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6.4 Graphing Sine and Cosine Functions Referring back to The Unit Circle, let’s create a table of angles and values of sine and cosine functions. We will consider θ in radian measure. angle θ 0 π/6 π/4 π/3 π/2 2π/3 3π/4 5π/6 π sin θ 0 ½ cos θ 1 3 /2 2 /2 3 /2 1 3 /2 2 /2 ½ 0 angle θ π 7π/6 sin θ 0 -½ cos θ -1 - 3 /2 2 /2 ½ 5π/4 4π/3 - 2 /2 -½ 0 -½ 3π/2 5π/3 - 2 /2 - 3 /2 -1 - 3 /2 - 2 /2 - 3 /2 -1 7π/4 11π/6 - 2 /2 -½ 2π 0 2 /2 3 /2 1 0 ½ By plotting (θ, sin θ), we create the graph representing one cycle of y = sin x. By plotting (θ, cos θ), we create the graph representing one cycle of y = cos x. To graph these functions on your calculator, make sure you are in radian mode and once you type in the equation be sure to select ZOOM-TRIG. This allows the x interval to be partitioned every π/2. Vertical stretching can be determined by a coefficient on the trigonometric function. This coefficient is multiplied to the y-values in the previous chart. Example Graph y = 3sin x on the interval [-2π, 2π]. Notice that the y-values are multiplied by 3. The important angles to note when graphing are 0, π/2, π, 3π/2, and 2π. Try the following: Graph on [-2π, 2π]. y = 2cos x y = 4sin x y = 5cos x y = ½ sin x Answers: Notice the pattern in these graphs. What is the length of one period? Sine and Cosine both have a period of 2π. Examine the graphs above, notice the absolute values of the minimum and maximum y-values correlate with the coefficient of the functions. What do you think a negative coefficient would do to the graphs of sine and cosine functions? Example Graph y = -2cos x on [0, 2π]. Notice this graph is a refection in the x-axis of the previous problem y = 2cos x. Try the following: Graph on [0, 2π]. y = -4sin x y = - ½ sin x y = -3/2 cos x Answers: How could we affect the length of the period of these graphs? Is there a way to have 2 cycles of sine on the interval [0, 2π]? We can establish this by having a multiplier to the x-value. For instance, y = sin 2x means we will graph sine function containing 2 cycles on the interval [0, 2π]. So what would happen to the graph of y = sin ( ½ x)? Only half of a cycle will occur on the interval [0, 2π]. So that means that one cycle will be on the interval [0, 4π]. Example Graph y = cos (2x) on [0, 2π]. State the minimum and maximum values of the graph. The minimum is at –1 when x = π/2, 3π/2 and the maximum is at 1 when x = 0, π, 2π. We could algorithmically calculate the length of one cycle, or period. 2 Period = , where b is the coefficient on x. b Try the following: Graph one period of the functions below. In addition, determine the length of a period and state the minimum and maximum values. y = 3sin ( ½ x) y = -2cos x y = -sin (3x) y = 3cos (2x) Answers: Min: -1 when x = 3π Max: 1 when x=π Min: -1 when x = 0 and 2π Max: 1 when x=π Min: -1 when x = π/6, 5π/6, & 3π/2 Max: 1 when x = π/2, 7π/6, & 11 π/6 Min: -1 when x = π/2 Max: 1 when x = 0 and π Note: The coefficient of the function is called the amplitude of the function.