# Inverse functions: Arcsine and SSA (Part 13)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like $\sin \theta = 0.8$. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of $\triangle ABC$, $a$ will be the length of the side opposite $\angle A$, $b$ will be the length of the side opposite $\angle B$, and $c$ will be the length of the side opposite angle $C$. Also $\alpha$ will be the measure of $\angle A$, $\beta$ will be measure of $\angle B$, and $\gamma$ will be the measure of $\angle C$. Modern textbooks tend not to use $\alpha$, $\beta$, and $\gamma$ for these kinds of problems, for which I have only one response:

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has exactly one solution:

Solve $\triangle ABC$ if $a = 5$, $c = 10$, and $\alpha = 30^\circ$.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

Notice that the red circle intersects the dashed black line at exactly one point. Therefore, we know that there will be exactly one solution for this case.

Of course, the reason that the dashed circle and line intersect at exactly one point is because $a = c \sin \alpha$, so that the triangle is a right triangle.

Of course, students should not be expected to make a picture this accurately when doing homework. Fortunately, this impossibility naturally falls out of the equation when using the Law of Sines:

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

$\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1/2}{5} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle 1 = \sin \gamma$

$90^\circ = \gamma$

The jump to the last step is only possible because there’s exactly one angle between $0^\circ$ and $90^\circ$ whose sine is equal to $1$. In the next couple posts in this series, we’ll see what happens when we get a step where $0 < \sin \gamma < 1$.

Anyway, for the problem at hand, from this point forward it’s easy to solve for the remaining pieces. We begin by finding $\beta$:

$\beta = 180^\circ - \alpha - \gamma = 60^\circ$

Then we can use the Law of Sines (or, in this case, the Pythagorean Theorem), to find $b$:

$\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \beta}{b}$

$\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin 60^\circ}{b}$

$\displaystyle \frac{1/2}{5} = \displaystyle \frac{\sqrt{3}/2}{b}$

$b = 5\sqrt{3}$

In the next few posts of this series, I’ll consider the other SSA cases — including the case where two solutions are possible.

# Inverse functions: Arcsine and SSA (Part 12)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like $\sin \theta = 0.8$. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of $\triangle ABC$, $a$ will be the length of the side opposite $\angle A$, $b$ will be the length of the side opposite $\angle B$, and $c$ will be the length of the side opposite angle $C$. Also $\alpha$ will be the measure of $\angle A$, $\beta$ will be measure of $\angle B$, and $\gamma$ will be the measure of $\angle C$. Modern textbooks tend not to use $\alpha$, $\beta$, and $\gamma$ for these kinds of problems, for which I have only one response:

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has no solution:

Solve $\triangle ABC$ if $a = 3$, $c = 10$, and $\alpha = 30^\circ$.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

The red dashed circle with center $B$ illustrates the dilemma: “side” $BC$ is simply too short to reach the horizontal dashed line to make the vertex $C$, dangling limply from the vertex $B$.

Of course, students should not be expected to make a picture this accurately when doing homework. Fortunately, this impossibility naturally falls out of the equation when using the Law of Sines:

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

$\displaystyle \frac{\sin 30^\circ}{3} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{1/2}{3} = \displaystyle \frac{\sin \gamma}{10}$

$\displaystyle \frac{5}{3} = \sin \gamma$

Since $\sin \gamma$ must like between $0$ and $1$ (said another way, $\sin^{-1} \frac{5}{3}$ is undefined), we know that this triangle cannot be solved.

In the next few posts of this series, I’ll consider the other SSA cases — including the case where two solutions are possible.

# Inverse Functions: Arcsine and Pedagogy (Part 11)

In yesterday’s post, we saw that restricting the domain of $g(x) = \sin x$ to $[-\pi/2,\pi/2]$ permits the definition of $g^{-1}(x) = \sin^{-1} x$, shown in the purple graph.

