In a right triangle:

sin(2x + 4) = cos (46)

What is the value of x?

The question above is from standardized tests for a geometry course. If my goal is to have students understand cofunctions, what benefit is there to force students to do algebra? It seems that we are making geometry harder than it needs to be.

And I guess this could probably apply to other fields of math?

I can understand analysis courses requiring topology but the topology is taught with regards to the analysis and not as pure topology.

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    $\begingroup$ Being able to understand, analyze, and solve such functional relationships is important in certain contexts. Perhaps the contexts comprised by the standardized tests are broader than the context you posit. (I don't know anything about the tests from which this problem comes.) $\endgroup$
    – Raciquel
    Sep 24 at 0:53
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    $\begingroup$ What’s a cofunction? $\endgroup$ Sep 24 at 7:31
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    $\begingroup$ Can you explain why you assume the question is about geometry and why an algebraic understanding of cofunctions is not valuable? In other words, what's not algebraic about cos(x) = sin(pi/2-x)? $\endgroup$ Sep 24 at 15:56
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    $\begingroup$ If my goal is to have students understand cofunctions --- The question seems to me more about being able to APPLY cofunction identities, not understanding them or regurgitating them. Also, while $2x+4$ is a bit artificial, this is similar to what one often encounters in practice (later math, physics, engineering, etc.) -- in applying something like $\sin \theta = \cos (\frac{\pi}{2} - \theta),$ it's often (mostly?) the case that the "$\theta$" isn't just $\theta.$ $\endgroup$ Sep 24 at 20:30
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    $\begingroup$ @user3840170 cosine, cosecant, and cotangent (along with their esoteric cousins coversin and excosecant) are collectively called the "cofunctions." Some teachers stress to students that the term "cosine" should remind one "sine of complement," (and so on, for all three/five pairs) and find differentiating the "natural" functions from the cofunctions useful. Others never comment on it. $\endgroup$
    – nitsua60
    Sep 24 at 23:21

5 Answers 5


Is your ultimate goal really just to teach cofunctions?

Or are you trying to teach cofunctions so that the students can apply them later?

I am speaking as a student rather than an educator, but math, just as most subjects, is taught incrementally with more advanced concepts building on simpler concepts. All mathematics was invented to solve problems in the real world. Showing that a student can apply their knowledge of different concepts in isolation has little practical use. The real world is not a controlled laboratory environment where specific disciplines can be controlled and used in isolation. Only the simplest real-world problems can be solved by applying a single concept by itself.

Anecdotally, when I was in high school, I was continually surprised by the number of my classmates that would excel in advanced math and science courses yet struggle tremendously when multiple math concepts needed to be applied in conjunction, particularly with word problems. Well meaning educators do test concepts in isolation in order not to confuse students or "obfuscate" the problem, but it's my opinion that this will only serve to set the student up for failure in the future.

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    $\begingroup$ (+1) so that the students can apply them later --- I scanned the answers before writing my comment above, so it's possible I stole "apply" from your answer, although as I was writing my comment I originally didn't mention "apply", which was introduced during some later editing of my comment, and while I wasn't specifically thinking of your answer at the time, it may have been in my unconscious. (My comment originally had the simple harmonic motion example of rewriting $A\cos \omega t + B\sin \omega t$ as $C\cos(\omega t + \delta),$ then I realized this involved some more sophisticated trig.) $\endgroup$ Sep 24 at 20:35
  • $\begingroup$ @DaveLRenfro we easily could have come up with the same word. I think it's the logical conclusion to make $\endgroup$
    – PC Luddite
    Sep 24 at 20:52
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    $\begingroup$ This is the kind of answer I would like to be articulate enough to post myself. I strongly object to the OP's use of "obfuscate" in the title, as if that's a bad thing. This kind intermixing of concepts of what educators SHOULD be doing. It's about problem solving. $\endgroup$ Sep 24 at 22:35
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    $\begingroup$ +1 I strongly agree with this answer except for the sentence "All mathematics was invented to solve problems in the real world". This is probably true (at least to a large extent) for the kind of mathematics that this question is about, but claiming this for all mathematics is certainly too strong. $\endgroup$ Sep 25 at 5:16
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    $\begingroup$ 100% on this answer. Especially US schools (thanks standardized testing), but very widely elsewhere too, teachers just teach the material and be done with it, but that is in most cases absolutely not the actual point of the class. $\endgroup$
    – Hobbamok
    Sep 26 at 11:38

Tests seek to measure ability. Math ability, like most other forms of ability (including athletic ability), isn't solely dependent on one's ability to execute individual skills in isolation -- it also depends on one's ability to strategically combine skills.

Your goal might just be for your students to understand cofunctions, but the goal of the test is to also measure students' abilities to strategically combine that skill with other skills.


