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Trigonometry has some complex stuff, but sometimes students have trouble with the beginning, the basics.

As an example, here's what seems to be a simple question:

$$2cos(x)=1$$ $$0°≤x≤360°$$ Find the possible values of $x$

For students, this question may not seem as simple to them. Is there a good way I can explain this question, and these classes of questions in general?

If you can, provide an example question and an example explanation.

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    $\begingroup$ What is the $cos^{-1}\times (\frac{7}{6})$ supposed to be? Using improper notation certainly would confuse the students. $\endgroup$ – kcrisman Nov 7 '16 at 13:53
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    $\begingroup$ I suspect many students will be confused when you say "find the value" when there is more than one value satisfying the constraints you gave. $\endgroup$ – Dave L Renfro Nov 7 '16 at 15:31
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    $\begingroup$ In case it helps, I found the following on an old trigonometry test of mine just now: Find all values of $\theta$ such that $\theta$ is in the interval $\left[0^{\text{o}},\,360^{\text{o}}\right)$ and $\cos \theta = \frac{1}{2}$. If there are no such values of $\theta$, then state this. $\endgroup$ – Dave L Renfro Nov 7 '16 at 15:38
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    $\begingroup$ To revisit this question (and since I have a few moments left in my lunch break), the way I used to explain how to see the "big picture" when solving these types of questions is to reduce the equation(s) to SINE = constant(s) and/or COSINE = constant(s), and then draw a unit circle with appropriate vertical and/or horizontal lines, a method that makes use of the fact that SINE corresponds to $y$-coordinates on the unit circle and COSINE corresponds to $x$-coordinates on the unit circle. Thus, for $\cos \theta = \frac{1}{2},$ sketch $x^2+y^2=1$ along with $x=\frac{1}{2}.$ $\endgroup$ – Dave L Renfro Nov 7 '16 at 19:12
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    $\begingroup$ The point of the unit circle is to see how many solutions and which quadrants they're in. The exact values come from knowing the values to the reference angles and how to make use of this in other quadrants. This isn't a good week for me --- very busy at work --- so I can't really say much more now, but I might come back to this in a week or two if no one else has posted much. $\endgroup$ – Dave L Renfro Nov 7 '16 at 22:35
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It seems like a problem of unclear or inapplicable definitions. As stated in the comments, it sounds like you've given the students the following definition: Given an angle $t$, we consider a right triangle with an angle of measure $t$, and define $\cos(t)$ to be the ratio of the side length of this triangle adjacent to the given angle to the length of the hypotenuse.

The problem is, this definition only makes sense when $0 < t < \pi/2$ (or, if you prefer degrees, $0^\circ < t < 90^\circ$. Indeed, a right triangle's other two angles have measure strictly between $0^\circ$ and $90^\circ$. But you've given the students a problem that assumes cosine has been defined for any $t$ between $0^\circ$ and $360^\circ$, which makes no sense with the definition they've been given!

I would recommend instead defining cosine as the $x$-coordinate on the unit circle. A simple diagram (drawing an appropriate right triangle inside the unit circle) shows this agrees with the other definition for angles in the first quadrant, but the unit circle definition also clearly makes sense for any angle. Plus, it makes it much easier to reason about this kind of question — just draw the line $x = 1/2$ and see where it intersects the unit circle, then use geometric reasoning to find the angles.

I'm still a bit confused about what you mean by "algebraic methods", by the way. Both the triangle definition and the unit circle definition are geometric, so you really can't compute values of cosine using these definitions without appealing to geometry.

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    $\begingroup$ @TheBitByte: I still don't understand how you expect your students to solve the problem (or even make sense of the problem statement) without appealing to the unit circle definition. Could you post a solution along the lines of what you'd want your students to give? $\endgroup$ – Daniel Hast Nov 10 '16 at 3:57
  • $\begingroup$ @TheBitByte Yeah... no. And not to mention radians to degrees will only confuse them more. $\endgroup$ – Simply Beautiful Art Nov 12 '16 at 13:55
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For the easy memorizing of special points:

$$\cos(0^\circ)=\frac{\sqrt4}2\\\cos(30^\circ)=\frac{\sqrt3}2\\\cos(45^\circ)=\frac{\sqrt2}2\\\cos(60^\circ)=\frac{\sqrt1}2\\\cos(90^\circ)=\frac{\sqrt0}2$$

Same thing for $\sin(\theta)$ but going backwards. Then it should be easy to see that $x=60^\circ$ is one answer. From there, I would try to show them the unit circle and how it is symmetric. Using that, it should become obvious that $300^\circ$ is the only other answer.

Beyond that, either you are teaching some pretty advanced trigonometry, it happens to be a special special point, like $\cos(15^\circ)$ using the half-angle theorem, or you give them calculators. I can't imagine it being any other way, unless you were teaching a numerical analysis type of class.

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    $\begingroup$ Why would you ask students to memorize values of cosine or sine? It seems like a pointless exercise that short-circuits the actual (and actually interesting) geometric reasoning that goes into finding those values. $\endgroup$ – Daniel Hast Nov 12 '16 at 16:06
  • $\begingroup$ @DanielHast Well, after learning what sine and cosine mean, it helps to memorize these values since they appear more than other values. And as to how much of an exercise this is, I think it's really easy. If you take a look, there is a pretty simple pattern, one that takes only moments to memorize. $\endgroup$ – Simply Beautiful Art Nov 12 '16 at 16:27
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    $\begingroup$ Why it does help to memorize those values? What's the point if you don't understand why cosine and sine have those special values? $\endgroup$ – Daniel Hast Nov 12 '16 at 16:35
  • $\begingroup$ @DanielHast Well, clearly by the time one gets to this, they should have a basic understanding of why cosine and sine have those special values. This answer doesn't serve to teach trigonometry, just to help others recall certain values more easily. Your answer does a wonderful job of that, and I applaud, but the OP still isn't satisfied. $\endgroup$ – Simply Beautiful Art Nov 12 '16 at 16:42
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    $\begingroup$ @danielhast for the same reason the OP avoids the unit circle, you'd end up having to remember how to construct those triangles, and as far as remembering goes, my answer is arguably simplest. $\endgroup$ – Simply Beautiful Art Nov 12 '16 at 19:08
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If students are struggling to understand a relatively simple trig equation like the one given, I find that graphs often help. We can sketch y=2cos(x) and y=1, and get a feeling for how many solutions to expect in the given interval. Moreover symmetries of the graphs can help determine the full set of solutions, much as the unit circle approach does. Interestingly, I believe in the UK the unit circle development of trig functions isn't such a common approach but many teachers use the (IMO awful) 'CAST' diagram for solving trig equations, devoid of deeper understanding.

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  • $\begingroup$ I feel like there's a logical gap there, similar to in the OP's attempt at explaining the problem: Your post makes it sound like the unit circle definition is optional, but how is cosine even defined for arbitrary angles if not by the unit circle? For the graph of cosine to have any symmetries in the first place, it has to be defined for angles outside the first quadrant. $\endgroup$ – Daniel Hast Jul 18 '17 at 15:09
  • $\begingroup$ I didn't intend for it to sound optional - I completely agree with you in terms of extending the meaning of cosine beyond acute angles. However, when it comes to the specific skill of finding multiple solutions from an equation, I've found that my students have typically preferred the graphical point of view over the unit circle representation. $\endgroup$ – sxpmaths Jul 23 '17 at 14:13

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