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I'm a trigonometry teaching assistant this semester and have a perhaps basic question about the motivation for using the circle in the study of trigonometry. I certainly understand Pythagorean Theorem and all that (I would hope so if I'm a teaching assistant!) but am looking for more clarification on why we need the circle, not so much that we can use it.

I'll be more specific- to create an even angle incrementation, it seems unfeasible, for example, to use the line $y = -x+1$, divide it into even increments, and then draw segments to the origin to this lattice of points, because otherwise $\tan(\pi/3)$ would equal $(2/3)/(1/3)=2$. But why mathematically can't we do this?

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There's no reason we can't do this-it just wouldn't define a useful function. –  Kevin Carlson Aug 19 '12 at 0:11
    
But if we could, the tangent function wouldn't be well-defined correct, given the above example? –  Erik Aug 19 '12 at 0:14
    
I guess what I'm getting at is why can't we do this and be able to yield the same outputs as we would when defining the trig functions in terms of the circle. –  Erik Aug 19 '12 at 0:15
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I'm not exactly sure what you're asking. You can define angles that way, but then you lose rotational invariance. –  copper.hat Aug 19 '12 at 0:33
    
@copper.hat that was what I was asking. And I was asking for more mathematical insight into that fact. –  Erik Aug 19 '12 at 2:33

4 Answers 4

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Given your comments, one of the biggest problems with the construction you've offered is that it can't define the trig functions for angles of all real values, or even for every angle between 0 and $2\pi$, since rays from the origin hit your line at angles between $-\frac{\pi}{4}$ and $\frac{3\pi}{4}$. So we'd want to at least propose some other closed curve, say a smooth one since the trig functions are smooth, as the place where we define our functions.

Perhaps the simplest reason that the circle is a natural place to define trigonometric functions is that angle is a measure of arc of a circle.

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I appreciate your feedback. But I would really like to know why the above construction doesn't yield the same tangent values as the definition in terms of the circle. –  Erik Aug 19 '12 at 0:23
    
I don't follow your computation, after looking closer. Are you drawing the right triangle with hypotenuse from the origin to the line $-x+1$, legs along the $x$-axis and parallel to the $y$-axis, and hypotenuse at angle $\pi/3$ from the $x$-axis? The end of the hypotenuse for this is at $(\frac{1}{1+\sqrt{3}},\frac{\sqrt{3}}{1+\sqrt{3}}),$ which would give the correct tangent of $\sqrt{3}$. –  Kevin Carlson Aug 19 '12 at 0:29
    
Kevin, your last comment got at my main question. Why must angle be defined to be the measure of arc of a circle? I'm quite confident you're correct but am unsure why this needs to be the case. –  Erik Aug 19 '12 at 0:44
    
What I did was not actually use an angle pi/3 but consider the angle formed by the positive x-axis and the line connecting the origin to the point (1/3, 2/3), which is two-thirds of the way from (1,0) to (0,1). I guess my question in this context might be stated: why is such an angle formed not 2/3 of pi/2? –  Erik Aug 19 '12 at 0:46
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@Erik: Because the whole thing arose more than two millenia ago when people were doing planet and star observations. The only measurable thing was angular distances between heavenly bodies, as observed from Earth. –  André Nicolas Aug 19 '12 at 6:01

Suppose that instead of parametrizing the circle by arc length $\theta$, so that $(\cos\theta,\sin\theta)$ is a typical point on the circle, one parametrizes it thus: $$ t\mapsto \left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right)\text{ for }t\in\mathbb{R}\cup\{\infty\}. \tag{1} $$ The parameter space is the one-point compactification of the circle, i.e. there's just one $\infty$, which is at both ends of the line $\mathbb{R}$, rather than two, called $\pm\infty$. So $\mathbb{R}\cup\{\infty\}$ is itself topologically a circle, and $\infty$ is mapped to $(-1,0)$.

