How to layout a solution to a trig equation?

Question

I am interested in how you would encourage students to layout their working to a trigonometric equation.

For instance, let's consider this problem:

Solve the equation $6\cos x - 8\sin x = 7$ for $0 < x < 2 \pi$

Then I might write $$6 \cos x - 8 \sin x = -10 \sin \left(x - 0.6435... \right)$$

and then solve $$-10 \sin(x - 0.6435...) = 7$$

This re-arranges to $$\sin(x-0.6435...) = -0.7$$ The principal value is $x - 0.6435... = -0.7753...$.

The values in range are $x-0.6435... = \pi + 0.7753...$ and $x - 0.6435... = 2\pi - 0.7753...$

And if you solve these you get $x=4.56$ and $x=6.15$ to two decimal places.

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However, this has a LOT of decimals and I find that it is not the clearest way for a student to lay out their working.

Of course we can work more exactly, but then that'd be quite difficult for lower ability high school students to understand rather than the formulaic approach.

So how would you advise a student to layout their working for solving a trigonometric equation to keep it as simple and neat as possible?

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Edit: in UK schools, solving equations of the form $a\cos x + b \sin x = c$ is almost always encouraged to be done by first writing $a \cos x + b \sin x$ in the form $R \sin (x + \alpha)$ or $R \cos (x + \alpha)$, depending on what is most suitable. That is why I skipped that step.

My main point is I guess a general question in how do you avoid the excessive use of decimals in a problem like this? Of course, as I said above, you would work exactly and for me and you as skilled mathematicians this is easy. However, for the lower ability students, I wondered if anybody else had a neater way to lay something like this out that doesn't up the demand that much.

Answer 1

In most mathematics classes, we don't actually care about the solution to an exercise. The point is to get students to practice with the concepts, and figure out how to communicate their thinking. The solution you present has a number of problems for which I would likely deduct points:

• The first step is to assert that $6 \cos(x) - 8 \sin(x) = -10 \sin(x - 0.6435\dotso)$. First off, why? How did you come to that conclusion? Why is this true? Second, it isn't really even true. The ellipses kind of save the day, but what you seem to have done is used a calculator to compute something, which means that there is some roundoff error somewhere. You don't have true equality.

• Farther along, you assert that "The principle value is $x - 0.6435\dotso = -0.7753\dotso$". Again, why? How did you get that?

• In general, I would delay calculator based approximations until as late as possible. If there are ugly expressions which you want to avoid working with, do as Andrew Chin suggests, and hide those expressions in a variable.

• Much of what you have written is in the first person singular, but you finish in the second person. It is also phrased somewhat passively. I suppose that this is a stylistic choice, but I tend to take off (small fractions of) points for style. Mathematics should be written actively, preferably in the imperative, and if one must resort to pronouns, use the first person plural (e.g. "We write...", "We solve...").

• Near the end of the problem, you get two solutions by taking advantage of the fact that $\sin(\theta) = \sin(\pi - \theta)$ for any angle. This should be made explicit.

If I were to write up a solution to this problem, I would likely write something similar to the following:

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Solve $6\cos(x)- 8\sin(x) = 7$ for $0 < x < 2\pi$.

