How to layout a solution to a trig equation?

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.

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?

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.

• What is the exact value that the $0.6435...$ is supposed to approximate? I would use that exact value or substitute that value with a letter to represent that constant. – Andrew Chin Aug 26 '20 at 13:46
• @Andrew Chin: At the risk of admitting ignorance well beyond what many here might expect from me, I couldn't figure out where $0.6435...$ came from. I thought maybe it was some way of rewriting $7$ in which a typo had been made (because $7$ doesn't vary when $x$ varies). It wasn't until I read Xander Henderson's answer that I realized what was being done, namely a rewriting of $6\cos x - 8\sin x$ performed independently of the given equation. FYI, I would probably not have even considered doing this, (continued) – Dave L Renfro Aug 26 '20 at 15:42
• as the most straightforward approach to me (and probably the only thing that I would have thought of unless I spent a few minutes thinking about other ways this might be solved) is to use $\sin^2 x + \cos^2 x = 1$ to express the left side of the equation quadratically in terms of one trig function, and then solve the corresponding quadratic equation for that trig function. But if I were conveying this to someone else (e.g. writing a homework or exam solution that would be graded), I would definitely do what Henderson says, and explain/justify within reason what I'm doing. – Dave L Renfro Aug 26 '20 at 15:53
• @Xander Henderson: I spent a few minutes working this out on paper after I wrote my comments, and I believe the reason for the OP's method of solution is to avoid extraneous solutions that arise when squaring (either when solving the radical equation by what I suggested, or when putting the $\sin x$ and $\cos x$ terms on opposite sides and squaring and applying $\sin^2 x + \cos^2 x = 1).$ For me, since checking for extraneous solutions when squaring is so ingrained, it's still simpler than having to remember a special "trig trick" device, but perhaps not for students for whom all this is new. – Dave L Renfro Aug 26 '20 at 16:31
• I have edited the main post with some more information about why I asked the question also. – Ben Derby Aug 26 '20 at 16:46

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:

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 $$x \approx 4.5605.$$

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.

• The ellipses kind of save the day --- In some situations I used notation like this when solving problems on the blackboard in class, but it was usually near the end when a numerical value was being obtained. However, for some students in lower level classes who are not so used to abstractly working with fractions, radicals, etc. I sometimes (earlier in the course) would put in the decimal approximations earlier, if I felt the students would better see what was going on (often this would be with individual students in office hours). But I still think the OP's write up was obscure. – Dave L Renfro Aug 26 '20 at 16:01

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 $$x = 6.151 \ldots$$.

• maybe this would be how I'd want students to present their work --- To be more accurate, this is how I might present the solution in class or in a handout. I wouldn't expect students to include as many words, but their solutions should still be sufficiently complete that (in my opinion) a good student in someone else's trig class could follow their reasoning. It should also be sufficiently complete that if they make a minor error along the way, I can still continue reading along with them for the purpose of awarding points for other aspects (i.e. award appropriate partial credit). – Dave L Renfro Aug 26 '20 at 18:34
• (+1) My only comment is that I really dislike it when students write things like $$\sin(\theta) = x \\ \theta = \arcsin(x) + \text{whatever}.$$ I strongly feel that there should be a "verb" which links those two lines together, e.g. $$\sin(\theta) = x \\ \implies \theta = \arcsin(x) + \text{whatever}.$$ I also worry about numerically checking for extraneous solutions, but graphing software can give one a better sense of how the problem is working. – Xander Henderson Aug 26 '20 at 19:01
• Oh, and I would use $\approx$ rather than $=$, and ditch the ellipses, but that's a pretty minor point. I really am just nitpicking, now. – Xander Henderson Aug 26 '20 at 19:02
• I used the ellipses to more strongly indicate truncation, and the purpose of giving a few digits like this was so that students could later check the calculations. In written handouts I did typically use the $\approx$ symbol, but on the blackboard I usually didn't because sometimes students don't notice it or don't reproduce it in their notes. One of the problems with using implies between the steps is that it's often not just "Step 6" implies "Step 7", but something like "(part of hypothesis) + (Step 1) + (Step 3) + (Step 6)" implies "Step 7". It also adds clutter at this level. – Dave L Renfro Aug 26 '20 at 19:12

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 $$0, which means the only ones we need are $$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 $$0. We see that $$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.)