrt{x^2 -1}} = \sec^{-1} 2 - \sec^{-1} \displaystyle \frac{2\sqrt{3}}{3}$

$= \cos^{-1} \left( \displaystyle \frac{1}{2} \right) - \cos^{-1} \left( \displaystyle \frac{\sqrt{3}}{2} \right)$

$= \displaystyle \frac{\pi}{3} - \frac{\pi}{6}$

$= \displaystyle \frac{\pi}{6}$

However, the orange signed area on the interval $[-2,-2\sqrt{3}/3]$  is incorrectly computed using this formula!

$\displaystyle \int_{-2}^{-2\sqrt{3}/3} \frac{dx}{x \sqrt{x^2 -1}} = \sec^{-1} \left( - \displaystyle \frac{2\sqrt{3}}{3} \right) - \sec^{-1} (-2)$

$= \cos^{-1} \left( -\displaystyle \frac{\sqrt{3}}{2} \right) -\cos^{-1} \left( \displaystyle \frac{1}{2} \right)$

$= \displaystyle \frac{5\pi}{6} - \frac{2\pi}{3}$

$= \displaystyle \frac{\pi}{6}$

This is patently false, as the picture clearly indicates that the above integral has to be negative. For this reason, careful calculus textbooks will often ask students to solve a problem like

$\displaystyle \int \frac{dx}{x \sqrt{x^2 -1}}, \qquad x > 1$

and the caveat $x > 1$ is needed to ensure that the correct antiderivative is used. Indeed, a calculus textbook that doesn’t include such caveats are worthy of any scorn that an instructor cares to heap upon it.

# Inverse Functions: Arcsecant (Part 26)

We now turn to a little-taught and perhaps controversial inverse function: arcsecant. As we’ve seen throughout this series, the domain of this inverse function must be chosen so that the graph of $y = \sec x$ satisfies the horizontal line test. It turns out that the choice of domain has surprising consequences that are almost unforeseeable using only the tools of Precalculus.

The standard definition of $y = \sec^{-1} x$ uses the interval $[0,\pi]$ — or, more precisely, $[0,\pi/2) \cup (\pi/2, \pi]$ to avoid the vertical asymptote at $x = \pi/2$. This portion of the graph of $y = \sec x$ satisfies the horizontal line test and, conveniently, matches almost perfectly the domain of $y = \cos^{-1} x$. This is perhaps not surprising since, when both are defined, $\cos x$ and $\sec x$ are reciprocals.

Since this range of $\sec^{-1} x$ matches that of $\cos^{-1} x$, we have the convenient identity

$\sec^{-1} x = \cos^{-1} \left( \displaystyle \frac{1}{x} \right)$

To see why this works, let’s examine the right triangle below. Notice that

$\cos \theta = \displaystyle \frac{x}{1} \qquad \Longrightarrow \qquad \theta = \cos^{-1} x$.

Also,

$\sec\theta = \displaystyle \frac{1}{x} \qquad \Longrightarrow \qquad \theta = \cos^{-1} \left( \displaystyle \frac{1}{x} \right)$.

This argument provides the justification for $0 < \theta < \pi/2$ — that is, for $x > 1$ — but it still works for $x = 1$ and $x \le -1$.

So this seems like the most natural definition in the world for $\sec^{-1} x$. Unfortunately, there are consequences for this choice in calculus, as we’ll see in tomorrow’s post.

# Inverse Functions: Arctangent and Angle Between Two Lines (Part 25)

The smallest angle between the non-perpendicular lines $y = m_1 x + b_1$ and $y = m_1 x + b_2$ can be found using the formula

$\theta = \displaystyle \tan^{-1} \left( \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \right)$.

A generation ago, this formula used to be taught in a typical Precalculus class (or, as it was called back then, analytical geometry). However, I find that analytic geometry has fallen out of favor in modern Precalculus courses.

Why does this formula work? Consider the graphs of $y = m_1 x$ and $y = m_1 x + b_1$, and let’s measure the angle that the line makes with the positive $x-$axis.