Today, I want to give some pedagogical thoughts on teaching this concept to Preaclculus students.

1. Notice that, if the purple graph was completed upward or downward, anything more than a half-period would violate the vertical line test and thus fail to be a function. Thinking back to the original function, that’s another way of saying that the original sine wave violates the horizontal line test.

2. Restricting the domain to $[-\pi/2,\pi/2]$ was a perfectly arbitrary decision. As shown above, there are plenty of other domains that would have worked acceptably. Only tradition requires us to choose $[-\pi/2,\pi/2]$.  (By the way, finding an expression for the restriction of $f$ to, say, $[\pi/2,3\pi/2]$ is a standard problem in a first course in real analysis.)

3. Since $\sin^2 x$ is typically used as shorthand for $(\sin x)^2$, some students will make the natural mistake of thinking that $\sin^{-1} x$ is just shorthand for $(\sin x)^{-1}$, or $\csc x$. So I like to address this head-on when introducing inverse trigonometric functions for the first time to my Precalculus students.

4. Unlike other inverse functions, it can be a little tricky for students to draw the graph of $y = \sin^{-1} x$ by hand because the line of reflection $y =x$ actually is the linearization of $y = \sin x$ at $x = 0$. In other words, $y = x$ is the first term of the Taylor series of $y = \sin x$ at $x = 0$. (For more details, see https://meangreenmath.com/2013/07/24/taylor-series-without-calculus-2/ or https://meangreenmath.com/2013/07/06/reminding-students-about-taylor-series-part-6/) Therefore, as seen in the picture, the line $y =x$ is very, very close to the graph of $y = \sin x$ for $x \approx 0$.

To assist students with accurately drawing by hand the graph of $y = \sin^{-1} x$, I point out that the original function $y = \sin x$ levels off horizontally at the points $(-\pi/2,-1)$ and $(\pi/2,1)$. Therefore, after reflecting through the line $y = x$, the graph of $y = \sin^{-1} x$ enters almost vertically through the points $(-1,-\pi/2)$ and $(1,\pi/2)$.

5. Most Precalculus students are not savvy enough to appreciate the nuances of domain and range in the above definitions. Therefore, after illustrating the importance of choosing an interval that satisfies the horizontal line test, I’ll give the following ways of remembering what $\sin^{-1} x$ means:

“Arcsine of $x$ is an angle. It is the angle whose sine is equal to $x$. And it’s the angle that lies between $-\pi/2$ and $\pi/2$.

OR

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

6. Since $g: [-\pi,2,\pi/2] \to [-1,1]$ and $g^{-1}: [-1,1] \to [-\pi/2,\pi/2]$ are inverse functions, it’s always true that $g(g^{-1}(x)) = x$ and $g^{-1}(g(x)) = x$. However, $g^{-1}$ and $f$ are not inverse functions, where $f: \mathbb{R} \to [-1,1]$ is the full sine function $f(x) = \sin x$. Therefore, it’s possible for $g^{-1}(f(x))$ to be something other than $x$. For example,

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

$\sin^{-1} \left( \sin \displaystyle \frac{5\pi}{6} \right) = \sin^{-1} \left(\displaystyle \frac{1}{2} \right) = \displaystyle \frac{\pi}{6} \ne \displaystyle \frac{5\pi}{6}$

This is analogous to our earlier observation involving the square root function, which was also defined by a restricted domain:

$\sqrt{(-3)^2} = \sqrt{9} = 3 \ne -3$.

# Inverse Functions: Arcsine (Part 10)

Let’s now switch from functions of the form $y= x^{m/n}$ to the trigonometric functions. We begin with $y = \sin x$. We will define the function $y = \sin^{-1} x$ using much of the same reasoning that defined $\sqrt{x}$.

We begin by looking at the graph of $y = \sin x$.