Solution: By the angle addition formula the sine function, $$ 7 = 6\cos(x)- 8\sin(x) = A \sin(x+\alpha) = A\sin(\alpha)\cos(x) + A \cos(\alpha)\sin(x). \tag{*}$$ Equating coefficients on the left and right, \begin{align} \begin{cases} 6 = A\sin(\alpha), \\ -8 = A\cos(\alpha) \\ \end{cases} &\implies A = -\frac{8}{\cos(\alpha)} \\&\implies 6 = -\frac{8}{\cos(\alpha)} \sin(\alpha) \\&\implies -\frac{3}{4} = \frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha). \end{align} This final equation is solved by $$ \alpha = \arctan\left(-\frac{3}{4}\right) + k\pi,\qquad k\in\mathbb{Z}.$$ With $k$ even, the angle $\alpha$ is in the fourth quadrant, and so corresponds to a right triangle with $\alpha$ at the origin, and vertices at $(4,0)$ and $(4,-3)$. This is a $3$-$4$-$5$ right triangle with $\alpha$ adjacent to the leg of length $4$, thus $$ \cos(\alpha) = \frac{4}{5}. $$ By similar reasoning, if $k$ is odd, then $\alpha$ corresponds to an angle in quadrant II, and so $$ \cos(\alpha) = -\frac{4}{5}. $$ Therefore $$ \cos(\alpha) = (-1)^k \frac{4}{5}, \qquad k\in\mathbb{Z}. $$ Hence $$ A = -\frac{8}{\cos(\alpha)} = -\frac{8}{(-1)^k (4/5)} = (-1)^{k+1} \frac{8\cdot 5}{4} = (-1)^{k+1} 10. $$ Substitute this into (*) and use the fact that $\arctan(-X) = -\arctan(X)$ to get $$ 7 = 6\cos(x)- 8\sin(x) = (-1)^{k+1} 10 \sin\left(x-\arctan\left(\frac{3}{4}\right) + k\pi \right), \qquad k\in \mathbb{Z}. $$ Isolate the sine term and solve to obtain solutions: \begin{align} &\sin\left(x-\arctan\left(\frac{3}{4}\right)+ k\pi \right) = (-1)^{k+1} \frac{7}{10} \\ &\qquad\implies x-\arctan\left(\frac{3}{4}\right)+ k\pi = \arcsin\left( (-1)^{k+1} \frac{7}{10} \right) \\ &\qquad\implies x = \arcsin\left( (-1)^{k+1} \frac{7}{10} \right) + \arctan\left(\frac{3}{4}\right)+ k \pi. \tag{**} \end{align} Observe that \begin{align} &x = \arcsin((-1)^{k+1} X) + k\pi \\ &\qquad \implies \pi - x = (k+1)\pi - \arcsin((-1)^{k+1} X) = (k+1)\pi + \arcsin((-1)^{k+2}X). \end{align} If the first statement corresponds to an even $k$, then the second corresponds to an odd $k$, and vice versa. Thus (**) gives all possible solutions to the original equation.

To obtain decimal approximations, note that when $k$ is even, the solution is $$ x = \arcsin\left( -\frac{7}{10} \right) + \arctan\left(\frac{3}{4}\right) + 2n \pi \approx -0.1319 + (6.2832)n, \qquad n\in\mathbb{Z}. $$ To obtain a solution between $0$ and $2\pi$, take $n=1$ to get $$ x \approx 6.1513.$$

When $k$ is odd, the solution is $$ x = \arcsin\left( \frac{7}{10} \right) + \arctan\left(\frac{3}{4}\right)+ (2n+1) \pi \approx 1.4189 + 6.2832(2n+1), \qquad n\in\mathbb{Z}.$$ To obtain a solution between $0$ and $2\pi$, take $n=0$ to get $$ x \approx 4.5605. $$

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I will note that my solution is quite long. There are likely a lot of places where it could be simplified and shortened—the above is, essentially, a stream of consciousness, which has not been edited much. However, this is kind of how I expect students to turn in work: stream of consciousness. The important thing is that each step of the computation is explained and justified.

Answer 2

This was written after the OP's Edit update, which in turn was made after Xander Henderson's answer.

Using $\;a \cos x + b \sin x \; = \;R \sin (x + \alpha)$

Assuming the student is allowed to use without verification the relevant formulas, maybe this would be how I'd want students to present their work. (Note: All decimal expansions are 3-digit truncations of the corresponding exact values.)

Since $\;6\cos x - 8\sin x = R\sin(x + \alpha)\;$ for $R = -\sqrt{6^2 + 8^2} = -10$ and $\alpha = \arctan \left(-\frac{3}{4}\right) = -0.643 \ldots,$ we have

$$-10 \sin \left[x \; + \; \arctan\left(-\frac{3}{4}\right)\right] \;\; = \;\; 7$$

$$\sin \left[x \; - \; \arctan\left(\frac{3}{4}\right)\right] \;\; = \;\; -0.7$$

Sine is negative only in Q3 and Q4, and because $\arcsin(-0.7) = -0.775 \ldots,$ the $(x$-axis) reference angle for $x - \arctan\left(\frac{3}{4}\right)$ in these quadrants is $0.775 \ldots$. Therefore,

$$ x \; - \; \arctan\left(\frac{3}{4}\right) \; = \; \pi + 0.775\ldots, \;\; 2\pi - 0.775\ldots $$

$$ x \; - \; \arctan\left(\frac{3}{4}\right) \; = \; 3.916 \ldots, \;\; 5.507 \ldots $$

$$ x \;\; = \;\; (3.916 \ldots) + (0.643 \ldots), \;\; (5.507 \ldots) + (0.643 \ldots) $$

$$ x \;\; = \;\; 4.560 \ldots, \;\; 6.151 \ldots $$

Since all solutions to $\;\sin (\text{stuff}) = -0.7\;$ are given by adding $2n\pi$ to the above two solutions (i.e. by adding integer multiples of $2\pi),$ and only $n=0$ gives values of $x$ such that $0 < x < 2\pi,$ the above two values are all solutions for $x$ such that $0 < x < 2\pi.$