The lines $y = m_1 x + b_1$ and $y = m_1 x$ are parallel, and the $x-$axis is a transversal intersecting these two parallel lines. Therefore, the angles that both lines make with the positive $x-$axis are congruent. In other words, the $+ b_1$ is entirely superfluous to finding the angle $\theta_1$. The important thing that matters is the slope of the line, not where the line intersects the $y-$axis.

The point $(1, m_1)$ lies on the line $y = m_1 x$, which also passes through the origin. By definition of tangent, $\tan \theta_1$ can be found by dividing the $y-$ and $x-$coordinates:

$\tan \theta_1 = \displaystyle \frac{m}{1} = m_1$.

We now turn to the problem of finding the angle between two lines. As noted above, the $y-$intercepts do not matter, and so we only need to find the smallest angle between the lines $y = m_1 x$ and $y = m_2 x$.

The angle $\theta$ will either be equal to $\theta_1 - \theta_2$ or $\theta_2 - \theta_1$, depending on the values of $m_1$ and $m_2$. Let’s now compute both $\tan (\theta_1 - \theta_2)$ and $\tan (\theta_2 - \theta_1)$ using the formula for the difference of two angles:

$\tan (\theta_1 - \theta_2) = \displaystyle \frac{\tan \theta_1 - \tan \theta_2}{1 + \tan \theta_1 \tan \theta_2}$

$\tan (\theta_2 - \theta_1) = \displaystyle \frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_2 \tan \theta_1}$

Since the smallest angle $\theta$ must lie between $0$ and $\pi/2$, the value of $\tan \theta$ must be positive (or undefined if $\theta = \pi/2$… for now, we’ll ignore this special case). Therefore, whichever of the above two lines holds, it must be that

$\tan \theta = \displaystyle \left| \frac{\tan \theta_1 - \tan \theta_2}{1 + \tan \theta_1 \tan \theta_2} \right|$

We now use the fact that $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$:

$\tan \theta = \displaystyle \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

$\theta = \tan^{-1} \left( \displaystyle \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \right)$

The above formula only applies to non-perpendicular lines. However, the perpendicular case may be remembered as almost a special case of the above formula. After all, $\tan \theta$ is undefined at $\theta = \pi/2 = 90^\circ$, and the right hand side is also undefined if $1 + m_1 m_2 = 0$. This matches the theorem that the two lines are perpendicular if and only if $m_1 m_2 = -1$, or that the slopes of the two lines are negative reciprocals.

# Inverse Functions: Arctangent and Angle Between Two Lines (Part 24)

Here’s a straightforward application of arctangent that, a generation ago, used to be taught in a typical Precalculus class (or, as it was called back then, analytical geometry).

Find the smallest angle between the lines $y= 3x$ and $y = -x/2$.

This problem is almost equivalent to finding the angle between the vectors $\langle 1,3 \rangle$ and $\langle -2,1 \rangle$. I use the caveat almost because the angle between two vectors could be between $0$ and $\pi$, while the smallest angle between two lines must lie between $0$ and $\pi/2$.

This smallest angle can be found using the formula

$\theta = \displaystyle \tan^{-1} \left( \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \right)$,

where $m_1$ and $m_2$ are the slopes of the two lines. In the present case,

$\theta = \tan^{-1} \left( \left| \displaystyle \frac{ 3 - (-1/2) }{1 + (3)(-1/2)} \right| \right)$

$\theta = \tan^{-1} \left( \left| \displaystyle \frac{7/2}{-1/2} \right| \right)$

$\theta = \tan^{-1} 7$

$\theta \approx 81.87^\circ$.

Not surprisingly, we obtain the same answer that we obtained a couple of posts ago using arccosine. The following picture makes clear why $\tan^{-1} 7 = \cos^{-1} \displaystyle \frac{1}{\sqrt{50}}$.

In tomorrow’s post, I’ll explain why the above formula actually works.

# Inverse Functions: Arccosine and Dot Products (Part 23)

The Law of Cosines can be applied to find the angle between two vectors ${\bf a}$ and ${\bf b}$. To begin, we draw the vectors ${\bf a}$ and ${\bf b}$, as well as the vector ${\bf c}$ (to be determined momentarily) that connects the tips of the vectors ${\bf a}$ and ${\bf b}$.

