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Pre-Calc Exam Notes 121

# Pre-Calc Exam Notes 121 - − 1(sin y = y for − π 2 ≤...

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Inverse Trigonometric Functions Section 5.3 121 The basic idea is that f 1 “undoes” what f does, and vice versa. In other words, f 1 ( f ( x )) = x for all x in the domain of f , and f ( f 1 ( y )) = y for all y in the range of f . We know from their graphs that none of the trigonometric functions are one-to-one over their entire domains. However, we can restrict those functions to subsets of their domains where they are one-to-one. For example, y = sin x is one-to-one over the interval bracketleftbig π 2 , π 2 bracketrightbig , as we see in the graph below: x y 0 1 1 π 2 π π 2 π y = sin x Figure 5.3.4 y = sin x with x restricted to bracketleftbig π 2 , π 2 bracketrightbig For π 2 x π 2 we have 1 sin x 1, so we can define the inverse sine function y = sin 1 x (sometimes called the arc sine and denoted by y = arcsin x ) whose domain is the interval [ 1 , 1] and whose range is the interval bracketleftbig π 2 , π 2 bracketrightbig . In other words:
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Unformatted text preview: − 1 (sin y ) = y for − π 2 ≤ y ≤ π 2 (5.2) sin (sin − 1 x ) = x for − 1 ≤ x ≤ 1 (5.3) Example 5.13 Find sin − 1 ( sin π 4 ) . Solution: Since − π 2 ≤ π 4 ≤ π 2 , we know that sin − 1 ( sin π 4 ) = π 4 , by formula (5.2). Example 5.14 Find sin − 1 ( sin 5 π 4 ) . Solution: Since 5 π 4 > π 2 , we can not use formula (5.2). But we know that sin 5 π 4 = − 1 r 2 . Thus, sin − 1 ( sin 5 π 4 ) = sin − 1 p − 1 r 2 P is, by de±nition, the angle y such that − π 2 ≤ y ≤ π 2 and sin y =− 1 r 2 . That angle is y =− π 4 , since sin ( − π 4 ) = − sin ( π 4 ) = − 1 r 2 . Thus, sin − 1 ( sin 5 π 4 ) = − π 4 ....
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