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Trigonometric Functions

I've got an idea for a great invention. I'm going to get a flat piece of wood and cut it out in the shape of a circle. In the centre of the circle, I'm going to drill a hole and stick a dowel of wood in there. I think I'm going to call my new invention a wheel.

Okay, okay, so I ran down to the patent office and found out that the wheel has already been invented... (Apparently, Bill Gates just bought the patent rights from Michael Jackson last year...) But, since it's such a great idea let's look at one anyway. Let's drill another hole out on the edge of the wheel and stick in a handle so that it looks like the contraption in Figure 1.3. If I turn the wheel, the handle goes around in circles.

Figure 1.3: Wheel rotating counterclockwise when viewed from the side of the handle that's sticking out on the right.

Now let's think of an animation of the rotating wheel. In addition, we'll look at the height of the handle relative to the centre of the wheel. As the wheel rotates, the handle will obviously go up and down, but it will follow a specific pattern over time. If that height is graphed in time as the wheel rotates, we get a nice wave as is shown in Figure 1.4.

Figure 1.4: A record of the height of the handle over time producing a wave that appears on the right of the wheel.

That nice wave tells us a couple of things about the wheel:

Firstly, if we assume that the handle is right on the edge of the wheel, it tells us the diameter of the wheel itself. The total height of the wave from the positive peak to negative trough is a measurement of the total vertical travel of the handle, equal to the diameter. The maximum displacement from 0 is equal to the radius of the wheel.

Secondly, if we consider that the wave is a plot of vertical displacement over time, then we can see the amount of time it took for the handle to make a full rotation. Using this amount of time, we can determine how frequently the wheel is rotating. If it takes 0.5 seconds to complete a rotation (or for the wave to get back to where it started half a second ago) then the wheel must be making two complete rotations per second.

Thirdly, if the wave is a plot of the vertical displacement vs. time, then the slope of the wave is proportional to the vertical speed of the handle. When the slope is 0 the handle is stopped. (Remember that slope = rise/run, therefore the slope is 0 when the ``rise'' or the change in vertical displacement is 0 - this happens at the peak and trough because the handle is finished going in one direction and is instantaneously stopped in order to start heading in the opposite direction.) Note that the handle isn't really stopped - it's still turning around the wheel - but for that moment in time, it's not moving in the vertical plane.

Finally, if we think of the wave as being a plot of the vertical displacement vs. the angular rotation, then we can see the relationship between these two as is shown in Figure 1.5. In this case, the horizontal (X) axis of the waveform is the angular rotation of the wheel and the vertical height of the waveform (the Y-value) is the vertical displacement of the handle.

Figure 1.5: Graphs showing the relationship between the angle of rotation of the wheel and the waveform's X-axis.

This wave that we're looking at is typically called a sine wave - the word sine coming from the same root as words like ``sinuous'' and ``sinus'' (as in ``sinus cavity'') - from the Latin word ``sinus'' meaning ``a bay''. This specific waveshape describes a bunch of things in the universe - take, for example, a weight suspended on a spring or a piece of elastic. If the weight is pulled down, then it'll bob up and down, theoretically forever. If you graphed the vertical displacement of the weight over time, you'd get a graph exactly like the one we've created above - it's a sine wave.

Note that most physics textbooks call this behaviour simple harmonic motion.

There's one important thing that the wave isn't telling us - the direction of rotation of the wheel. If the wheel were turning clockwise instead of counterclockwise, then the wave would look exactly the same as is shown in Figure 1.6.

Figure 1.6: Two wheels rotating at the same speed in opposite directions resulting in the same waveform.

So, how do we get this piece of information? Well, as it stands now, we're just getting the height information of the handle. That's the same as if we were sitting to the side of the wheel, looking at its edge, watching the handle bob up and down, but not knowing anything about it going from side to side. In order to get this information, we'll have to look at the wheel along its edge from below. This will result in two waves - the sine wave that we saw above, and a second wave that shows the horizontal displacement of the handle over time as is shown in Figure 1.7.

Figure 1.7: Graphs showing the relationship between the angle of rotation of the wheel and the vertical and horizontal displacements of the handle.

As can be seen in this diagram, if the wheel were turning in the opposite direction as in the example in Figure 1.6, then although the vertical displacement would be the same, the horizontal displacement would be opposite, and we'd know right away that the wheel was tuning in the opposite direction.

