Derive the Trigonometric Functions 
How can the Sine Function be derived?
Given $\angle{A}$ as input, derive a function that would give $\frac{a}{c}$ as output.
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How can the Cosine Function be derived?
Given $\angle{A}$ as input, derive a function that would give $\frac{b}{c}$ as output.
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How can the Tangent Function be derived?
Given $\angle{A}$ as input, derive a function that would give $\frac{a}{b}$ as output.
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I am looking for either of the following:


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*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.
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Side note:
I am aware of the Taylor-series expansion of the trigonometric functions. $$$$

I am also aware of the exponential definition of the trigonometric functions.$$$$
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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.
 A: 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.
A: 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.