Using the usual rules for adding vectors, we see that ${\bf a} + {\bf c} = {\bf b}$, so that ${\bf c} = {\bf b} - {\bf a}$

We now apply the Law of Cosines to find $\theta$:

$\parallel \! \! {\bf c} \! \! \parallel^2 = \parallel \! \! {\bf a} \! \! \parallel^2 + \parallel \! \! {\bf b} \! \! \parallel^2 - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

$\parallel \! \! {\bf b} - {\bf a} \! \! \parallel^2 = \parallel \! \! {\bf a} \! \! \parallel^2 + \parallel \! \! {\bf b} \! \! \parallel^2 - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

We now apply the rule $\parallel \! \! {\bf a} \! \! \parallel^2 = {\bf a} \cdot {\bf a}$, convert the square of the norms into dot products. We then use the distributive and commutative properties of dot products to simplify.

$( {\bf b} - {\bf a} ) \cdot ({\bf b} - {\bf a}) = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

${\bf b} \cdot ({\bf b} - {\bf a}) - {\bf a} \cdot ({\bf b} - {\bf a}) = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

${\bf b} \cdot ({\bf b} - {\bf a}) -{\bf a} \cdot ({\bf b} - {\bf a}) = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

${\bf b} \cdot {\bf b} - {\bf a} \cdot {\bf b} - {\bf a} \cdot {\bf b} + {\bf a} \cdot {\bf a} = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

${\bf b} \cdot {\bf b} - 2 {\bf a} \cdot {\bf b} + {\bf a} \cdot {\bf a} = {\bf a} \cdot {\bf a} + {\bf b} \cdot {\bf b} - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

We can now cancel from the left and right sides and solve for $\cos \theta$:

$- 2 {\bf a} \cdot {\bf b} = - 2 \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel \cos \theta$

$\displaystyle \frac{ {\bf a} \cdot {\bf b} }{ \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel } = \cos \theta$

Finally, we are guaranteed that the angle between two vectors must lie between $0$ and $\pi$ (or, in degrees, between $0^\circ$ and $180^\circ$). Since this is the range of arccosine, we are permitted to use this inverse function to solve for $\theta$:

$\cos^{-1} \left( \displaystyle \frac{ {\bf a} \cdot {\bf b} }{ \parallel \! \! {\bf a} \! \! \parallel \parallel \! \! {\bf b} \! \! \parallel } \right) = \theta$

The good news is that there’s nothing special about two dimensions in the above proof, and so this formula may used for vectors in $\mathbb{R}^n$ for any dimension $n \ge 2$.

In the next post, we’ll consider how this same problem can be solved — but only in two dimensions — using arctangent.

# Inverse Functions: Arccosine and Dot Products (Part 22)

Here’s a straightforward application of arccosine, that, as far as I can tell, isn’t taught too often in Precalculus and is not part of the Common Core standards for vectors and matrices.

Find the angle between the vectors $\langle 1,3 \rangle$ and $\langle -2,1 \rangle$.

This problem is equivalent to finding the angle between the lines $y = 3x$ and $y = -x/2$. The angle $\theta$ is not drawn in standard position, which makes measurement of the angle initial daunting.

Fortunately, there is the straightforward formula for the angle between two vectors ${\bf a}$ and ${\bf b}$:

$\theta = \cos^{-1} \left( \displaystyle \frac{ {\bf a} \cdot {\bf b} }{ \parallel \!\! {\bf a} \!\! \parallel \parallel \!\! {\bf b} \!\! \parallel } \right)$

We recall that ${\bf a} \cdot {\bf b}$ is the dot product (or inner product) of the two vectors ${\bf a}$ and ${\bf b}$, while $\parallel \!\! {\bf a} \!\! \parallel = \sqrt{ {\bf a} \cdot {\bf a} }$ is the norm (or length) of the vector ${\bf a}$.

For this particular example,

$\theta = \cos^{-1} \left( \displaystyle \frac{\langle 1,3 \rangle \cdot \langle -2,1 \rangle }{ \parallel \!\!\langle 1,3 \rangle \!\! \parallel \parallel \!\!\langle -2,1 \rangle \!\! \parallel } \right)$

$\theta = \cos^{-1} \left( \displaystyle \frac{ (1)(-2) + (3)(1) }{ \sqrt{ (1)^2 + (3)^2} \sqrt{ (-2)^2 + 1^2 }} \right)$

$\theta = \cos^{-1} \displaystyle \frac{1}{\sqrt{50}}$

$\theta \approx 81.87^\circ$

In the next post, we’ll discuss why this actually works. And then we’ll consider how the same problem can be solved more directly using arctangent.