This second waveform is called a cosine wave (because it's the compliment of the sine wave). Notice how, whenever the sine wave is at a maximum or a minimum, the cosine wave is at 0 - in the middle of its movement. The opposite is also true - whenever the cosine is at a maximum or a minimum, the sine wave is at 0. The four points that we talked about earlier (regarding what the sine wave tells us) are also true for the cosine - we know the diameter of the wheel, the speed of its rotation, and the horizontal (not vertical) displacement of the handle at a given time or angle of rotation.

Keep in mind as well that if we only knew the cosine, we still wouldn't know the direction of rotation of the wheel - we need to know the simultaneous values of the sine and the cosine to know whether the wheel is going clockwise or counterclockwise.

Now then, let's assume for a moment that the circle has a radius of 1. (1 centimeter, 1 foot... it doesn't matter so long as we keep thinking in the same units for the rest of this little chat.) If that's the case then the maximum value of the sine wave will be 1 and the minimum will be -1. The same holds true for the cosine wave. Also, looking back at Figure 1.5, we can see that the value of the sine is 1 when the angle of rotation (also known as the phase angle) is $90^{\circ}$. At the same time, the value of the cosine is 0 (because there's 0 horizontal displacement at $90^{\circ}$). Using this, we can complete Table 1.1.

Table 1.1: Values of sine and cosine for various angles

Phase	Vertical displacement	Horizontal displacement
(degrees)	(Sine)	(Cosine)
$0^{\circ }$	0	1
$45^{\circ}$	0.707	0.707
$90^{\circ}$	1	0
$135^{\circ }$	0.707	-0.707
$180^{\circ}$	0	-1
$225^{\circ}$	-0.707	-0.707
$270^{\circ}$	-1	0
$315^{\circ}$	-0.707	0.707

In fact, if you get out your calculator and start looking for the Sine (``sin'' on a calculator) and the Cosine (``cos'') for every angle between 0 and $359^{\circ }$ (no point in checking 360 because it'll be the same as 0 - you've made a full rototation at that point...) and plot each value, you'll get a graph that looks like Figure 1.8.

Figure 1.8: The relationship between Sine (blue) and Cosine (red) for angles from $0^{\circ }$ to $359^{\circ }$.

As can be seen in Figure 1.8, the sine and cosine intersect at $45^{\circ}$ (with a value of 0.707 or $\frac{1}{\sqrt{2}}$) and at $215^{\circ}$ (with a value of -0.707 or $-\frac{1}{\sqrt{2}}$).^2.2 Also, you can see from this graph that a cosine is essentially a sine wave, but $90^{\circ}$ earlier. That is to say that the value of a cosine at any angle is the same as the value of the sine $90^{\circ}$ later. These two things provide a small clue as to another way of looking at this relationship.

Look at the first $90^{\circ}$ of rotation of the handle. If we draw a line from the centre of the wheel to the location of the handle at a given angle, and then add lines showing the vertical and horizontal displacements as in Figure 1.7, then we get a triangle like the one shown in Figure 1.9.

Figure 1.9: A right triangle within the rotating wheel. Notice that the value of the sine wave is the green vertical leg of the triangle, the value of the cosine is the red horizontal leg of the triangle and the diameter of the wheel (and therefore the peak values of both the sine and cosine) is the hypotenuse.

Now, if the radius of the wheel (the hypotenuse of the triangle) is 1, then the vertical line is the sine of the inside angle indicated with a red arrow. Likewise, the horizontal leg of the triangle is the cosine of the angle.

Also, we know from Pythagoreas that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides (remember $a^{2} + b^{2} = c^{2}$ where c is the length of the hypotenuse). In other words, in the case of our triangle above where the hypotenuse is equal to 1, then the sin of the angle squared + the cosine of the angle squared = 1 squared... This is a rule (shown below) that is true for any angle.

$$(\sin \alpha)^{2} + (\cos \alpha)^{2} = 1$$ (2.10)

This is usually written as

$$\sin^{2} \alpha + \cos^{2} \alpha = 1$$ (2.11)

where $\alpha$ is any angle.

Since this is true, then when the angle is $45^{\circ}$, then we know that the right triangle is isoceles - meaning that the two legs other than the hypotenuse are of equal length (take a look at the graph in Figure 1.8). Not only are they the same length, but, their squares add up to 1. Remember that $a^{2} + b^{2} = c^{2}$ and that $c^{2} = 1$. Therefore, with a little bit of math, we can see that the value of the sine and the cosine when the angle is $45^{\circ}$ is $\frac{1}{\sqrt{2}}$ because it's the square root of $\frac{1}{2}$ and $\sqrt{\frac{1}{2}} = \frac{\sqrt{1}}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.