Stitz-Zeager_College_Algebra_e-book

# 87 k sin arcsin087 2k sin arcsin087 087

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Unformatted text preview: 2 a 2x = a x 0 2x = 0 0 π 2 2x = π 2 π 2x = π 3π 2 π 4 π 2 3π 4 2π 2x = 2π π 3π 2 2x = 10.5 Graphs of the Trigonometric Functions 685 Next, we substitute these x values into f (x). If f (x) exists, we have a point on the graph; otherwise, we have found a vertical asymptote. In addition to these points and asymptotes, we have graphed the associated cosine curve – in this case y = 1 − 2 cos(2x) – dotted in the picture below. Since one cycle is graphed over the interval [0, π ], the period is π − 0 = π . y x 0 f (x) (x, f (x)) −1 (0, −1) π 4 π 2 3π 4 undeﬁned π −1 3 2 1 3 π 2,3 π 4 −1 π 2 3π 4 x π undeﬁned (π, −1) One cycle of y = 1 − 2 sec(2x). 2. Proceeding as before, we set the argument of cosecant in g (x) = quarter marks and solve for x. a π − πx = a csc(π −πx)−5 3 equal to the x 0 π − πx = 0 1 π 2 π − πx = π 2 1 2 π π − πx = π 0 3π 2 3π 2 −1 2 2π π − πx = 2π −1 π − πx = Substituting these x-values into g (x), we generate the grap...
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