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Lecture 5: Inverse Trigonometric Functions 5.1 The inverse sine function The function f (x) = sin(x) is not one-to-one on (−∞, ∞), but is on − π2 , π2 . Moreover, f still has range [−1, 1] when restricted to this interval. Hence it is reasonable to restrict π π f to − 2 , 2 to obtain an inverse for the sine function. Definition We call the inverse of the sine function, when restricted to − π2 , π2 , the arcsine function, with value at a point x denoted by arcsin(x) or sin−1 (x). In other words, π π ≤y≤ . 2 2 Note that the arcsine function has domain [−1, 1] and range − π2 , π2 . y = sin−1 (x) if and only if sin(y) = x and − Example We have sin−1 (0) = 0, π , 2 π sin−1 (−1) = − , 2 sin−1 (1) = and −1 sin 1 − 2 π =− . 6 Note that sin(sin−1 (x)) = x for any −1 ≤ x ≤ 1 and sin−1 (sin(x)) = x for any − π2 ≤ x ≤ π 2 . However, the latter equality does not hold for other values of x. For example, −1 sin sin 7π 6 = sin −1 1 − 2 π =− . 6 Since for x in [−1, 1] we have sin(sin−1 (x)) = x, it follows that d d sin(sin−1 (x)) = x. dx dx Hence cos(sin−1 (x)) d sin−1 (x) = 1, dx 5-1 Lecture 5: Inverse Trigonometric Functions and so 5-2 d 1 sin−1 (x) = . dx cos(sin−1 (x)) Now if y = sin−1 (x), then − π2 ≤ y ≤ π 2 and sin(y) = x. Thus cos2 (y) = 1 − sin2 (y) = 1 − x2 . Since cos(y) ≥ 0 for − π2 ≤ y ≤ π 2, we have cos(sin−1 (x)) = cos(y) = p 1 − x2 . Thus we have the following proposition. Proposition Example d 1 sin−1 (x) = √ for all −1 < x < 1. dx 1 − x2 We have d 6x sin−1 (3x2 ) = √ . dx 1 − 9x4 Note that f (x) = sin−1 (x) is an increasing function on [−1, 1], with graph as shown below. -1.5 -1 1.5 1.5 1 1 0.5 0.5 0.5 -0.5 1 1.5 -1.5 -1 0.5 -0.5 -0.5 -0.5 -1 -1 -1.5 -1.5 Graphs of y = sin(x) and y = sin−1 (x) We now have the integration formula Z 1 √ dx = sin−1 (x) + c. 2 1−x 1 1.5 Lecture 5: Inverse Trigonometric Functions Example 5-3 Using the substitution u = 2x, we have Z 0 1 4 Z 1 2 1 du 1 − u2 0 12 1 −1 = sin (u) 2 0 1 1 1 = sin−1 − sin−1 (0) 2 2 2 π = . 12 1 1 √ dx = 2 1 − 4x2 √ 5.2 The inverse tangent function Note that the function f (x) = tan(x) is one-to-one if its domain is restricted to − π2 , π2 . Moreover, f still has range (−∞, ∞) when restricted to this interval. Hence it is reasonable π π to restrict f to − 2 , 2 to obtain an inverse for the tangent function. Definition We call the inverse of the tangent function, when restricted to − π2 , π2 , the arctangent function, with value at a point x denoted by arctan(x) or tan−1 (x). In other words, y = tan−1 (x) if and only if tan(y) = x and − π π <y< . 2 2 Note that the arctangent function has domain (−∞, ∞) and range − π2 , π2 . Example We have tan−1 (0) = 0, tan−1 (1) = and π , 4 √ π tan−1 (− 3) = − . 3 Since tan(tan−1 (x)) = x for all x in (−∞, ∞), we have d d tan(tan−1 (x)) = x. dx dx Hence sec2 (tan−1 (x)) d tan−1 (x) = 1, dx Lecture 5: Inverse Trigonometric Functions and so 5-4 d 1 tan−1 (x) = . 2 dx sec (tan−1 (x)) Now if y = tan−1 (x), then − π2 < y < π 2 and tan(y) = x. Thus sec2 (y) = 1 + tan2 (y) = 1 + x2 . Thus we have the following proposition. Proposition Example d 1 tan−1 (x) = . dx 1 + x2 If f (x) = 4 tan−1 x 2 , then 0 f (x) = 1+ 4 x 2 1 8 = . 