Trigonometry Lecture Notes_part1-1

# • the range consists of all numbers from-1 1 •

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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 x-intercepts, 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 x-intercepts, the maximum point, and the minimum point. Start with the value of x where the cycle begins and add quarter-periods – 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 x-intercepts, 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 x-values for each of the key points. The five x-values 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 ....
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• The range consists of all numbers from-1 1 • The...

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