Precursor: Vertical lines are numerical representations of various multiples of p .

As expected we see a periodic function of period 2p and amplitude 1.

The period of a function is the time it takes for the function to repeat the same value; the

wavelength or distance between two waves crests (top crests or bottom crests). In the above

case, the sin graph repeats at every interval of 2p (the vertical line) therefore itís period is

said to be 2p .

The amplitude of a function is the height of that function (the range of the function divided

by two). In the above case we can see that the sin graph ranges from +1 to -1 which is a

distance of 2 then divided by 2 yields 1. So itís amplitude is 1. This sets the appropriate

definitions for the other cases of sin.

Note on the origins of sin: It is important to note that the sin (specifically the above) function is defined for the values of triangles inscribed inside the unit circle. Where sin = opposite side (from q ) / hypotenuse. Picking a point on the unit circle (call * ) and then creating a triangle by the vector from the origin (0, 0) and the line between the point and the x-axis.

Now as * moves around the unit circle we will arrive at different values of sin because we will be generating different triangles (also different angles q ). The path of * is circular so we will eventually wind up back at our starting point, hence the sin functions periodicity. For other values of a, b, & c this will be akin to inscribing triangles inside circles (or ellipses) of various sizes. This analogy will be omitted to but further exploration of this direct relationship is encouraged for a greater understanding.

Interesting, it seems that our period has decreased by a factor of 1/2 (our amplitude remains unaffected, a = 1). That is, the sin function now repeats every period of p instead of 2p . This leads us to speculate as to what would happen if we increased b.

There seems to be a direct correlation between the parameter b and the period of this function. With b = 3 our period is

(2p )/3. It seems that the period of our function in a general case will be:

Lets try the b = 6 case. Our period should be (2p )(1/6) = p /3. Which means that our sin function will repeat every

interval of p /3.

Yes! Period is p /3 and we have 6 cycles of sin x before we hit 2p . So b is related to the period of our function.

Yes, our period has doubled. This makes sense, because 1/(1/b) = b. So we are going to be stretching our graph. Letís

look at one more, say f(x) = sin (1/6 x). This should stretch our graph by a factor of 6, that is, the graph will complete

one period every 12p.

The above graph is for a period of 6p. So our period is 12p, because the sine function is symmetrical for our chosen

parameters. The following is the picture of sin (1/6 x) on a period of 6p £ x £ 12p:
 

Now let us investigate what happens if b is negative. When I think of negative, I usually think of "reverse".

Wow, our graph has flipped! It looks like it is reflected about the x-axis. HmmmÖ maybe I can make a few observations

and speculations. Then we will test them.

I conjecture that b will have the same effect as discussed previously for 0 < b < 1, and b<1 (except that our graph will be flipped about the x-axis for negative b).

For 0 < b < 1 this will stretch our sine graph by the established relation,

and for b <- 1 the period of sin x will shrink following the relation Period = (2p )(1/b).

Let's try a few examples:

So we have investigated what the parameter b does to a sin (bx + c) with a=1, c=0.

For all of the above cases our amplitude has remained the same at a value of one. Note also if b = 0 then sin (0x) = sin 0

and our period is meaningless. Now itís time to see what a and c do.

Let us look at c first. The graph of f(x) = sin (x + p/2):

HmmÖ our graph seems to have moved either to the left or the right by p/2 units. Which one is it? With my concept of

negative as being reverse let me try f(x) = sin(x - p/2):

This clears things up a little. It appears that f(x) = sin (x + p/2) shifts the graph of sin(x) to the left by p/2 units, and the graph of f(x) = sin (x - p/2) moves the sin x graph to the right by p/2 units.

Letís try a conjecture at this point.

The parameter c causes a shift in the sin x graph, either to the left for positive c, or to the right for negative c. This phase shift moves the graph by c units.

Let us try another example: f(x) = sin (x - (p - 2.14159)).

Since p = 3.14159 (roughly) then p - 2.14159 = 1, so f(x) = sin (x - 1).

This should cause a phase shift in our graph by 1 unit to the right.

The below graph has integer values on the x-axis so this is easy to see. The vertical line corresponds to 2p:
 

Since this is addition and we can only have either positive or negative numbers (c=0 reduces f to f(x) = a sin bx) then

these are the only two effects possible for the c parameter (*please see closing when x is combined with the b parameter).

Note again that the amplitude has remained the same at a value of 1.

Now lets examine the behavior of a.

Let us break this down into cases for easier study:

Case I: a > 1 ( a =1 makes f(x) = sin (bx + c), which does not tell us what a does).

Case II: 0 < a < 1 (a = 0 reduces f to 0, because (any # times zero is zero).

Case III: a < -1 ( a = -1 makes f(x) = - sin (bx + c), which just reflects the effects of b & c over the x-axis).

Case IV: -1 < a < 0.

In this instance our amplitude is 3. But it may be too early to say that there exists a direct correlation between a and the amplitude. Let us look at the other cases:

In the second case a is a number between 0 and 1. Lets try a = 1/2, so

There seems to be a direct correlation between a and the amplitude of this function. Let us try a really small a,that is,

let us try choosing a close to zero. Say a = .0001. What do you think will happen in this case to the sin (x) graph?

In the above graphs, a did not change the period of our function at all but it did make the graph higher (larger amplitude)

for a > 1 and smaller height for 0 < a < 1. So for a really small our sin x graph should resemble a horizontal line.

Let's see the picture below

This totally makes sense because as a approaches zero then a ( sin (x)) is close to y = 0 * sin (x), which is y = 0!

Excellent. Let us look at case III and IV before finalizing this.

Case III: a<-1

First let us see what happens when a = -2 and a = -3. Remember our concept of negative, which shall remain as an

abstract term meaning reverse.

It seems that our relationship between a and the amplitude remains unchanged, but the negative sign has flipped our

graph about the x-axis. So, the coefficient a in front of sin (bx + c) is the amplitude [if a is negative we take the

absolute value of a].

So our amplitude is 1/2 and the negative sign flips the sin(x) graph.

Now, we can combine different a, b, & c to produce different effects on the sin (x) graph:

Lets try a = 3, b = 1/2, and c = -2.

We know the a parameter causes the amplitude to be 3. There is no negative sign so the graph is not flipped.

We know the b parameter affects the period. Since the Period = 2 p((1/[1/2]) =4p, then our sine graph will complete one cycle after a 4p period.

We also know that our c parameter shifts the graph to the right by 4(*) units because c is negative in this case.

There is something to note that when we have an equation like

then our period is still computed the way discussed. When examining the shift to the left or right we must adjust c to the period, that is if we multiply by n in the case of sin (1/n x + c) our total c shift will be nc to the right. This only happens when the b parameter is 0 < b < 1 or -1 < -b < 0.

There are so many different variations and wonderful things that can happen with the sine function. I encourage you to explore more with it. Try it with complex numbers or compose it with functions for the parameters a, b, & c.

As with everything,  possibilities are infinite!

The following mess corresponds to f(x) =

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