Now do some geometry: let $t$ be the $y$-coordinate of $(0,t)$, and draw a straight line through $(-1,0)$ and $(0,t)$, and look at the point where that line intersects the circle. That point of intersection is just the point to which $t$ is mapped in $(1)$.

Later edit: an error appears below. I just noticed I did something dumb: the mapping between the circle and the line $y=1-x$ that associates a point on that line with a point on that circle if the line through them goes through $(0,0)$ is not equivalent to the one in $(1)$ because the center of projection is the center of the circle rather than a point on the circle.
end of later edit

This mapping is in a sense equivalent to the one you propose: I think you can find an affine mapping from $t$ to your $x$ on the line $y=-x+1$, such that the point on the circle to which $t$ is mapped and the point on the circle to which $x$ is mapped are related by linear-fractional transformations of the $x$- and $y$-coordinates.

The substitution $$ \begin{align} (\cos\theta,\sin\theta) & = \left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right) \\[10pt] d\theta & = \frac{2\,dt}{1+t^2} \end{align} $$ is the Weierstrass substitution, which transforms integrals of rational functions of sine and cosine, to integrals of simply rational functions. I'm pretty sure proposed mapping from the $(x,y=-x+1)$ to the circle would accomplish the same thing.

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I appreciate the sophisticated response and the time you took to come up with it. I'll give it some thought. Unfortunately I can't give anybody an up-vote because I just joined the site. Otherwise, I'd give you a few for your two answers if I could. –  Erik Aug 19 '12 at 3:30
    
@Erik : There is a mistake in the later part of this answer. See the "later edit" above. –  Michael Hardy Aug 21 '12 at 22:01

The reason we use the unit circle instead of your idea is that $\cos$ and $\sin$ parametrize the unit circle very naturally in terms of our familiar coordinates. I mean, $x^2+y^2=1$ is parametrized as $x=\cos(\theta),$ $y= \sin(\theta)$. So,using the unit circle, you can just read off the values of sine and cosine from the $x,y$ coordinates of the points on the unit circle. This is a useful thing, both computationally and conceptually.

While I suppose one could--in theory--parametrize the line $y=1-x$ using the Cosine and Sine, the parametrization will be nowhere near as neat or useful as it was above. In fact, it will be very ugly, so you can no longer read the values of sine and cosine off from the coordinates as we could in the prettier example of using the unit circle.

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A polar parametrization of a line is useful for converting angular momentum to linear momentum (or vise versa). –  Baby Dragon Aug 22 '12 at 1:04

Norman Wildberger's book Rational Trigonometry shows that one can do an immense amount of trigonometry and applications to geometry without any parametrization of the circle by arc length. He treats triangles largely without mentioning circles.

Notice that the squares of the sine and cosine are rational functions of the slopes of two lines meeting at an angle. One can deal with those rational functions without dealing with any parametrization of the circle by arc length. Wildberger doesn't deal with sines and cosines, but with their squares. In an $n$-dimensional space, the angle between two vectors depends on the equivalence classes to which they belong, where two vectors are equivalent if one is a scalar multiple of the other. If you call such an equivalence class a "slope" then you still get the squares of the sine and cosine as rational functions of the slope.

Of course, one thereby gives up the ability to deal with Fourier sine- and cosine-series. So there's a trade-off: some efficiency is gained and the ability to do some things is lost.

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I'll have to give your response some thought. I appreciate it. Maybe either you, Michael, or Kevin have answered my essential question, but I'm not sure. My main question is referenced in the example in the original question - if we draw a line from the origin to (1/3, 2/3), which is two-thirds of the way from (1,0) to (0,1), and reason that the angle formed is pi/3 since pi/2 times 2/3 is pi/3, we now find that the "tangent" value is 2, inconsistent with the actual value tan(pi/3)=√3. –  Erik Aug 19 '12 at 0:37
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My answer is of course in a sense a partial answer, dealing with a point of view that is not the usual one. –  Michael Hardy Aug 19 '12 at 1:49
    
....and now I've posted another partial answer, looking at the question from yet another point of view. –  Michael Hardy Aug 19 '12 at 3:20

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