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When teaching trigonometric identities, I found the students had trouble proving them. All the students taking turns to ask almost all the questions of an exercise embarrased me somewhat. In order to make it easier, I told them that these questions are asked not to test your algebraic ability, but to test how you use the reciprocal, quotient and Pythagorean relations we studied.

What should I mention to make it easier for them to solve those problems? Are there fundamental things we should consider while proving any kind of identities?

As for an example: $\frac{\cos \left(\theta \right)}{1+\sin \left(\theta \right)}+\tan \left(\theta \right)=\sec \left(\theta \right)$

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Which kind of education level are we talking about? Can you add further information on the course you were teaching? – Roland May 15 '14 at 12:22
I am talking about secondery level education. I tried to mention this in the tags but it said i cannot create one. – idiosincrasia23 May 15 '14 at 12:26
To apply those relations requires familiarity with algebraic manipulation. Can't do without, perhaps you should review that with them, possibly in the setting of simplifying some horribly-looking expressions with an assortment of trigonometric functions? – vonbrand May 15 '14 at 15:20
I don't have time to write an answer now, but the method I describe in this 26 October 2010 AP-Calculus post at Math Forum may be worth considering. The method cuts down on the creativity needed by the students while still giving them practice in algebraic manipulation and using the various trig. identities. – Dave L Renfro May 15 '14 at 20:16
I'd tend to echo @KevinO'Bryant 's comment in his answer below that these issues are anachronisms. Except as inherited tradition, "proving trig identities" under various rule systems is artificial make-work. If/when I need a trig identity, I convert the issue to exponentials, using complex numbers, just as KevinO'B. indicated. I recognize that many of us have no control of curriculum, but it's a pity that kids have to face fake difficulties when they could be learning something that works better, is easier, and connects to other things. – paul garrett Jul 16 '14 at 22:02

5 Answers 5

Trigonometric identities test students' algebraic ability. I don't think it helps to tell students otherwise or tell them that that's not the point.

If you'd rather give trigonometry problems that don't rely so heavily on algebraic ability, you can try numerical ones. Plot $\sin(x)-x$: for what values of $x$ is this less than 0.1? Plot some temperature data for your city on a reasonable axis: what simple function has the same maximum and minimum in roughly the same places? If the earth is a sphere and the equator has a circumference of 40,000 km, what is the circumference of the Arctic circle, at latitude 66 degrees?

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It seems as though your students can't cross the barrier of trying out algebraic manipulations. This is a general problem, that needs to be addressed some time and rather early than late, but at the moment it helps, if you give them an enumerated list of manipulations to apply, that covers most easy examples:

  1. Replace all trigonometric functions by $\sin$ and $\cos$.

  2. Apply addition theorems until only simple arguments ($x$, $y$, $z$, …) appear.

  3. Remove all roots and fractions by reverse operations. (should already be a known heuristic step to them)

  4. Factorize to get factors of the type $\sin^2x +\cos^2x$ and apply Trigonometric Pythagorean.

  5. Use other standard algebraic tools to proove identity.

This algorithm might create really long terms, but should work in most easy examples. If exercised some, students might observe occurences of shortcuts, that were left out (Trigonometric Pythagorean, Addition Theorems, standard algebraic tools, …). They should then apply these shortcuts whereever possible. After some more exercise, they might get the idea, that there is no enumerated list that always yields the shortest, easiest and correct result, but only a set of possible manipulations that can be applied and the rest is trial-and-error with some experience.

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I work as a math aide, students (grade 9-12) come to see me when they need help. I recently had a number of student come in, all looking for help on this exact issue. It's not teacher specific, three teachers' classes are in sync, the same chapter in the Trig book.

Your example is exactly an example of one that I needed to address. The fact that tan(θ) is sin(θ)/cos(θ) was the first thing I tried to point out. That was fine, but then it took more coaxing to move forward even a bit. The process of multiplying the denominators to create a common denominator (and then multiplying the numerators to even things out, of course, before adding) was a struggle. I was relying on the assumption that these algebraic manipulations would come easily, but this wasn't the case. It took me solving a proof from beginning to end, then spelling out the steps for the next problem. Finally, on the 3rd or 4th problem, the student would work completely independently.

There are a number of distinct skills required to get through this type of exercise, and if I were in front of a classroom, I'd do my best to evaluate what knowledge gaps the class had. If the lower 2/3 of the class needs a quick arithmetic review (add 1/3 + 1/6, or 4/5 + 5/8) it might be worth it.

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+1 for "If the lower 2/3 of the class needs a quick arithmetic review (add 1/3 + 1/6, or 4/5 + 5/8) it might be worth it." I've seen first-hand too many teachers (and professors, etc.) win the battle and lose the war on issues such as this. Just review it quickly and move on -- this takes less time than a diatribe about what they should know, which also tends to alienate the students. – Dave L Renfro May 16 '14 at 20:15

A trig "identity" is true if and only if the algebraic ``identity'' obtained by replacing $$\sin(k\theta) \mapsto \frac{z^k-z^{-k}}{2i}, \cos(k\theta) \mapsto \frac{z^k + z^{-k}}{2}$$ is true. We already teach them algebra, and trig identities are no more than algebra (with imaginary numbers). Add in $$d\theta \mapsto \frac{dz}{i z}$$ and the method of partial fractions, and you can do all of the trig integrals, too.

