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I am trying to write a very very basic trig primer, from scratch.

Early on, I wish to discuss slopes (say, the slope of a line through the origin), and I would like to give the 'right' hints on how the slope of a line is related to the angle of the line with the x-axis.

How would you go? Are there families of cases (slope values) which provide easier to grasp insights (besides, say, slope=1)?

Note, the examples should be used to introduce arctan(), and so a definition of arctan() should not be implied or preassumed.

Sorry to be quite vague.

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  • $\begingroup$ Which level of education is this for and what kind of background knowledge do the pupils/students have? $\endgroup$
    – Tommi
    Jan 31, 2022 at 8:56
  • $\begingroup$ actually, this is to discuss how much background you need to introduce very basic trig insights (the less the better) $\endgroup$
    – mario
    Jan 31, 2022 at 11:05
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    $\begingroup$ I use the general word "direction" to describe the property that both angles and slopes describe. "Angles and slopes are two different ways of assigning a number value to direction. However, it's really difficult to convert from one to the other. Any ideas why or how?" I find that this kind of wording to introduce the motivation for this topic is very useful to get students on the same page. $\endgroup$
    – CosmoVibe
    Jan 31, 2022 at 19:18
  • $\begingroup$ @CosmoVibe I am curious about which kind of answers you get, and what you do explain next. Thanks for your comment. $\endgroup$
    – mario
    Jan 31, 2022 at 20:00
  • $\begingroup$ Usually, students will not understand why the conversion is difficult. "For a given angle t equivalent to slope m (aka $\tan{t} = m$), if we double the angle we just double the slope, right?" is basically what they think. Use a slope of $1$ as an example to show why that doesn't hold. Go through a couple more examples, and they will quickly see that the relationship between the two is not straightforward. Basically, follow the other answers given by other people here. :) Once they appreciate the difficulty of the problem, they will at least see the value in studying the $\tan$ function. $\endgroup$
    – CosmoVibe
    Jan 31, 2022 at 22:52

3 Answers 3

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You want to convince students that an angle in a right triangle depends on the slope of the line (forming the hypotenuse).

At this point, they are likely already familiar with calculating the slope of a line. I wouldn't worry about choosing a nice ratio of sides for the slope, but rather I would give students a protractor and have them draw (probably on graph paper) different right triangles whose hypotenuses have the same slope. I'd have them measure the corresponding interior angles to confirm that they are all equal in measure. If you did this with enough different slopes (i.e. each student gets a different slope to give their hypotenuses), then you could show them their compiled data in a table (or dots on a coordinate plane), announce that this must be a function, and then give them its proper name and show them that their calculator has it built-in.

You could extend the idea to the other trig ratios, all using the pictures they already drew.

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  • $\begingroup$ Thanks for your help, but I don't want to use a calculator (at least for a while). Actually, I am looking for hints about how to introduce the design of an algorithm slope:angle, maybe for a limited number of cases. For example, say I consider lines with slope 1, 1/2, 1/3, 1/5. Is there an easy to show way to calculate the corresponding angles? Are there a families of lines/slopes such that that is easier? $\endgroup$
    – mario
    Jan 31, 2022 at 8:17
  • $\begingroup$ It seems I misinterpreted your question. It's possible that your question is better suited for Mathematics Stack Exchange as you've got an idea in mind for "how" to introduce the topic, but you're looking for particular examples and an algorithm that you can use to determine an angle from a ratio of sides. This sounds like a good math question for MSE. $\endgroup$
    – Nick C
    Jan 31, 2022 at 14:44
  • $\begingroup$ Ok, thanks, I am thinking it is too basic for MSE, but I may try. $\endgroup$
    – mario
    Jan 31, 2022 at 15:44
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    $\begingroup$ some good reply at MSE $\endgroup$
    – mario
    Jan 31, 2022 at 18:20
  • $\begingroup$ "different right triangles whose hypotenuses have the same slope" - should probably prove that similar triangles have the same measures of corresponding angles. $\endgroup$
    – Rusty Core
    Jan 31, 2022 at 19:30
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If you don't want to use a calculator, I'd adapt Nick C's answer as follows:

Use $\frac y x$ as as slope, with $x=10$ and $y = 1, 2, 3, 4, 5, 10, 20$; have the students draw the lines and measure the angle. I assume that the students are aware that the slope does not depend on the choice of $x$, but you could also vary this.

If you're looking for ways to determine the angle without measuerment, the only useful values are probably

  • slope $\frac{1} {\sqrt{3}}$ for an angle of $\frac{\pi}{6}$
  • slope $1$ for an angle of $\frac{\pi}{4}$
  • slope $\sqrt{3}$ for an angle of $\frac{\pi}{3}$.

The plot of the angle vs. slope is easily obtained and shows the relation you're looking for.

One could also discuss the behavior for $y \to \infty$ or $\alpha \to 90^\circ$.

