Oh my god
xd Poor Victorian Doll. I think you all are going to confuse her ^^;; I read through your explanations and while it makes perfect sense to someone who has done trig for years, that must have been complete nonsense to her
sweatdrop We didn't even explain why sine and a circle are related.
OK- first, let us have a simple trigonometry lesson. Is everyone fine with starting here? We'll present trigonometry first in its most basic form, and then in the form of a huge problem.
3nodding The most basic form (which you have probably seen before): the triangle.
http://www.mathsrevision.net/gcse/trig.gif (yes- it's a crude drawing. I like it, though, because it was simple. This drawing courtesy of a google image search)
In this picture, you see an angle A, and you see a right triangle (it's
extremely important that this triangle is a right triangle). Mathematicians learned something very amazing. That is this: if angle A is always the same, then the ratios of any two sides will always be the same. Since a ratio is nothing but a fancy word for a division problem, that means that so long as A doesn't change, you can divide any one side by any one other side and there is a certain number you can expect to get. Let me try and explain with two basic triangles that follow this pattern.
Triangle 1 will have an opposite side of 3, an adjacent side of 4, and a hypotenuse of (do you remember Pythagorean theorem?) 5. This makes sense. In this triangle, angle A would have a value of about 36.87 degrees
Triangle 2 will have an opposite side of 9, an adjacent side of 12, and a hypotenuse of (you guessed it!) 15. In this triangle, angle A also happens to be about 36.87 degrees (if you don't believe me, get a ruler and protractor and try to create these triangles).
You learn in geometry that if two triangles have 3 congruent angles then their sides all have a similar ratio, so this pattern is exactly what is taught in geometry. Now let us take the sine (opposite divided by hypotenuse) of each triangle. The first one yields 3/5. The next yield 9/15. Each of these has the same value of .6! Let us now take the tangent (opposite divided by adjacent) of each. The first yield 3/4, and the second 9/12. Again- each is .75
So we can see that the ratios of the sides will be the same regardless of their exact value. For this reason, we can state that the sine of 36.87 degrees IS ALWAYS .6 (regardless of the lengths of the sides). If you don't believe me, take a graphing calculator and type sin(36.87°)
Since an angle will always have the same sine, cosine, and tangent- we can therefore use these known values to solve common problems. Suppose we have a right triangle and we know one side and we know one angle (other than the right angle)- yet we know no other sides. Let us set up a sample problem and see just how easy this is to solve:
Triangle ABC consists of lines AB, BC, and CA as well as angles A, B, and C. Angle A is our right angle. Angle C has a known value of 30°. The length of BC (the hypotenuse) is known to be 16. We are asked to find the length of AB.
If we look at our trig triangle from earlier, we will see that line AB is
opposite of angle C. Thus we are trying to find the length of the opposite side, and we know the length of the hypotenuse. Sine is defined as opposite divided by hypotenuse, and so we know that the sine of angle A is something (the length of the opposite) divided by 16 (the length of the hypotenuse). To write this algebraically:
sin(30°) = x/16
The thing is- we proved earlier that sin(30°) is a constant value! sin(30°) actually happens to be 1/2. So our algebra now looks like this:
1/2 = x/16
This problem should be child's play to you. x is clearly 8. We now know the value of AB. Why would this information be useful? What if BC (the hypotenuse) was a ladder of known length 16 feet and it had to be placed at a 30 degree angle in order to reach a tree branch? How else could we possibly know the height of this branch?
That is trig in it's simplest form. Given an angle and a right triangle we can determine the ratio of any two sides. Also, given one of these two sides we can determine the other since we know the ratio.
Now let us turn trig into a very annoying and difficult problem. Let us look at a circle with a radius of 1, and attempt to find any point on this circle.
At first, this problem seems impossible. How can we possible know where a single point is along a circle? We know that every point is a distance of 1 from the center, but we cannot know how far up it is without knowing how far out it is, and we cannot know how far out it is without knowing how far up it is. We're at a loss.
We solve this problem by defining points around the circle with angles.
http://www.wjagray.co.uk/maths/Lectures/Images/TrigCircle.GIFThe variable θ represents any angle (perhaps 30 degrees, perhaps 45 degrees, perhaps 52.839 degrees). As you can see, given an angle we can locate any one point along this circle. All we know about this point, however, is that it is 1 unit away from the center (since the circle has a radius of 1, all points on the circle are 1 unit from the center). We learn more about this point by drawing a vertical line straight down from it to touch the x-axis like so:
http://www.hyper-ad.com/tutoring/math/trig/images/trig_angl467.gifOur radius, our vertical line, and the x-axis now form a triangle. Also- since the x-axis is horizontal and our line was drawn vertical, our line creates a right angle and we now have a right triangle. Since we have a right triangle, we know that the sine, cosine, and tangent of θ are constant. Our x coordinate (how far out our point is along the x-axis) can be determined by the length of the
adjacent side. The y coordinate (how far out our point is along the y-axis) can be determined by the length of the
opposite side. Knowing the length of the hypotenuse (1) and our angle (θ) we can use a problem similar to the branch and ladder one earlier to find our x and y.
However, we can simplify this just a bit. First, let us look at the equation we have been using prior:
Sine = opposite / hypotenuse
or:
sin(θ) = y / 1
(recall that the length of our opposite side is the same as our y coordinate)
When we divide by 1, the number will always remain the same. Thus:
sin(θ) = y
Using similar logic:
cos(θ) = x
So we can now find any point on a circle with radius 1 knowing only the angle. Given any angle θ, the coordinates of a point on the circle are:
(x, y) --> (cos(θ), sin(θ))
What do we do if the radius is more than 1, however? What if we have a circle of radius 5? Or 5000? Well, let us return to our algebra problem.
sin(θ) = opposite / hypotenuse
so:
sin(θ) = y / 5000
Multiply both sides by 5000:
5000sin(θ) = y
You'll find that for any radius, merely multiply this radius by the sine of θ and you'll receive the proper y coordinate. Multiply this radius by the cosine of θ and you'll receive the proper x coordinate.
Now we come to graphing. When we graph something, just what does the graph show? The graph shows what the value of one variable is at every given value of another variable. We have one point for every possible x showing what y would be given that x. These points join together to form a line of some form.
In this case, we wish to graph sin(x)
As you'll recall from earlier, sin(θ) = y for a point on a circle of radius 1. And so sin(x) is equal to
the point on a circle of radius 1 at angle x.
This explains the strange drawing process of the website that you found. It drew the graph by moving to the right (increasing x) and as it did, it showed how high a point on the circle would be (the y of a point on a circle of radius 1) at that given angle. So at 90 degrees, we have a y of 1 (you're going straight up and touching as high as possible on this circle). At 270 degrees we have a y of -1 (you're touching as low as possible). At 0 and 180, you are on the x-axis and have neither gone up nor down. You have a y of 0. This explains why the sine graph comes back to cross over the x-axis.
In truth, though, a sine graph goes on forever. 360 degrees is an entire circle, however 720 degrees is two entire circles. At 810 degrees you will have gone two entire laps around the circle and now be at a value of 1. So a real sine graph goes up and down in a wave shape forever.
As was said earlier, if a pitch is higher, then you go around this circle faster, yes? That would mean that you go up faster, and then down faster, and then back up faster. This would mean that when you draw it out, your waves would be scrunched up close together. If you have learned about sound waves in any physical science class, this makes sense- as it would cause the sine wave to have a higher frequency (as a result of more
frequently going a full lap around the circle)
Please tell me if you're confused. I tried to simplify a half-semester's worth of learning as best as I could ^^;;
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