# Inverse Functions: Arccosine and SSS (Part 21)

Arccosine has an important advantage over arcsine when solving for the parts of a triangle: there is no possibility ambiguity about the angle.

Solve $\triangle ABC$ if $a = 16$, $b = 20$, and $c = 25$.

When solving for the three angles, it’s best to start with the biggest angle (that is, the angle opposite the biggest side). To see why, let’s see what happens if we first use the Law of Cosines to solve for one of the two smaller angles, say $\alpha$:

$a^2 = b^2 + c^2 - 2 b c \cos \alpha$

$256 = 400 + 625 - 1000 \cos \alpha$

$-769 = -1000 \cos \alpha$

$0.769 = \cos \alpha$

$\alpha \approx 39.746^\circ$

So far, so good. Now let’s try using the Law of Sines to solve for $\gamma$:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin 39.746^\circ}{16} \approx \displaystyle \frac{\sin \gamma}{25}$

$0.99883 \approx \sin \gamma$

Uh oh… there are two possible solutions for $\gamma$ since, hypothetically, $\gamma$ could be in either the first or second quadrant! So we have no way of knowing, using only the Law of Sines, whether $\gamma \approx 87.223^\circ$ or if $\gamma \approx 180^\circ - 87.223^\circ = 92.777^\circ$.

For this reason, it would have been far better to solve for the biggest angle first. For the present example, the biggest answer is $\gamma$ since that’s the angle opposite the longest side.

$c^2 = a^2 + b^2 - 2 a b \cos \gamma$

$625 = 256 + 400 - 640 \cos \gamma$

$-31 =-640 \cos \gamma$

$0.0484375 = \cos \gamma$

Using a calculator, we find that $\gamma \approx 87.223^\circ$.

We now use the Law of Sines to solve for either $\alpha$ or $\beta$ (pretending that we didn’t do the work above). Let’s solve for $\alpha$:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}$

$\displaystyle \frac{\sin \alpha}{16} \approx \displaystyle \frac{\sin 87.223}{25}$

$\sin \alpha \approx 0.63949$

This equation also has two solutions in the interval $[0^\circ, 180^\circ]$, namely, $\alpha \approx 39.736^\circ$ and $\alpha \approx 180^\circ - 39.736^\circ = 140.264^\circ$. However, we know full well that the answer can’t be larger than $\gamma$ since that’s already known to be the largest angle. So there’s no need to overthink the matter — the answer from blindly using arcsine on a calculator is going to be the answer for $\alpha$.

Naturally, the easiest way of finding $\beta$ is by computing $180^\circ - \alpha - \gamma$.

# Inverse Functions: Arccosine and SSS (Part 20)

The Law of Cosines also recognizes when the purported sides of a triangle are impossible.

Solve $\triangle ABC if$latex a = 16$, $b = 20$, and $c = 40$. Hopefully students would recognize that $c > a + b$, thus quickly demonstrating that the triangle is impossible. However, this also falls out of the Law of Cosines: $c^2 = a^2 + b^2 - 2 a b \cos \gamma$ $1600 = 256 + 400 - 640 \cos \gamma$ $944 =-640 \cos \gamma$ $-1.475 = \cos \gamma$ Since the cosine of an angle can’t be less than -1, we can conclude that this is impossible. Stated another way, we have the implications (since $a$, $b$, and $c$ are all positive) $c > a + b \Longleftrightarrow c^2 > (a+b)^2$ $\Longleftrightarrow a^2 + b^2 - 2 a b \cos \gamma > a^2 + 2 a b + b^2$ $\Longleftrightarrow -2 a b \cos \gamma > 2 a b$ $\Longleftrightarrow \cos \gamma < -1$ Since the last statement is impossible, so is the first one. # Inverse Functions: Arccosine and SSS (Part 19) Arccosine has an important advantage over arcsine when solving for the parts of a triangle: there is no possibility ambiguity about the angle. Solve $\triangle ABC if$latex a = 16$, $b = 20$, and $c = 25$.