Trig identities such as the one in the OP are an anachronism. Since we don't sail ships by hand anymore, it is pretty rare for anyone to need to be facile with trigonometry past the right-triangle and unit circle interpretations of $\sin(\theta),\cos(\theta)$, and their immediate consequences.

And while I'm on the subject of anachronisms, $$\sin( \tfrac{\pi}{4}) = \frac{1}{\sqrt 2}.$$ It's not $\frac{\sqrt 2}{2}$. Of course they are equal, but which expression is simpler? The only justification for the old $\sqrt{2}/2$ (how ugly is that?) is if you use a table of square roots to handle conversions to decimal form. Our parents did that, and we learned how to do it, but I don't see a point to teaching our children how to use tables.

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"Trig identities such as the one in the OP are an anachronism" - I'm not in a position to disagree with this sentiment, but it seems this doesn't answer the question. The identities, or whatever you'd like to call them, are part of the Trig curriculum. The topic was just covered at my school, and I am still working on the tricks and tips to offer the students that come in with this assignment. – JoeTaxpayer May 30 '14 at 17:36
That's a relevant point, and it wasn't clear to me that the OP has a set curriculum s/he has to follow. As a uni professor, I have a hand in setting our curriculum, and for anyone in my position it does answer the question, particularly the "fundamental" part. And moreover, my first paragraph gives a simple to use algorithm guaranteed to always work. How much "easier for them" can it be made? – Kevin O'Bryant Jun 1 '14 at 0:28
Writing $\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$ has the nice effect, that it combines so well with the other "base" values: $\sin(0)=\frac{\sqrt{0}}{2}$, $\sin(\frac{\pi}{6})=\frac{\sqrt{1}}{2}$, $\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$, $\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$, $\sin(\frac{\pi}{2})=\frac{\sqrt{4}}{2}$. Apart from that, rationalizing the denominator is indeed an anachronism. – Toscho Jun 1 '14 at 16:41
Fair point, Toscho. I suppose one could also remember them as $\sqrt{0/4}, \sqrt{1/4}, \sqrt{2/4}, \sqrt{3/4}, \sqrt{4/4}$. – Kevin O'Bryant Jun 2 '14 at 16:38
Another justification for $\sqrt{2}/2$, or rationalizing any radical, is that a standardized form for answers makes grading quicker. (It's an important practical consideration for teachers in many school systems in the US, unfortunately.) – Michael E2 Sep 2 '14 at 3:21

As a response to Kevin's answer, and for those who might better understand what the OP is required to teach as part of a high school trig class. OP did not explicitly it, but often, the instruction given is to manipulate the left side of the equation until it's clearly equal to the right side, therefore no multiplying both sides, cross multiplying, etc. Some teachers ask students to use a two column proof style, justifying each manipulation along the way. This is how I walked students through the OP's example. The first transition from (1) to (2) was the one that took the most discussion even though it seems the most obvious (to me) thing to do.

(1) $\frac{\cos \left(\theta \right)}{1+\sin \left(\theta \right)}+\tan \left(\theta \right)=\sec \left(\theta \right)$

(2) $\frac{\cos (\theta )}{1+\sin (\theta )}+\frac{\sin (\theta )}{\cos (\theta )}$

(3) $\frac{\cos (\theta )}{1+\sin (\theta )}\left(\frac{\cos \theta }{\cos \theta }\right)+\frac{\sin (\theta )}{\cos (\theta )}\left(\frac{1+\sin (\theta )}{1+\sin (\theta )}\right)$

(4) $\frac{\left(\cos ^2\theta +\sin ^2\theta +\sin \theta \right)}{\left(1+\sin (\theta )\right)\cos \theta }$

(5) $\frac{\left(1+\sin (\theta )\right)}{\left(1+\sin (\theta )\right)\cos \theta }$

(6) $\frac{1}{\cos \theta }$

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Cross multiplying looks so much easier. And why ever bother with $\sec$? Even $\cot$ is not really needed. – Toscho Jun 1 '14 at 16:33
No doubt. Perhaps AaKASH is working from the same book my student are, Algebra and Trigonometry: Functions and Applications, by Paul Foerster, or the exercise is a common one, but I'm not in a position to dispute the rules. – JoeTaxpayer Jun 1 '14 at 17:20
Why? A rule to solve or proove something with only one method is bullshit. One can give guidelines, but nothing more. – Toscho Jun 2 '14 at 13:22
When there are multiple ways to solve a problem, teachers will often teach each one independent of the others. In this case, the goal was to teach it under these constraints. As a math aide, my job is to help, not dispute pedagogy. – JoeTaxpayer Jun 2 '14 at 13:34
What does "math aide" mean? – Toscho Jun 2 '14 at 13:49

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