Unformatted text preview: = 0, the formula holds trivially, since tan (sin − 1 0) = tan 0 = = 1 − 2 . Now suppose that 0 < x < 1. Let θ = sin − 1 x . Then θ is in QI and sin θ = x . Draw a right triangle with an angle θ such that the opposite leg has length x and the hypotenuse has length 1, as in Figure 5.3.10 (note that this is possible since < x < 1). Then sin θ = x 1 = x . By the Pythagorean Theorem, the adjacent leg has length r 1 − x 2 . Thus, tan θ = x 1 − x 2 . If − 1 < x < 0 then θ = sin − 1 x is in QIV. So we can draw the same triangle except that it would be “upside down” and we would again have tan θ = x 1 − x 2 , since the tangent and sine have the same sign (negative) in QIV. Thus, tan (sin − 1 x ) = x 1 − x 2 for − 1 < x < 1....
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 Fall '11
 Dr.Cheun
 Calculus, Trigonometry, Real Numbers, Pythagorean Theorem, Right triangle, Hypotenuse, Tan, triangle

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