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Right angled triangle

How can the Sine Function be derived?

Given $\angle{A}$ as input, derive a function that would give $\frac{a}{c}$ as output.

$$$$ How can the Cosine Function be derived?

Given $\angle{A}$ as input, derive a function that would give $\frac{b}{c}$ as output.

$$$$

How can the Tangent Function be derived?

Given $\angle{A}$ as input, derive a function that would give $\frac{a}{b}$ as output.

$$$$

I am looking for either of the following:

  • The historical way to calculate the trigonometric functions as well as a proof that it works for a right-angled triangle
  • Any other way to calculate the trigonometric functions as well as a proof that it works for a right-angled triangle

In other words, an algorithm on its own would not be enough, you have to prove that it works for a right-angled triangle. $$$$ Side note:

I am aware of the Taylor-series expansion of the trigonometric functions. $$$$ Taylor Series

I am also aware of the exponential definition of the trigonometric functions.$$$$ enter image description here $$$$

If you could geometrically prove how any of these trigonometric identities work for a right-angled triangle, that would answer my question as well.

Another side note

I do not believe this question belongs in The History of Science and Mathematics-Stack Exchange. That forum focuses on where and when certain Mathematical concepts were created, which is not my question.

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    $\begingroup$ You should ask this question to the forum $History\; of \;Science\; and\; Mathematics$ $\endgroup$ – user249332 Sep 24 '15 at 19:14
  • $\begingroup$ Thanks, I will. I was not aware of that community, but I am going to keep the question on this forum as well, since it still overlaps. $\endgroup$ – Paul Sep 24 '15 at 19:17
  • $\begingroup$ But that was specially made for this type of questions. $\endgroup$ – user249332 Sep 24 '15 at 19:19
  • $\begingroup$ Yes, but there are 2000 users of that forum and 200 000 users of this forum. And this question still is relevant to this site. $\endgroup$ – Paul Sep 24 '15 at 19:20
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    $\begingroup$ This question was already asked and answered on the HSM Stackexchange site. $\endgroup$ – David K Sep 24 '15 at 21:02
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I'll just make a side note that I find it more natural to think about trigonometric functions as functions on unit circle. Once you know that $\theta \mapsto e^{i \theta}$ is surjective mapping of $\mathbb R$ in unit circle you can geometrically define $\cos \theta$ as projection of the point $e^{i \theta}$ on $x$ axis and $\sin \theta$ as projection on $y$ axis. This is way of thinking that I think is most natural and useful in practice (for example in calculations involving trig functions.) It's also geometric, which as I understand is whole point here. Once you have this definitions you can easily see the relation of these trig functions with right-angled triangles. This has great advantage that once you have that you can easily evaluate your functions in terms of exponentials (which means that you also automatically have Taylor expansion.) In this way you not only obtain a way to calculate these rations in right-angled triangle (i.e. answer original question) but also establish beautiful and profound correspondence between simple geometry and mathemathical analysis.

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  • $\begingroup$ I never actually tried to visualise $\frac{e^{i\theta}+e^{-i\theta}}{2}$ before. It actually makes perfect sense now, $\endgroup$ – Paul Sep 25 '15 at 10:22
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    $\begingroup$ You might also enjoy the fact that actually whole complex analysis has very geometric interpretation: a function is complex differentiable if and only if it is conformal mapping (i,e, preserving angles) of plane into itself and preserves orientation (or "handendness".) My opinion is that complex numbers should be thought of as a neat representation of geometry of Euclidean plane. Almost every basic formula and operation related to complex numbers can be interpreted in geometric terms. $\endgroup$ – Blazej Sep 25 '15 at 10:33
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    $\begingroup$ I made a slight mistake in my comment. What I wrote is only true away from the points where derivative vanishes. But otherwise it holds. $\endgroup$ – Blazej Sep 26 '15 at 10:59
  • $\begingroup$ To nit-pick: That $\theta \mapsto e^{i\theta}$ is surjective isn't quite sufficient; you also need to know that it's a constant-speed map, or equivalently that this "angle" in the exponent is linearly proportional to the geometric angle. Otherwise I don't see anything stopping us from defining $\cos \theta := \operatorname{Re} e^{i\theta^2}$, for example. $\endgroup$ – epimorphic Sep 27 '15 at 21:57
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    $\begingroup$ @Paul Pretty easy, actually. The "velocity" of the function is $\frac{d}{d\theta} e^{i\theta} = i e^{i\theta}$, whose magnitude is just $1$. This shows that the arc length spanned from angle $\theta_1$ to $\theta_2 \geq \theta_1$ is $\int_{\theta_1}^{\theta_2} \left\lvert \frac{d}{d\theta} e^{i\theta} \right\rvert d\theta = \theta_2 - \theta_1$, just as we would expect from classical geometry. Variations like $\theta \mapsto e^{i\theta^3}$ don't satisfy this. $\endgroup$ – epimorphic Oct 7 '15 at 14:34
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This is an expansion of a previous answer by Raj (an answer which was deleted), which I thought was note-worthy enough to keep. But it is by no means a complete.

1) Assume there exists a function that satisfies the definition of $\sin(A)$. (Call this function $\sin(A)$). In other words, there exists a function that takes $\angle{A}$ as input and gives $\frac{a}{c}$ as output.

2) Assume the same for $\cos(A)$.

3) Prove that $\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$

4) Prove that $\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$

5) Prove the half angle formula and the double angle formula for $cosA$ and $sinA$

6) Physically measure that $\sin30^\circ$ = $\frac{1}{2}$. Take this as your reference point.

7) Using the half-angle, double-angle, and sum formula, it would now be possible to calculate $\sin(A)$ using the reference point.

8) $\cos60^\circ$ = $\frac{1}{2}$ could be the reference point for the cosine function.

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    $\begingroup$ Use a backslash with trig. func. names — it makes them $\LaTeX$ symbols, so that they appear in upright font: \sin, \cos render as $\sin$ and $\cos$ instead of $sin$ (which looks like a multiplication $s\cdot i\cdot n$). $\endgroup$ – CiaPan Sep 25 '15 at 9:38
  • $\begingroup$ Apparently there is no 'degree' symbol available, but you can use a superscript circle for it: 60^\circ results in $60^\circ$. $\endgroup$ – CiaPan Sep 25 '15 at 9:39
  • $\begingroup$ @Paul I don't understand how this answer O.P's question. $\endgroup$ – Babai Sep 27 '15 at 19:40
  • $\begingroup$ @Babai To me, it seems like a decent attempt at the second bullet point in the question (though perhaps it should address how to define the function for angles that are not dyadic rational multiples of $60^\circ$, self-consistency, etc). The only really bad part I see is the physical measurement, but that's easily replaceable. Note also that the author of the answer is OP himself. $\endgroup$ – epimorphic Sep 27 '15 at 21:12
  • $\begingroup$ @epimorphic I did not notice that the author of the answer is the O.P himself. I feel defining the functions all kinds of angles is the main point. $\endgroup$ – Babai Sep 27 '15 at 21:16

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