This multi-step question requires students to understand and apply multiple concepts or strategies to solve the problem. The goal of a standardized test is not to provide a correctly-sequenced list of questions that leads students to conceptual understanding; instead, tests need to include more difficult questions in order to distinguish between two students with above-average levels of ability. Ideally, students would have learned these concepts before approaching this problem. If a student is not able to solve the related equation $\mathrm{sin}(y) = \mathrm{cos}(46)$, then he or she is not ready to solve this one. In fact, one "strategy" to solving this problem would be to first solve the simpler equation and then assuming $y = 2x+4$, solve for $x$. This strategy effectively separates the issues for the student.

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    $\begingroup$ Indeed, I think many students would be able to solve $\sin(y) = \cos(46)$ by blindly copying a procedure from their textbook with only a superficial understanding, and the "obfuscation" of asking them to solve $\sin(2x+4) = \cos(46)$ will help them get a deeper understanding of what they're doing. $\endgroup$
    – Stef
    Sep 26 at 14:33

I have been on committees that write questions for standardized tests and placement tests. In this role, I have reviewed results of many trigonometry questions that were piloted and then revised for re-piloting, removed for future piloting all together, or approved for inclusion as a graded test item.

If this were piloted in the form of a multiple-choice question, I would expect the point-biserial correlation test score would raise lots of red flag!

The first thing that made me cringe with this question is probably just professional baggage: I got hung up quickly on the 46. I spent too much mental energy wondering if somebody meant $46^{\circ}$ instead of 46 radians. After all, $46$ is close to $45$ and $45^{\circ}$ is nice. The strangeness of the integer 46 was a big distraction for me. I got over it, and said to myself that in the spirit of the Douglas Adams answer to the universe, too bad it was not 42. Oh well.

But I did not get over it gracefully! The darn integer 46 and $\cos(46)$ made me start thinking about $\arcsin(\cos(46))$. Embarrassing. I thought to myself, "I mean really, $\cos(46)$ has no prospects for a future."

Eventually I realized I should think in terms of $\sin(\theta)=\cos(46)$ and then $$\sin(\theta)=\cos(\theta+\pi/2)=\cos(46)$$ and then I knew I was on the right path, finally. I was still annoyed with the manifest radian context with the $\pi$ on the left side and that thing I thought might be a misprint for $46$ degrees on the right.

Then the results are still messy and kind of tricky for most students:

$$\theta=2k\pi+\frac{27\pi}{2}+46$$ or $$\theta=2k\pi+\frac{29\pi}{2}-46$$ for $k\in{\bf Z}$

Yikes! And I still have to come back and fiddle with the substitution $\theta=2x+4$.

My guess is that if this question were piloted on a standardized test, it would come back with very poor results and beg for some revisions and extensive repiloting before being allowed on the exam for grading.

Observation: if the 46 was indeed supposed to be $46^{\circ}$ then the solutions for $\theta$ (in radians) would be cleaner:


or $$\theta=2k\pi+\frac{11\pi}{45}$$ for $k\in{\bf Z}$.

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    $\begingroup$ I just assumed everything was supposed to be in degrees, which is even more likely now that we know this was in a geometry section (maybe before radians are even introduced). In that case I don't think there's any reason to switch to radians for the answer. Since this was a standardized test though, we're only seeing part of the question. I wonder what the answer choices were. $\endgroup$
    – Thierry
    Sep 26 at 0:00
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    $\begingroup$ @Thierry you are probably correct. In degrees we would have $\theta=360k+136$ or $\theta=360k+44$ which seems brighter. And this plays nicely with the substitution $\theta=2x+4$. $\endgroup$
    – user52817
    Sep 26 at 0:13
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    $\begingroup$ In context of a right triangle, the only valid value for k is zero, $0<2k\pi +\frac{34\pi}{45} <\frac\pi2$ $\endgroup$
    – nickalh
    Sep 26 at 19:31
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    $\begingroup$ @nicalh yes but the original question did not include the context of a right triangle. Originally it just read “solve $\sin(2x + 4) = \cos (46)$.” $\endgroup$
    – user52817
    Sep 26 at 20:54
  • $\begingroup$ That's definitely a downside of stackExchange. The problems often evolve sometime invalidating answers. I've been known to do that myself- post a question and over a day or two revise the question, explaining more thoroughly what I understand, don't understand, etc. In my case, I try to caution others, the question will be revised. $\endgroup$
    – nickalh
    Sep 26 at 22:16

The goal IS NOT "to teach cofunctions". Trig functions are taught as a package and the relationships between them are usually taught as Pythagorean identities AND NOT as defining a cofun in terms of the original fun. Standardized tests are not always competently designed -- the "in a right triangle" context has no bearing on the problem or its solution -- sin and cos retain an inherently Pythagorean relationship to each other even when they are defined by other methods. Moreover, the algebra in 2x+4 isn't there to obscure anything -- "Algebra 1" is normally taught in the semester immediately preceding geometry, and a standardized test can be trying to confirm the progress through one to the other -- it can be testing whether you know how to solve 2x-4 = whatever, AND whether you know how to compute whatever from 46°. Also, the fact the question is quoted without SAYING ° opens a completely separate can of worms, which becomes a barrel if the test itself didn't clarify that in the original.


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