Depending on the setting, you could also use the relation to form pairs (or groups) of students to work together: hand out cards with either angle or slope (maybe different fractions for the same ratio) and have the students find their partners who have the "same line".

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  • $\begingroup$ Thanks. "draw the lines and measure the angle". Say, I do not want to measure the angle by a protractor, but just show by some basic geometry that if the slope is n/m, then the corresponding angle is such and such fraction of $\pi$. My ignorance, are there values for which it is possible? $\endgroup$
    – mario
    Jan 31, 2022 at 11:07
  • $\begingroup$ I have seen your edit. Definitely a lack of knowledge on my side. If I have lines with slopes 1,1/2,1/3,1/4, ... (or maybe, for an N, 1/N, 2/N, 3/N,...,1; or for just a particularly lucky family of lines) is there any way to calculate the corresponding angles, maybe to some degree of approximation? $\endgroup$
    – mario
    Jan 31, 2022 at 13:50
  • $\begingroup$ i don't know any more than given in my answer. why don't you just go with drawing and measuring? $\endgroup$
    – Jasper
    Jan 31, 2022 at 14:20
  • $\begingroup$ Ok, let us say I assign to draw a family of lines, in terms of easy rise/run slopes. Then, I ask to measure corresponding angles by a protractor. Indeed that could help to give a good insight on how values are connected. Next, could I show how to extract a rule about that? How? Or, could I show how we could have calculated the angles without a (direct/by hand) measure? $\endgroup$
    – mario
    Jan 31, 2022 at 15:42
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I'd probably show them a picture that shows the relationship. If $\theta$ is measured in radians, then in the unit circle figure below, the area of the two triangles is $\tan \theta = m$ and the area of the intersection with the circle is $\theta$. Since they are areas, they are easy to compare. It is easy to see that $\tan\theta > \theta$ for $0<\theta<\pi/2$, and it is easy to see roughly how the difference grows as $\theta$ increases.

enter image description here

If you want to compute $\theta$ for given values of $m$, it is possible to do so approximately, but it takes a lot of work and I wouldn't advise it.

However, it's not too hard to make a first approximation. First, it is easy to see that $\tan\theta = \theta\ +$ the two parts of the triangles outside the circle. If we approximate these by triangles, replacing the circular arc by a straight line, then we can find the area of each little triangle to be $$ {1\over2}\left(1 - {1 \over \sqrt{1+m^2}}\right)\,m \,, $$ by dropping a vertical line from the vertex on the circle and using similar triangles to derive the height from the base $m$. From this, we get that $$\theta = {m \over \sqrt{1+m^2}}$$ plus the amount cut off by straightening the arc.

For $m = 0.5$, we get an approximation of $\theta \approx 0.447$ where actually $\theta = 0.464$.

A slightly more complicated way we can calculate $\theta$ in terms of $m$ is to recursively bisect the angle and inscribe triangles inside the sector. This will require the half-angle formulas to get the approximation in terms of $m$, so that might be an impediment for a motivating example at the beginning of trigonometry.

After $k$ bisections, the area of the inscribed triangles add up to $$2^k \sin (\theta/2^k) \,.$$ We then apply the half-angle formula for sine (once) and the half-angle formula for cosine $k-1$ times, until we arrive at an argument of $\theta$. Finally we substitute for $\cos \theta$, the formula in terms of $m=\tan\theta$, namely $\cos\theta = 1/\sqrt{1+m^2}$. It makes the work easier if we write the half-angle formulas as substitutions in the form $$\eqalign{ \sin\alpha &\mapsto {1\over2}\sqrt{\mathstrut2-2\cos 2\alpha} \cr 2\cos\alpha &\mapsto \sqrt{\mathstrut2+2\cos 2\alpha} \cr }$$ Then a nice recursive pattern emerges.

For $k=4$, the steps are as follows: $$\eqalign{ & 2^4 \sin \big(\theta/2^4\big) \cr & 2^3 \sqrt{2 - 2 \cos \big(\theta/2^3\big)} \cr & 2^3 \sqrt{2 - \sqrt{2 + 2 \cos \big(\theta/2^2\big)}} \cr & 2^3 \sqrt{2 - \sqrt{2 + \sqrt{2 + 2 \cos \big(\theta/2\big)}}} \cr & 2^3 \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{\mathstrut2 + 2 \cos \theta}}}}, \quad \cos\theta = 1/\sqrt{1+m^2} \cr }$$ This gives an approximation of $\theta \approx 0.463583$ compared to the correctly rounded value of $0.463648$.

The general pattern in terms of $k$ is $$2^k \sqrt{2 - \sqrt{2 + \sqrt{2 + \cdots \sqrt{2 + 2\big/\sqrt{1+m^2}}}}} \,,$$ where the total number of square-roots is $k$. It has a linear convergence rate of $1/4$, that is, the error asymptotically decreases as $1/4^k$.

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