Using $\;\sin^2 x + \cos^2 x = 1$

$$ 6\cos x \; = \; 7 + 8\sin x $$

$$ (6\cos x)^2 \; = \; (7 + 8\sin x)^2 $$

Because we've squared both sides of an equation, we will need to check for extraneous solutions at the end.

$$ 36\cos^2 x \;\; = \;\; 49 + 112\sin x + 64\sin^2 x $$

$$ 36(1 - \sin^2 x) \;\; = \;\; 49 + 112\sin x + 64\sin^2 x $$

$$ 100\sin^2 x + 112\sin x + 13 \; = \; 0 $$

$$ \sin x \;\; = \;\; \frac{-112 \; \pm \; \sqrt{{112}^2 \; - \; 4(100)(13)}}{200} $$

$$ \sin x \;\; = \;\; \frac{-112 \; \pm \; \sqrt{7344}}{200} $$

$$ \sin x \;\; = \;\; -0.131 \ldots, \;\; -0.988 \ldots $$

Sine is negative only in Q3 and Q4, and because $\arcsin(-0.131 \ldots) = -0.131 \ldots$ and $\arcsin(-0.988 \ldots) = -1.418 \ldots,$ the $(x$-axis) reference angles for $x$ in these quadrants are $0.131 \ldots$ and $1.418 \ldots$. Therefore,

$$ x \;\; = \;\; \pi + 0.131 \ldots, \;\; \pi + 1.418 \ldots, \;\; 2\pi - 0.131 \ldots, \;\; 2\pi - 1.418 \ldots $$

$$ x \;\; = \;\; 3.273 \ldots, \;\; 4.560 \ldots, \;\; 6.151 \ldots, \;\; 4.864 \ldots $$

Since all solutions to $\;\sin (\text{stuff}) = \text{constant}\;$ are given by adding $2n\pi$ to the above four solutions (i.e. by adding integer multiples of $2\pi),$ and only $n=0$ gives values of $x$ such that $0 < x < 2\pi,$ the above four values include all solutions for $x$ such that $0 < x < 2\pi.$

Checking for extraneous solutions, we find that $3.273 \ldots$ and $4.864 \ldots$ do not satisfy the original equation. Therefore, the solutions to the given equation such that $0 < x < 2\pi$ are $ x = 4.560 \ldots$ and $x = 6.151 \ldots$.

Answer 3

The two solutions already posted suggest using "arctan(-3/4)" in place of the lengthy decimal throughout. I suggest something slightly different, which is a valuable strategy whenever you have an annoying repetition: give the repeated thing a name. So, something like this (where I have made a few incidental choices in different ways from the original solution; obviously you can use $\sin(x+\theta)$ instead of $\cos(x+\theta)$ and obviously you can use $\alpha$ instead of $\theta$ or whatever). I've actually found it convenient to give names to two different quantities that arise in the solution

We begin by writing $6\cos x-8\sin x=R\cos(x+\theta)$. Following the usual procedure we find $R=\sqrt{6^2+8^2}=10$ and $\theta=\tan^{-1}(8/6)=0.927...$ .

Now we must solve $10\cos(x+\theta)=7$ or, equivalently, $\cos(x+\theta)=0.7$. One solution to this is $x+\theta=\cos^{-1}(0.7)=0.795...$; calling the RHS $y$, all the solutions are $x=(2\pi n\pm y)-\theta$ for integer $n$. We are interested in solutions with $0<x<2\pi$, which means the only ones we need are $2\pi+y-\theta=6.151...$ and $2\pi-y-\theta=4.560...$ .

In practice, a student might do well to be more explicit about figuring out which integer multiples of $2\pi$ and which choices of sign yield solutions in the given range. Something like this:

... We are interested in solutions with $0<x<2\pi$. We see that $y-\theta$ is a little smaller than 0, so this is too small but $2\pi+y-\theta$ is in range (and $2\pi$ more than that is too large); and $-y-\theta$ is very roughly -2, so again this is too small but adding $2\pi$ brings it into range (and adding $4\pi$ makes it too large). So our solutions are $2\pi+y-\theta=6.151...$ and $2\pi-y-\theta=4.560...$ .

(Most students will probably need to compute the actual values rather than just saying "a little smaller than 0" and "very roughly -2", but if I were writing a model answer then I might leave phrases like those in it, to encourage students to learn to do that sort of rough estimate in their heads. Contrariwise: Some students will need to write down three numerical values for each choice of sign, one "too small", one "just right", and one "too large". There's something to be said for doing that in a model answer, too.)