To solve for, say, the angle $\gamma$, we employ the Law of Cosines:

$c^2 = a^2 + b^2 - 2 a b \cos \gamma$

$625 = 256 + 400 - 640 \cos \gamma$

$-31 =-640 \cos \gamma$

$0.0484375 = \cos \gamma$

Using a calculator, we find that $\gamma \approx 87.2^\circ$. And the good news is that there is no need to overthink this… this is guaranteed to be the angle since the range of $y = \cos^{-1} x$ is $[0,\pi]$, or $[0^\circ, 180^\circ]$ in degrees. So the equation

$\cos x = \hbox{something}$

is guaranteed to have a unique solution between $0^\circ$ and $180^\circ$. (But there are infinitely many solutions on $\mathbb{R}$. And since an angle in a triangle must lie between $0^\circ$ and $180^\circ$, the practical upshot is that just plugging into a calculator blindly is perfectly OK for this problem. This is in stark contrast to the Law of Sines, for which some attention must be paid for solutions in the interval $[0^\circ,90^\circ]$ and also the interval $[90^\circ, 180^\circ]$.

From this point forward, the Law of Cosines could be employed again to find either $\alpha$ or $\beta$. Indeed, this would be my preference since the sides $a$, $b$, and $c$ are exactly. However, my experience is that students prefer the simplicity of the Law of Sines to solve for one of these angles, using the now known pair of $c$ (exactly known) and $\gamma$ (approximately known with a calculator).

# Inverse Functions: Arccosine and Arctangent (Part 18)

In this series, we’ve seen that the inverse of function that fails the horizontal line test can be defined by appropriately restricting the domain of the function. For example, we now look at the graph of $y = \cos x$:

As with the graph of $y = \sin x$, we select a section that satisfies the horizontal line test and ignore the rest of the graph. (I described this in more rigorous terms when I considered arcsine, so I will not repeat the rigor here.) There are plenty of choices that could be made; by tradition, the interval $[0,\pi]$ is chosen.

Reflecting only the half-wave of the cosine graph on the interval $[0,\pi]$ through the line $y = x$ produces the graph of $y= \cos^{-1} x$. Again, to assist my students when graphing this function, I point out that the graph of cosine has horizontal tangent lines at the points $(0,1)$ and $(\pi,-1)$. Therefore, after reflecting through the line $y = x$, we see that the graph of $\cos^{-1} x$ has vertical tangent lines at $(1,0)$ and $(-1,\pi)$.

The same logic applies when defining the arctangent function. By tradition, the interval $(-\pi/2,\pi/2)$ is chosen as the section of the graph of $y = \tan x$ that satisfies the horizontal line test.

Reflecting only the half-wave of the cosine graph on the interval $[0,\pi]$ through the line $y = x$ produces the graph of $y= \tan^{-1} x$. Like the (more complicated) logistic growth function, this function has two different horizontal asymptotes that govern the behavior of the function as $x \to \pm \infty$.

So here are the rules that I want my Precalculus students to memorize:

$y = \sin^{-1} x$ means that $x = \sin y$ and $\displaystyle -\frac{\pi}{2} \le y \le \displaystyle \frac{\pi}{2}$

$y = \cos^{-1} x$ means that $x = \cos y$ and $0 \le y \le \pi$

$y = \sin^{-1} x$ means that $x = \sin y$ and $\displaystyle -\frac{\pi}{2} < y < \displaystyle \frac{\pi}{2}$

Students using forget that the range of arccosine is different than the other two, and I’ll usually have to produce the graph of $y = \cos x$ to explain and re-explain why this one is different.

Because these functions are defined on restricted domains, the usual funny things can happen. For example,

$\cos^{-1} (\cos 2\pi) = \cos^{-1} 1 = 0 \ne 2 \pi$

$\tan^{-1} \left( \tan \displaystyle \frac{3\pi}{4} \right) = \tan^{-1} (-1) = -\displaystyle \frac{\pi}{4} \ne \displaystyle \frac{3\pi}{4}$