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I am teaching a Trigonometry class, and every year we get to this point my students start asking a lot of questions. And for good reason.

Here is my issue: We teach that $\arccos\frac{1}{2}=\frac{\pi}{3}$ and this is the only answer because our range is restricted to $0\le{y}\le{\pi}$ We've been over that cosine and sine are not one to one functions which is why we restrict the domain to find the inverses.

However, when we start solving trigonometric equations and solve problems such as:

Find the solutions of $2\cos x=1$ in the interval [$\pi,2\pi]$

We solve by dividing by 2, and we have $\cos x=\frac{1}{2}$. Now, the book states that we take $\arccos(\frac{1}{2})$ and we find that equals $\frac{5\pi}{3}$. The problem I run into is that students object that $\arccos(\frac{1}{2})=\frac{5\pi}{3}$ because it does not fit within the range of $0\le{y}\le{\pi}$. Is it incorrect notation to use arccos in this situation as the book does? Because I agree with students that this does not fall within the range of arccos. It confuses students every year.

Instead, is it more proper to leave the equation $\cos x=\frac{1}{2}$ and find the value of $x$ without showing the arccos step?

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  • $\begingroup$ As Michael E2's comment on Benoît Kloeckner's answer points out, the best answer may depend on how your students have been previously indoctrinated. If they have a religious objection to multivalued functions, does that objection extend to statements such as "every positive number has two square roots?" In any case, it's easy to describe the solution set of $x^2=4$. Then you can build up in difficulty to $\tan x=0$, $\sin x=0$, $\cos x=1$, and then examples whose solution sets are more complicated to describe, such as $\cos x=0$ and $\cos x=1/2$. $\endgroup$ – Ben Crowell Jan 28 '17 at 2:19
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You use arcos to get pi/3. Then you use that as a reference angle, and find the angle in your range with the same reference angle. If the angle is not from a special triangle (30-60-90 or 45-45-90), you will definitely need the arccos step.

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  • $\begingroup$ Ahh, makes sense. I will start using this in my language about arccos. Thanks $\endgroup$ – MathGuy Jan 26 '17 at 20:12
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I certainly disagree strongly with your book: there should be one unique and consistent definition of each function. As is usual, you defined $\arccos$ as the map sending $x\in[-1,1]$ to the unique $\theta\in[0,\pi]$ such that $\cos\theta=x$, and this definition should not vary even by a nanometer.

To solve your problem, you could use (and prove) that the unique solution $\phi\in[\pi,2\pi]$ of $\cos\phi=x$ is $2\pi-\arccos x$ (it is easily checked that this number is in the correct range, and that its cosine is indeed $x$; then uniqueness is basically a change of coordinates). Using a trigonometric circle as mentioned by Michael E2 also has an important part of understanding issue, you don't want your students to memorize a bunch of raw formulas without any understanding or way to check them.

(Oh, and by the way: $\arccos(1/2)$ is $\pi/3$, not $2\pi/3$.)

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    $\begingroup$ I agree with what you say, but it reminds me that I was taught that there were two kinds of inverse trigonometric symbols, one denoting a "multivalued function" (lower case "arc") and another denoting the principal value (upper case "Arc"); the principal value is the standard definition nowadays, whatever the symbolism, it seems. However, I still get a few students (in college calculus, in the US) who have been taught that the inverse trig. functions are multivalued. $\endgroup$ – user1527 Jan 27 '17 at 11:23
  • $\begingroup$ Ah yes, typo there... arccos(1/2) is pi/3 $\endgroup$ – MathGuy Jan 27 '17 at 16:30
  • $\begingroup$ @MichaelE2: you must be right, some other convention must exist. The main point is: once you've fixed a convention, you stick to it; here I used the convention which seemed to have been introduced by the asker. Also, I would feel quite difficult to use multivalued functions without writing a misleading statement on the board (should we then write $\arccos(1/2)$ as a set? or change the very meaning of $=$? $\endgroup$ – Benoît Kloeckner Jan 28 '17 at 9:32
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I like to draw the analogy to the square-root function vs. solving a square.

If we want to calculate $\sqrt{9}$ then there is only one answer (especially if we're using a calculator) and it is $3$. We choose to take the positive root. However, when solving $x^2=9$, we typically want all $x$ solutions and so we say $x=\pm\sqrt{9}=\pm 3$. It may help to remind them that the graph of $y=\sqrt{x}$ is only half of the graph of $y^2=x$.

Likewise, when we want to calculate $\arccos(1/2)$ there is only one answer (and this is the one the calculator gives) and it is $\pi/3$. This, too, is a choice. So, again, when solving $\cos(x)=1/2$, we want all those choices. For $\arccos$ we have more than just the $\pm$, we also have to consider $x=\pm \pi/3+2\pi k$ (for $k$ an integer). (And $\arcsin$ gets even more difficult since $\pm$ isn't the right thing.) Again, remember that the graph of $y=\arccos(x)$ is only a small piece of $\cos(y)=x$ that gets repeated over and over again.

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It's really not the same function, unless you start talking about Riemann surfaces and multivalued functions (which I assume you won't be). Computer algebra systems distinguish between solving an equation and evaluating inverse trig functions, too. Personally, I would avoid using the arccosine language when solving, other than the specific step. Inverse trig functions are their own beasts, useful in their own ways (notably solving useful integrals).

Instead, when solving, draw a picture of the function (here, $2\cos(x)$ or $\cos(x)$) and see what they can expect in terms of the symmetry. For my money, solving to get $\pm \pi/3$ and then pointing out the $2\pi$ repetition, and then finally narrowing to the interval in question is the best bet. However, that requires enough time to have a fairly leisurely discussion, unless you feel like they really "get" that periodicity. Since your students are complaining that it's wrong instead of just writing things down numbly, you may be in good shape to have these discussions!

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  • $\begingroup$ Good idea. I can add this to my discussion in future years. $\endgroup$ – MathGuy Jan 26 '17 at 20:41

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