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Pre-Calc Exam Notes 125

Pre-Calc Exam Notes 125 - = 0 the formula holds trivially...

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Inverse Trigonometric Functions Section 5.3 125 Thus, y = tan 1 x is a function whose domain is the set of all real numbers and whose range is the interval ( π 2 , π 2 ) . In other words: tan 1 (tan y ) = y for π 2 < y < π 2 (5.6) tan (tan 1 x ) = x for all real x (5.7) Example5.17 Find tan 1 ( tan π 4 ) . Solution: Since π 2 π 4 π 2 , we know that tan 1 ( tan π 4 ) = π 4 , by formula (5.6). Example5.18 Find tan 1 (tan π ). Solution: Since π > π 2 , we can not use formula (5.6). But we know that tan π = 0. Thus, tan 1 (tan π ) = tan 1 0 is, by definition, the angle y such that π 2 y π 2 and tan y = 0. That angle is y = 0. Thus, tan 1 (tan π ) = 0 . Example5.19 Find the exact value of cos ( sin 1 ( 1 4 )) . Solution: Let θ = sin 1 ( 1 4 ) . We know that π 2 θ π 2 , so since sin θ =− 1 4 < 0, θ must be in QIV. Hence cos θ > 0. Thus, cos 2 θ = 1 sin 2 θ = 1 parenleftbigg 1 4 parenrightbigg 2 = 15 16 cos θ = radicallow 15 4 . Note that we took the positive square root above since cos θ > 0. Thus, cos ( sin 1 ( 1 4 )) = radicallow 15 4 . Example5.20 Show that tan (sin 1 x ) = x 1 x 2 for 1 < x < 1. radicallow 1 x 2 x 1 θ Figure 5.3.10
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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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