**Unformatted text preview: **− π ∪ − π , π ∪ π , π . Returning to arctan(2t), we note the double angle
2
4
44
42
2
identity tan(2t) = 1−tan(t()t) , is valid for values of t under consideration, hence we get
tan2
2
2
tan(2 arctan(x)) = tan(2t) = 1−tan(t()t) = 1−x 2 . To ﬁnd where this equivalence is valid we
x
tan2
ﬁrst note that the domain of arctan(x) is all real numbers, so the only exclusions come
from the x values which correspond to t = ± π , the values where tan(2t) is undeﬁned.
4
2
Since x = tan(t), we exclude x = tan ± π = ±1. Hence, tan(2 arctan(x)) = 1−x 2
4
x
5 for (−∞, −1) ∪ (−1, 1) ∪ (1, ∞).
holds (b) We let t = arccot(2x) so that 0 < t < π and cot(t) = 2x. In terms of t, cos(arccot(2x)) =
cos(t), and our goal is to express the latter in terms of x. Since cos(t) is always deﬁned,
there are no additional restrictions on t, and we can begin using identities to get exprest)
sions for cos(t) and cot(t). The identity cot(t) = cos(t) is valid for t in (0, π ), so if...

View
Full
Document