844 CHAPTER 8 Applications of Trigonometry EXAMPLE 5 Graphing a Polar Equation (Rose) Graph r = 3 cos 2u. SOLUTION Because the argument is 2u, the graph requires a greater number of points than when the argument is just u. We complete the table using selected angle measures through 360° in order to see the pattern of the graph. Approximate values in the table have been rounded to the nearest tenth. Plotting these points in order gives the graph of a four-leaved rose. Note in Figure 68(a) how the graph is developed with a continuous curve, beginning with the upper half of the right horizontal leaf and ending with the lower half of that leaf. As the graph is traced, the curve goes through the pole four times. This can be seen as a calculator graphs the curve. See Figure 68(b). U 2U cos 2U r =3 cos 2U U 2U cos 2U r =3 cos 2U 0° 0° 1 3 120° 240° -0.5 -1.5 15° 30° 0.9 2.6 135° 270° 0 0 30° 60° 0.5 1.5 180° 360° 1 3 45° 90° 0 0 225° 450° 0 0 60° 120° -0.5 -1.5 270° 540° -1 -3 75° 150° -0.9 -2.6 315° 630° 0 0 90° 180° -1 -3 360° 720° 1 3 NOTE To sketch the graph of r = 3 cos 2u in polar coordinates, it may be helpful to first sketch the graph of y = 3 cos 2x in rectangular coordinates. The minimum and maximum values of this function may be used to determine the location of the tips of the leaves, and the x-intercepts of this function may be used to determine where the polar graph passes through the pole. The equation r = 3 cos 2u in Example 5 has a graph that belongs to a family of curves called rose curves. r =a sin nU and r =a cos nU Equations of roses • The graph has n leaves if n is odd, and 2n leaves if n is even. • The absolute value of a determines the length of the leaves. S Now Try Exercise 51. Figure 68 −4.1 −6.6 4.1 6.6 Polar graphing mode (b) 180° 0° 90° 270° 1 6 4 7 5 2 8 3 3 Start Finish r = 3 cos 2U (a)
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