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**Unformatted text preview: **(x) tan(π )
1 − (tan(x))(0) which tells us the period of tan(x) is at most π . To show that it is exactly π , suppose p is a
positive real number so that tan(x + p) = tan(x) for all real numbers x. For x = 0, we have
tan(p) = tan(0 + p) = tan(0) = 0, which means p is a multiple of π . The smallest positive multiple
of π is π itself, so we have established the result. We take as our fundamental cycle for y = tan(x)
the interval − π , π , and use as our ‘quarter marks’ x = − π , − π , 0, π and π . From the graph, we
22
2
4
4
2
see conﬁrmation of our domain and range work in Section 10.3.1.
It should be no surprise that K (x) = cot(x) behaves similarly to J (x) = tan(x). Plotting cot(x)
over the interval [0, 2π ] results in the graph below.
y x
0 cot(x) (x, cot(x))
undeﬁned π
4
π
2
3π
4 0 π
4,1
π
2,0 −1 3π
4 , −1 π undeﬁned 5π
4
3π
2
7π
4 1
−1 2π undeﬁned 1 0 5π
4 ,1
3π
2 ,0
7π
4 , −1 1 π
4 π
2 3π
4 π 5π
4 3π
2 7π
4 2π x −1 The graph of y = cot(x) over [0, 2π ].
From these data, it clear...

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