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**Unformatted text preview: **ertical asymptotes.
y x
−π f (x) (x, f (x))
undeﬁned −π
2 2 0 1 (0, 1) π
2 0 π
2,0 π 2 −π, 2
2 undeﬁned 1
−π −π
2 π
2 π x −1
−2 One cycle of y = 1 − tan x
2 . We see that the period is π − (−π ) = 2π .
π
2. The ‘quarter marks’ for the fundamental cycle of the cotangent curve are 0, π , π , 34 and π .
42
π
π
To graph g (x) = 2 cot 2 x + π + 1, we begin by setting 2 x + π equal to each quarter mark
and solving for x. a
0
π
4
π
2
3π
4 π π
2x + π = a
π
2x + π = 0
π
π
2x + π = 4
π
π
2x + π = 2
π
3π
2x + π = 4
π
2x + π = π x
−2
3
−2 −1
1
−2 0 690 Foundations of Trigonometry We now use these x-values to generate our graph.
y x
−2 g (x)
undeﬁned −3
2 3 3
−2, 3 −1 1 (−1, 1) −1
2 −1 0 3 (x, g (x)) undeﬁned 2 − 1 , −1
2 1
−2 x −1
−1 One cycle of y = 2 cot π
x
2 + π + 1. We ﬁnd the period to be 0 − (−2) = 2.
As with the secant and cosecant functions, it is possible to extend the notion of period, phase shift
and vertical shift to the tangent and cotangent functions as we did for the cosine and sine functions
in Theorem 10.23. Since the number...

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