What is the Connection Between \(\theta\) and Trigonometric Function Outputs?

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The discussion explores the relationship between the angle \(\theta\) and the outputs of trigonometric functions, particularly focusing on sine, cosine, and tangent. It highlights that calculators often use Taylor expansions for these calculations, but questions whether there is a direct algebraic link between \(\theta\) and the function outputs, such as the ratio of adjacent to hypotenuse. Historical references are made to Ptolemy and Indian mathematicians who created trigonometric tables before the advent of calculus. The inquiry seeks alternative methods for calculating trigonometric values, particularly for specific angles like \(\frac{\pi}{9}\), without relying solely on manual calculations. The discussion ultimately emphasizes the quest for a deeper understanding of the connection between angles and their trigonometric ratios.
brendan_foo
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I believe that calculators use Taylor expansions to compute sines, cosines and tan's based upon the argument \theta (in radians of course). However, my question is, aside from these expansions, is there some sort of link between \theta and the output of the function itself.

I mean I know that \cos{\theta} = \frac {adj}{hyp} and the other trig ratios, but was this just worked out by hand, pencil and paper and kept in a tabular form before the Taylor expansion was devised? Is there a direct link between (\frac{adj}{hyp}) and \theta.

Get me?!
 
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Trig values were, yes, worked out by hand ()

One of Ptolemy's major contributions to Greek maths was his trig tables.
The Indian mathematicians did the same, but independently of the Greeks.

EDIT:
Hmm..now that I reread your question, it seems you were after something else..
 
No that's a great answer.. just curious. I know the Maclaurin series for trig functions takes the parameter and manipulates it to get a solution. However I wanted to know if there was some other relation between the argument and the answer. Say I had the angle \frac{\pi}{9} and I wanted to know the cosine of it, that is the ratio of the adjacent to the hypotenuse, then was there some algebraic manipulation you could do with the value \frac{\pi}{9} to yield the solution.

Aside from doing it by hand, I was curious whether it could be done another way before the days of calculus.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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