Of course, we can’t find an inverse for this function as stated. Colloquially, the graph of $f$ fails the horizontal line test. More precisely, there exist two numbers $x_1$ and $x_2$ so that $x_1 \ne x_2$ but $f(x_1) = f(x_2)$. Indeed, there are infinitely many such pairs.

So how will we find the inverse of $f$? Well, we can’t. But we can do something almost as good: we can define a new function $g$ that’s going look an awful lot like $f$. As before, we will restrict the domain of this new function $g$ so that $g$ satisfies the horizontal line test.

When we did this for the square root function, there were two natural choices for the restricted domain: either $[0,\infty)$ or $(-\infty,0]$. However, for the sine function, there are plenty of good options from which to choose. Indeed, here are four legitimate options just using the two periods of the sine function shown above. The fourth option is unorthodox, but it nevertheless satisfies the horizontal line test (as long as we’re careful with $\pm 2\pi$.

So which of these options should we choose? Historically, mathematicians have settled for the interval $[-\pi/2, \pi/2]$. Why did they settle on this interval? That I can answer with one word: tradition.

So we use the following bijective function (or, using the language that I used when I was a student, a one-to-one and onto function):

$g: [-\pi/2,\pi/2] \to [0,1]$ defined by $g(x) = \sin x$.

Since this is a bijection, it has an inverse function

$g^{-1}: [0,1] \to [-\pi/2,\pi/2]$

This inverse function is usually denoted as $y = \sin^{-1} x$ or $y = \arcsin x$. The graph of $y = \sin^{-1} x$ can be produced by reflecting through the line $y=x$, producing the purple graph below.

One other important note: since $g$ and $g^{-1}$ are inverse functions, it’s always true that $g(g^{-1}(x)) = x$ and $g^{-1}(g(x)) = x$. However, $g^{-1}$ and $f$ are not inverse functions, and so it’s possible for $g^{-1}(f(x))$ to be something other than $x$. This parallels the observation that, say, $\sqrt{(-3)^2}$ is not equal to $3$. We’ll discuss this further in future posts.

# Inverse Functions: Arcsine (Part 9)

I’d like to discuss a common mistake students make in trigonometry… as well as the one-liner that I use to (hopefully) help students not make this mistake in the future.

Question. Find all solutions (rounded to the nearest tenth of a degree) of $\sin x = 0.8$.

Erroneous Solution. Plugging into a calculator, we find that $x \approx 53.1^o$.

I offer a thought bubble if you’d like to think about why this answer is wrong.

The student correctly found the unique angle $x$ between $-90^o$ and $90^o$ so that $\sin x = 0.8$. That’s the definition of the arcsine function. However, there are plenty of other angles whose sine is equal to $0.7$. This can happen in two ways.

First, if $\sin x > 0$, then the angle $x$ could be in either the first quadrant or the second quadrant (thanks to the mnemonic All Scholars Take Calculus). So $x$ could be (accurate to one decimal place) equal to either $53.1^o$ or else $180^o - 53.1^o = 126.9^o$. Students can visualize this by drawing a picture, talking through each step of its construction (first black, then red, then brown, then green, then blue).

However, most students really believe that there’s a second angle that works when they see the results of a calculator.

Second, any angle that’s coterminal with either of these two angles also works. This can be drawn into the above picture and, as before, confirmed with a calculator.

So the complete answer (again, approximate to one decimal place) should be $53.1^{\circ} + 360n^o$ and $126.9 + 360n^{\circ}$, where $n$ is an integer. Since integers can be negative, there’s no need to write $\pm$ in the solution.

Therefore, the student who simply answers $53.1^o$ has missed infinitely many solutions. The student has missed every nontrivial angle that’s coterminal with $53.1^o$ and also every angle in the second quadrant that also works.

Here’s my one-liner — which never fails to get an embarrassed laugh — that hopefully helps students remember that merely using the arcsine function is not enough for solving problems such as this one.

You’ve forgotten infinitely many solutions. So how many points should I take off?