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Why does the Unit Circle work? |
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| Mar9-12, 12:58 AM | #1 |
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Why does the Unit Circle work?
In my Trig class, we learned about the unit circle and its relationship to the various trig functions (sin, cos, etc.).
I am curious to know why the unit circle works the way it does, and the how it was "derived" so to speak. More specifically, why does radius of the circle have to be 1 for the circle to work. Please try to keep your explanations as mathematically friendly as possible, as I am only in precalculus. Thank you |
| Mar9-12, 01:00 AM | #2 |
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All of those functions we're defined based on the unit circle. It works because we designed it to work.
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| Mar9-12, 02:33 AM | #3 |
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Since the hypotenuse will be the radius of the unit circle, this has to be 1 to make the length of the side opposite to angle A equal to sin(A) |
| Mar9-12, 03:41 AM | #4 |
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Why does the Unit Circle work?Thanks for the explanation,,, |
| Mar9-12, 06:21 AM | #5 |
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The idea begins When Babylonians have developed a system for measuring angles "degree measure", They discovered some trigonometric theorems but not explicitly , and then the sin function was first developed in India . But trigonometry wasn't translated to Europe through the Hindus , but through Muslims " arab " . Islamic Arabian mathematicians had taken the ideas from the hindus and expanded them where they have known the six main trigonometric function , they were Calculating them using triangles . So the trigonometry was invented depending on triangles , but it was seen after that its relation with unit circle which willem2 had mentioned . All I can say you that with that curiosity and when you take integral Calculus , I advise you to read spivak's construction for trigonometric function which is even more rigorous than defining them in terms of unit circle , where you will see why .
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| Mar9-12, 01:01 PM | #6 |
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if you are asking about the connection between the "circular" functions (sine and cosine as defined using the unit circle) and the "trigonometric functions) (sine and cosine as defined using right triangles), from a point, (x, y), on the unit circle, draw the line from (x, y) to (0, 0), the line from (x, y) to (x, 0), and the line from (x, 0) to (0, 0).
That gives a right triangle which has "near side" of length x, "opposite side" of length y, and hypotenuse of length 1. Therefore, [itex]sin(\theta)=[/itex]"opposite side over hypotenuse= y/1= y and [itex]cos(\theta)= [/itex]"near side over hypotenuse"= x. (Since the angles in a triangle can only be between 0 and 90 degrees, we really have to assume that (x, y) is in the first quadrant. The circle definition extends that to all real numbers.) |
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| trig equation, unit cirlce |
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