This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: • The range consists of all numbers from [1, 1] • The period is 2 π . • This function is an odd function which can be seen from a graph by observing symmetry with respect to the origin. Graphing Variations of y=sinx To graph variations of y = sinx by hand, it is helpful to find xintercepts, maximum points, and minimum points. There are three intercepts, so we can find these five points by using the following scheme. Let 1 x = the x value where the cycle begins. Then for all i in [2,5] the i th point can be found using: 1 4 i i period x x = + . We can stretch or shrink the graph by adding a coefficient other than one to the sine graph: sin y A x = This is the graph of y = sinx, compare it to the graph below of y = 2 sinx: In general the graph of y = Asinx ranges between A and A . We call A the amplitude of the graph. If it is A > 1 the graph is stretched, if A < 1 it is shrunk. Graphing Variations of y=sinx • Identify the amplitude and the period. • Find the values of x for the five key points – the three xintercepts, the maximum point, and the minimum point. Start with the value of x where the cycle begins and add quarterperiods – that is, period/4 – to find successive values of x . • Find the values of y for the five key points by evaluating the function at each value of x from step 2. • Connect the five key points with a smooth curve and graph one complete cycle of the given function. • Extend the graph in step 4 to the left or right as desired. Example 24 Determine the amplitude of y = 1/2 sin x . Then graph y = sin x and y = 1/2 sin x for 0 < x < 2 π . Step 1 Identify the amplitude and the period. The equation y = 1/2 sin x is of the form y = A sin x with A = 1/2. Thus, the amplitude  A  = 1/2. This means that the maximum value of y is 1/2 and the minimum value of y is 1/2. The period for both y = 1/2 sin x and y = sin x is 2 π . Step 2 Find the values of x for the five key points. We need to find the three xintercepts, the maximum point, and the minimum point on the interval [0, 2 π ]. To do so, we begin by dividing the period, 2 π , by 4. Period/4 = 2 π /4 = π /2 We start with the value of x where the cycle begins: x = 0. Now we add quarter periods, π / 2 , to generate xvalues for each of the key points. The five xvalues are x = 0, x = π /2, x = π , x = 3 π /2, x = 2 π Step 3 Find the values of y for the five key points. We evaluate the function at each value of x from step 2. (0,0), ( π /2, 1/2), ( π ,0), (3 π /2, 1/2), (2 π , 0) Step 4 Connect the five key points with a smooth curve and graph one complete cycle of the given function. The five key points for y = 1/2sin x are shown below. By connecting the points with a smooth curve, the figure shows one complete cycle of y = 1/2sin x . Also shown is graph of y = sin x . The graph of y = 1/2sin x shrinks the graph of y = sin x ....
View
Full Document
 Fall '12
 lipsh
 Trigonometry, Cos, Inverse function, Inverse trigonometric functions

Click to edit the document details