2 4 + x2 2 Note that f (x) = tan−1 (x) is an increasing function on (−∞, ∞) with lim tan−1 (x) = x→∞ and π 2 π lim tan−1 (x) = − . x→−∞ 2 Hence the lines y = π 2 and y = − π2 are horizontal asymptotes for the graph of f . Moreover, f 00 (x) = −(1 + x2 )−2 (2x) = − 2x , (1 + x2 )2 from which it follows that the graph of f is concave upward on (−∞, 0) and concave downward on (0, ∞). The graph is shown below. 1.5 4 1 2 -1.5 -1 0.5 0.5 -0.5 -2 1 1.5 -4 2 -2 -0.5 -1 -4 -1.5 Graphs of y = tan(x) and y = tan−1 (x) 4 Lecture 5: Inverse Trigonometric Functions 5-5 We now have the integration formula Z 1 dx = tan−1 (x) + c. 2 1+x Example Z Using the substitution u = x2 , we have 1 1 dx = 2 4+x 4 Z 1 1 dx = 2 2 1 + x2 Z 1 1 1 −1 −1 x du = tan (u) + c = tan + c. 1 + u2 2 2 2 5.3 The inverse secant function Note that the function f (x) = sec(x) is one-to-one if its domain is restricted to h π 3π 0, ∪ π, . 2 2 Moreover, f still has range (−∞, −1] ∪ [1, ∞) when restricted to this domain. Hence it is reasonable to restrict f to this domain to obtain an inverse for the secant function. Definition We call the inverse of the secant function, when restricted to h π 3π 0, ∪ π, , 2 2 the arcsecant function, with value at a point x denoted by arcsec(x) or sec−1 (x). In other words, y = sec−1 (x) if and only if sec(y) = x and either 0 ≤ y < Note that the arcsecant function has domain (−∞, −1] ∪ [1, ∞) and range h Example π 3π 0, ∪ π, . 2 2 We have sec−1 (1) = 0, sec−1 (−1) = π, 3π π or π ≤ y < . 2 2 Lecture 5: Inverse Trigonometric Functions and sec−1 (−2) = 5-6 4π . 3 Since sec(sec−1 (x)) = x, we have d d sec(sec−1 (x)) = x. dx dx Hence sec(sec−1 (x)) tan(sec−1 (x)) and so d sec−1 (x) = 1, dx d 1 sec−1 (x) = . dx x tan(sec−1 (x)) Now if y = sec−1 (x), then tan2 (y) = sec2 (y) − 1 = x2 − 1. Since tan(y) ≥ 0 for y in the range of the arcsecant function, we have tan(sec−1 (x)) = √ x2 − 1. Thus we have the following proposition. Proposition For all x ≥ 1 or x ≤ −1, d 1 sec−1 (x) = √ . dx x x2 − 1 The previous proposition gives us the integration formula Z 1 √ dx = sec−1 (x) + c. 2 x 1−x The graph of f (x) = sec−1 (x) is shown below. Notice how the vertical asymptotes of g(x) = sec(x) are transformed into horizontal asymptotes for f . 10 4 7.5 5 3 2.5 -2.5 1 2 3 2 4 1 -5 -7.5 -10 -10 -5 Graphs of y = sec(x) and y = sec−1 (x) 5 10 Lecture 5: Inverse Trigonometric Functions 5-7 5.4 The remaining inverse trigonometric functions The remaining inverse trigonometric functions are not used as frequently as the three we have considered above. They are defined as follows: y = cos−1 (x) if and only if cos(y) = x and 0 ≤ y ≤ π, y = cot−1 (x) if and only if cot(y) = x and 0 < y < π, and y = csc−1 (x) if and only if csc(y) = x and 0 < y ≤ π 3π or π < y ≤ . 2 2 One may show that d 1 cos−1 (x) = − √ , dx 1 − x2 d 1 cot−1 (x) = − , dx 1 + x2 and d 1 csc−1 (x) = − √ . dx x x2 − 1 Note that these are just the negations of the derivatives of the inverses of the corresponding co-functions. In particular, they do not give us any new integration formulas. Moreover, this implies that these functions differ by only a constant from the inverses of their corresponding co-functions.