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Sine and Cosine Exploring what periodic changes do to their graphs

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The BIG Question Did you prepare for today? Write yes or no on today’s date on your celeration chart. If so, estimate the amount of time and write it on your celeration chart.

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We have already covered the periods of sine and cosine. Who recalls their period? They both have a period of 2π. 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 2π. That is, the period is how long it takes for the graph to return to the same place it started on the interval 0 to 2π. If we start with y = sin (x), it takes the graph a cycle of 2π to get back to where it started at y = 0.

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

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What do you think happens when we mess with the inside of the sine or cosine function? That is, what will multiplying x by a real number B do to the graph of the functions? If we take the y = sin (x) and y= cos (x) and change them to: y=sin(Bx) and y=cos(Bx).

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Quick Algebra Review 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.

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y = 2x² y = x² y = x²/4

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y=sin(Bx) and y=cos(Bx) Do you think that multiplying the sine and cosine functions by a real number, B, on the inside will have a similar effect? Yes, because we are altering the function’s period on the interval 0 to 2π. 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(x) y = sin (2x) y = sin(x/2) 2π y = sin (x/2) so cycles half as fast. y = sin (2x) so cycles twice as fast. y = sin (x)

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What does horizontal shrinking imply with respect to cycling or period? The period will be less than 2π so the function will repeat faster, like the affect we saw when we multiplied by a number bigger than one to the quadratic function. Grows t wice as fast. Cycles twice as fast

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What does horizontal stretching imply with respect to cycling or period? The period will be greater than 2π so the function will repeat slower, like the affect we saw when we multiplied by a proper fraction to the quadratic function. Cycles slower Grows slower

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Effect on Period 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

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Example 1 Discuss the period of y = sin (2x). Solution : Since B = 2, This means that the graph completes a cycle at a period of π and at 2π 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 2π. 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 π and another at 2π.

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IMPORTANT question/inquiry/analysis: Why is 2π in the period divided by B and not multiplied by B? 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. 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. 2π half a cycle done 4π one cycle done. Thus it is cycling SLOWER! BUT if we multiplied the period of 2π by 2, then the period would be 4π. This tells me the graph would complete one cycle at 4π NOT 2 cycles by 2π, but half a cycle at 2π, giving it a horizontal stretch. To fix this problem, we divide 2π by B, Hence, the function y = sin(2x) will complete one cycle at π and a second one at 2π. Thus the function then has a horizontal shrinking. MORAL: A fractional pie cycles faster than a whole pie!!!! You try: 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? **The period for y = sin(x) and y = cos (x) is 0 ≤ x ≤ 2π, then the period for y =sin(Bx) and y =cos(Bx) is 0 ≤ Bx ≤ 2π. Thus dividing all parts by B, in which B > 0 we have, 0 ≤ x ≤ 2π/B. But there is a graphical way to see this as well.

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Example 2 Discuss the period of y = cos(x/2). Solution: Since B = ½, How did we simplify to get 4π? This means that the graph completes a cycle at a period of 4π. Moreover, at 2π the graph will have completed ½ a cycle. We know this since B = ½ the graph should complete half a cycle at 2π. Thus the graph of the equation y = cos(x/2) will have a horizontal stretch completing one cycle at 4π.

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You try Discuss the period of y = sin(x/6). Strategy: 1.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 2π. 3. Period is then B = 2π/(1/6) = 12π. 4. This means that the function will complete one cycle at 12π. 5. Thus, the graph of y = sin(x/6) will have a horizontal stretch completing one cycle at 12 π. Hence, one complete cycle in (0, 12π).

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EXAMPLE 3 Which graph goes with the equation y = sin (x/2)? Think: which graph has completed B =1/2 cycles on the interval 0 to2π? This is the only graph that completes half of a cycle by 2π.

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You Try Which graph goes with y = cos(2x)? Think: which graph has completed B = 2 cycles on the interval 0 to2π? This is the only graph that completes two full cycles.

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Example 5 Graph the equation y = cos(10x). Strategy: 1. Recall what the parent cosine function looks like. 2. Here B = 10, thus this graph should cycle10 times between 0 and 2π. 3. This tells me that this graph must have a horizontal shrinking. 4. The period = 2π/B = 2π/10 = π /5 which implies one cycle is completed at π /5. Here one cycle is completed so this must be π/5. Here is the tenth cycle, so this is where 2π is.

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Can You? Find 2π/5, 3π/5, 4π/5, etc. on the previous graph? Graph y = cos(x) over the previous graph ? Second cycle complete so here is where 2π/5 Fourth cycle complete so this is where 4π/5 y=cos(2π)

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Application: 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 http://www.intmath.com/trigonometric- graphs/2-graphs-sine-cosine-period.php This is the picture at the web site.

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Activity 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? 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?

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Screen Shot 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?

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Screen Shot 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?

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Screen Shot 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?

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Screen Shot 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?

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Thank – You for your attention and participation!

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Copyright © Cengage Learning. All rights reserved. 4 Trigonometric Functions.

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