Why does Sin represent Y on unit circle

[bmed90]

As the title inquires, I am curious as to how or why the Sin function represents y coordinate on the unit circle.

 

[mfb]

This can be used as a definition of the sine, and it was indeed the historical way to introduce the function.
How can it be surprising if it is the definition?

 

[bmed90]

I guess what I am asking is why is sin defined as the y coordinate?

 

[phinds]

I guess what I am asking is why is sin defined as the y coordinate?

Do you understand what a "sin" function is? Can you DRAW a unit circle and see what you would get if you took the sin of some arbitrary radius line?

 

[mfb]

I guess what I am asking is why is sin defined as the y coordinate?

The y coordinate is interesting in some mathematical calculations, so this function got its own name ("sine"). Other parameters got other names (like cosine, tangent, ...).

 

[bmed90]

Do you understand what a "sin" function is? Can you DRAW a unit circle and see what you would get if you took the sin of some arbitrary radius line?

I don't think you realize what I am asking. Hopefully this will clarify my question even further.

As you imply in your statement Sin(0) is 0 because if you look at the coordinate (0,1) and "take the sin" you get 0 which is the y coordinate.

My question is WHY does "taking the sin" of this coordinate mean to automatically pick the y coordinate. Why is this? I hope my question is crystal clear. I know how to draw the unit circle. Thanks.

 

[SteamKing]

The sine of the acute angle of a right triangle by definition is the ratio of the length of the side opposite the acute angle divided by the length of the hypotenuse. If you draw a right triangle inscribed in a unit circle, as has been suggested, and identify the pertinent parts of the triangle, you will see the relation.

By convention, the angular measure of angles inscribed in triangles assume that the positive x-axis is zero degrees and the positive y-axis is 90 degrees (or the equivalent radian measure).

 

[willem2]

Consider the triangle formed by the origin, a point on the unit circle in the first quadrant at an angle A, and the projection of this point on the x-axis. This is a right-angled triangle, with hypothenuse 1. since sin A = (opposite side of A)/(hypothenuse), the opposite side is sin A. This is also the y-coordinate of the point on the unit circle.

Reflection around the x- and y-axis will prove this for the other quadrants as well.

 

[bmed90]

Consider the triangle formed by the origin, a point on the unit circle in the first quadrant at an angle A, and the projection of this point on the x-axis. This is a right-angled triangle, with hypothenuse 1. since sin A = (opposite side of A)/(hypothenuse), the opposite side is sin A. This is also the y-coordinate of the point on the unit circle.

Reflection around the x- and y-axis will prove this for the other quadrants as well.

Thanks