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I don't agree with the perspective taken here; I don't see any reason to think that the definition in terms of triangles is the simplest or most natural, and I
by gjm11 3y ago
I don't agree with the perspective taken here; I don't see any reason to think that the definition in terms of triangles is the simplest or most natural, and I suspect the author feels that way just because it's how they were first taught about trigonometrical functions.
Of course I do agree that definitions in terms of power series and differential equations are less natural and require heavier mathematical machinery.
But: the "right-angled triangle" definitions have the severe drawback of only applying to a restricted range of angles, and their relationship to those higher-tech definitions is more indirect than necessary.
Instead, I claim that the One True Definition of the trig functions is: if you start with the point (1,0) and rotate it through an angle t (anticlockwise, which is conventionally the "positive" direction in maths) about (0,0), then the point where it ends up is (cos t, sin t).
This is just as simple, and just as geometrical, as the "triangle" definition. It leads directly to the differential-equation characterization (which in turn leads easily to the power series). For angles between 0 and pi/2, it's obviously equivalent to the "triangle" definition.
Can you get the addition theorems easily from this definition? Yes, and (again) without the severe drawback of only applying when all the angles involved are between 0 and pi/2 as for the "triangle" definition.
Start with a diagram showing the points (0,0), (cos t, 0), (0, sin t), (cos t, sin t). Now rotate the whole thing through an angle u about the origin. Obviously (0,0) stays where it is. (cos t, 0) just goes to cos t times (cos u, sin u), i.e., to (cos t cos u, cos t sin u). If you turn your head through 90 degrees it becomes clear that (0, sin t) similarly goes to (- sin t cos u, sin t sin u). And of course (cos t, sin t) goes to (cos t+u, sin t+u) since we have rotated it through an angle t and then through an angle u.
And now we're done, because the "vector addition" property the original diagram had remains true after rotation, so adding (cos t cos u, cos t sin u) to (- sin t cos u, sin t sin u) has to give you (cos t+u, sin t+u). And that's exactly the addition theorems.
(That may be hard to follow with no diagrams, but I think it's easier to follow than the triangle-stacking proof would be without diagrams, and with diagrams everything is pretty transparent.)
The right-angled-triangle definitions are traditional because historically trigonometry came before coordinates. But now we have coordinates and generally learn about them before we learn trigonometry, and at that point the point-on-the-unit-circle definitions are simpler, more general, and better suited for proving other things.
- enizor2 3y agoCould you expand on getting the differential equation from your definition? I don't really where to start from it.
- outop 3y agoDraw a line of length 1 which makes an angle of p with the positive x axis. The x and y coordinates of a, the point at the end of the line give cos(p) and sin(p). Now think about what happens if you increase p by a tiny bit. a moves tangentially. (This works very similar to how the tangent to a curve gives the change in y for a change in x.) So the vector of cos'(p), sin'(p) is given by a vector starting from a, at a right angle to 0a, and pointing in the positive direction. Since the point a moves through 2pi distance while p goes from 0 to 2pi (the definition of measuring angles in radians) the speed of the point is 1, and so the vector of derivatives has length 1. You can check easily that this makes cos'(p) = -sin(p) and sin'(p) = cos(p).
- gjm11 3y agoYup! (If I were trying to present this stuff in a maximally-elegant order without too much regard for what order human brains like to learn things in, the order of things would be: complex numbers, calculus, trigonometry. Then we define something that we might initially call e(t) to satisfy the differential equation de/dt = ie, and observe that having e and e' at right angles means that |e| remains constant, which means that |e'| also remains constant, so if we start with e(0)=1 then we have a point moving at unit speed around the unit circle, etc. Keep the linkage between the geometrical and formal points of view there at all times. But I suspect this wouldn't be great paedagogically for the majority of students.)
- lupire 3y agoIf you allow negative side lengths, as you should, much of geometry is unified and simplified, such as this situation. Negative (and imaginary) measurements make math better-behaved. (See also: quantum mechanics.) Rectangular coordinates are suboptimal because they are arbitrary in a way that hides some of the symmetry of mathematics. Your vector argument looks like a lot of algebra noise "without a diagram" because it relies on... triangles... for intuitive justification.