College Trigonometry, 2011a

830 Foundations of Trigonometry
which this equivalence is valid, we look back at our original substution, t = arccsc(4x).
Since the domain of arccsc(x) requires its argument x to satisfy |x| ≥1, the domain of
arccsc(4x) requires|4x| ≥1. Using Theorem 2.4, we rewrite this inequality and solve
to get x ≤− 1 4 or x ≥ 1 4
. Since we had no additional restrictions on t, the equivalence
√
cos(arccsc(4x)) = 16x 2 −1
4|x|
holds for all x in ( −∞, − 1 [
4]
∪
1
4 , ∞) .
10.6.2 Inverses of Secant and Cosecant: Calculus Friendly Approach
In this subsection, we restrict f(x) = sec(x) to [ 0, π ) [ )
2 ∪ π,
3π
2
y
y
3π
2
1
π
−1
π
2
π
3π
2
x
π
2
f(x) = sec(x) on [ 0, π 2
) [ )
∪ π,
3π
2
and we restrict g(x) = csc(x) to ( 0, π ] ( ]
2 ∪ π,
3π
2 .
reflect across y = x
−−−−−−−−−−−−→
switch x and y coordinates
−1 1
f −1 (x) = arcsec(x)
x
y
y
3π
2
1
π
−1
π
2
π
3π
2
x
π
2
g(x) = csc(x) on ( 0, π 2
] ( ]
∪ π,
3π
2
reflect across y = x
−−−−−−−−−−−−→
switch x and y coordinates
−1 1
g −1 (x) = arccsc(x)
x
Using these definitions, we get the following result.