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**Unformatted text preview: **= arcsin(x) so that t lies in the
interval − π , π with sin(t) = x. We aim to express cos (2 arcsin(x)) = cos(2t) in terms
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of x. Since cos(2t) is deﬁned everywhere, we get no additional restrictions on t. We
have three choices for rewriting cos(2t): cos2 (t) − sin2 (t), 2 cos2 (t) − 1 and 1 − 2 sin2 (t),
each of which is valid for t in − π , π . Since we already know x = sin(t), it is easiest to
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use the last form. We have cos (2 arcsin(x)) = cos(2t) = 1 − 2 sin2 (t) = 1 − 2x2 . Since
arcsin(x) is deﬁned only for −1 ≤ x ≤ 1, the equivalence cos (2 arcsin(x)) = 1 − 2x2 is
valid on [−1, 1].
A few remarks about Example 10.6.1 are in order. Most of the common errors encountered in
dealing with the inverse circular functions come from the need to restrict the domains of the
original functions so that they are one-to-one. One instance of this phenomenon is the fact that
arccos cos 11π = π as opposed to 11π . This is the exact same phenomenon discussed in...

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