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1 January, 14:46

Let f (x) = cos (arctanx). What is the range of f?

(a) {x| - (pi/2) (a) {x|0 (a) {x|0 = (a) {x|-1 (a) {x|-1=

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  1. 1 January, 15:34
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    -π/2 < arctan (x) < π/2

    So cos (π/2) < cos (arctan (x)) < cos (0)

    0 < cos (arctan (x)) < 1
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