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Sine and Cosine

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Sine and Cosine

Exploring what periodic changes do to their graphs

Did you prepare for today? Write yes or no on todayâ€™s date on your celeration chart.

If so, estimate the amount of time and write it on your celeration chart.

If we start with y = sin (x), it takes the graph a cycle of 2Ï€ to get back to where it started at y = 0.

We have already covered the periods of sine and cosine.

Who recalls their period?

They both have a period of 2Ï€.

Who recalls what that means?

The period means to make one complete cycle of the graph. In other words, the period tells us how often the graph goes one complete repetition around the unit circle or on the interval 0 to 2Ï€ . That is, the period is how long it takes for the graph to return to the same place it started.

Here is y = cos(x) and again it takes a cycle of 2Ï€ to get back to where it started at y = 1.

What do you think happens when we mess with the inside of the sine or cosine function?

If we take the y = sin (x) and y= cos (x) and change them to: y=sin(Bx) and y=cos(Bx).

That is, what will multiplying x by a real number B do to the graph of the functions?

Do you recall what happens in algebra to the graph of the quadratic function when we multiplied its equation by a real number?

If we multiplied by a number greater than one, it made the graph grow faster or taller and skinnier.

If we multiplied by a proper fraction, it slowed the growth of the graph down or made it wider.

What is a proper fraction?

It is a fraction in which the numerator is smaller

than the denominator.

y = 2xÂ²

y = xÂ²

y = xÂ²/4

Do you think that multiplying the sine and cosine functions by a real number, B, on the inside will have a similar effect?

YES

y = sin (x/2) so cycles half as fast.

y = sin (2x) so cycles twice as fast.

How?

By multiplying by a number bigger than one, B > 1, this will make the function cycle faster or give it a horizontal shrinking.

Multiplying it by a proper fraction, 0 < B < 1 this will make it cycle slower or give it a horizontal stretch.

y = sin (2x)

y = sin(x/2)

y = sin(x)

2Ï€

y = sin (x)

What does horizontal shrinking imply with respect to cycling or period?

The period will be less than 2Ï€ so the function will repeat faster, like the affect we saw when we multiplied by a number bigger than one to the quadratic function.

Cycles twice as fast

Grows twice as fast.

What does horizontal stretching imply with respect to cycling or period?

The period will be greater than 2Ï€ so the function will repeat slower, like the affect we saw when we multiplied by a proper fraction to the quadratic function.

Grows slower

Cycles slower

How does this affect the period?

It changes it to: period =

In the equations we looked at previously, y = sin(x) and y = cos(x), what is B?

1

Discuss the period of y = sin (2x).

Solution:Since B = 2,

- This means that the graph completes a cycle at a period of Ï€ and at 2Ï€ it will have completed two cycles, or B many cycles.

- Thus, B tells us how many completed cycles the graph will make within the interval from 0 to 2Ï€. Here it will be two.

- The graph of the equation y = sin (2x) will have a horizontal shrink in which it completes one cycle at Ï€ and another at 2Ï€.

To fix this problem, we divide 2Ï€ by B, Hence, the function y = sin(2x) will complete one cycle at Ï€ and a second one at 2Ï€. Thus the function then has a horizontal shrinking.

If we did not divide by B in the period, when we multiply x by B>1 in the equations

y= sin(Bx) and y= cos(Bx), the graph of the sine/cosine function would not cycle faster. In fact, it would cycle slower contradicting what we have seen happen to graphs of functions when they are multiplied by a number greater than one.

You try a similar analysis for the value of B when it is a proper fraction. Can you figure out why when you divide by a proper fraction that gives the graph a horizontal stretch?

Example: y = sin(2x).

Here we would know that the 2 tells us that the sine function should cycle twice as fast as the parent sine function.

BUT if we multiplied the period of 2Ï€ by 2, then the period would be 4Ï€. This tells me the graph would complete one cycle at 4Ï€ NOT2 cycles by 2Ï€, but half a cycle at 2Ï€, giving it a horizontal stretch.

2Ï€ half a cycle done

4Ï€ one cycle done. Thus it is cycling SLOWER!

How did we simplify to get 4Ï€?

Discuss the period of y = cos(x/2).

Solution: Since B = Â½,

- This means that the graph completes a cycle at a period of 4Ï€.

- Moreover, at 2Ï€ the graph will have completed Â½ a cycle.
- We know this since B = Â½ the graph should complete half a cycle at 2Ï€.

- Thus the graph of the equation y = cos(x/2) will have a horizontal stretch
- completing one cycle at 4Ï€.

Discuss the period of y = sin(x/6).

Strategy:

- Identify B. In this example it is B = 1/6, which is a proper fraction.

2. This means the function will cycle slower so will complete

1/6th of the cycle in the interval 0 to 2Ï€.

3. Period is then B = 2Ï€/(1/6) = 12Ï€.

4. This means that the function will complete one cycle at 12Ï€.

5. Thus, the graph of y = sin(x/6) will have a horizontal stretch completing one cycle at 12 Ï€. Hence, one complete cycle in (0, 12Ï€).

Which graph goes with the equation y = sin (x/2)?

Think: which graph has completed B =1/2 cycles on the interval 0 to2Ï€?

This is the only graph that completes half of a cycle by 2Ï€.

Think: which graph has completed

B = 2 cycles on the interval 0 to2Ï€?

Which graph goes with y = cos(2x)?

This is the only graph that completes two full cycles.

Here is the tenth cycle, so this is where 2Ï€ is.

Graph the equation y = cos(10x).

Strategy:

1. Recall what the parent cosine function looks like.

Here one cycle is completed so this must be Ï€/5.

2. Here B = 10, thus this graph should cycle10 times between 0 and 2Ï€.

3. This tells me that this graph must have a

horizontal shrinking.

4. The period = 2Ï€/B = 2Ï€/10 = Ï€ /5 which implies one cycle is completed at Ï€ /5.

Find 2Ï€/5, 3Ï€/5, 4Ï€/5, etc. on the previous graph?

Graph y = cos(x) over the previous graph?

Fourth cycle complete so this is where 4Ï€/5

Second cycle complete so here is where 2Ï€/5

y=cos(2Ï€)

The piston movement in a piston engine can be modeled using the sine function.

Go to the web site:

http://www.intmath.com/trigonometric-graphs/2-graphs-sine-cosine-period.php

This is the picture at the web site.

- Put in different values of B, specifically, B = .2, .5, 4, 6.
- Play with how different values of B affect the cycle of the sine function and how it affects the motion of the piston.
- Write a short summary of the number of cycles completed by the sine function for each of the given values and how the piston was affected for each given value of B.
- What is the interval covered by the graph?

Here is the screen shot for B = .2.

How many cycles would the sine function complete for this value of B?

How would this effect the rate of rotation in the piston?

Here is the screen shot for B = .5.

How many cycles would the sine function complete for this value of B?

How would this effect the rate of rotation in the piston?

Here is the screen shot for B = 4.

How many cycles would the sine function complete for this value of B?

How would this effect the rate of rotation in the piston?

Here is the screen shot for B = 6.

How many cycles would the sine function complete for this value of B?

How would this effect the rate of rotation in the piston?

Thank â€“ You for your attention and participation!