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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 22 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 22 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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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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