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Inverse Trigonometric Functions and Their Derivatives None of the trigonometric functions satisfies the horizontal line test, so none of them has an inverse. The inverse trigonometric functions are defined to be the inverses of particular parts of the trigonometric functions; parts that do have inverses. y = sin(x) π π − ≤ x≤ 2 2 sin −1(x) = arcsin(x) y = cos(x) 0 ≤ x ≤π cos−1(x) = arccos(x) y = tan(x) π π − ≤ x≤ 2 2 tan−1(x) = arctan(x) y = sec(x) 0 ≤ x ≤π x ≠π 2 sec−1(x) = arcsec(x) Evaluating inverse trigonometric functions The equation y = sin −1(x) is equivalent with the equation x = sin( y). Thus sin −1(x) should be thought of as the angle whose sine is x. However, many angles have sine equal to x, π and in this case, we want the angle that lies between and 2 π . A similar idea holds for all the other inverse − 2 trigonometric functions. Each is an angle, but you must choose the particular angle that satisfies the restriction appropriate to that function. Example. Find sin−1(−1) and cos−1(1/2) Solution. 1. Let y = sin−1(−1). Then sin(y) = −1, so y =−π 2 2. Let y = cos−1(1/2). Then cos(y) = 1/2, so y =π 3 We can also find identities involving the inverse trigonometric functions, by using ordinary trigonometric identities. Example. Show that cos(sin−1(x)) = 1− x 2 1 The result is obvious from the diagram: sin−1( x) 1− x 2 x Example. Clearly sin(sin−1(x)) = x. Is it also true that sin−1(sin (x)) = x? Actually, this is not true because of the restrictions on the inverse sine. The graph of sin−1(sin (x)) is shown below. 5π 1 then sin(x) = . But the angle in 4 2 the range −π ≤ x ≤ π having that sine is π not 5π . 4 4 2 2 For example, if x = All such expressions (inversetrigfn(trigfn) = ) must be treated with care. Derivatives of Inverse Trigonometric Functions Theorem. d sin −1(x) = dx 1 1− x2 Proof. Let y = sin −1(x) so sin( y) = x. Then cos( y) dy =1 dx dy = 1 = dx cos( y) 1 1−sin 2( y) = 1 1− x 2 In a similar way we can show that d cos−1( x) = dx Theorem. −1 1− x 2 d tan −1(x) = 1 dx 1+ x2 Proof. Let y = tan −1(x) so tan( y) = x. Then sec2 ( y) dy =1 dx dy = 1 = 1 = 1 dx sec2 ( y ) 1+ tan 2( y ) 1+ x 2 Theorem. d sec −1(x) = dx 1 | x| x 2 −1 Proof. Let y = sec−1( x) so sec( y) = x. Then sec( y)tan( y) dy =1 dx dy = 1 = cos( y) cos( y) dx sec( y) tan( y) sin( y) x x 2 −1 y 2( y) 1 cos = = sin( y) x2 x = 1 x2 −1 | x| x2 −1 1 We have now shown the following rules for differentiation. Basic Rule 1. d sin −1( x) = dx Generalized Rule 1 1− x 2 2. d tan −1(x) = 1 dx 1+ x 2 3. d sec−1(x) = dx 1 | x| x 2 −1 d sin −1(u) = dx 1 1−u du 2 dx d tan −1(u ) = 1 du dx 1+u 2 dx d sec−1(u) = dx 1 du dx |u| u 2 −1 The other three inverse trig functions have derivatives that are the negatives of their respective cofunctions. Miscellaneous Problems 1 − 1 Problem. Suppose that θ = cos . Find the exact values 2 of sin(θ ),tan(θ ),cot(θ ),sec(θ ),csc(θ ). Solution. Construct the following triangle. 2 3 θ 1 so that cos(θ ) = 1 2 The other functions can be read directly from that triangle. sin(θ ) = 3 2 tan(θ ) = 3 sec(θ ) = 2 2 cot(θ ) = 1 3 csc(θ ) = 2 3 3 θ 1 − 1 3 sec Problem. Compute sin . − 4 Solution. Construct the following triangle. 7 θ −3 4 Then sec(θ) = 4 7 Problem. Complete the identity tan cos −1.(x) = ? Solution. Construct the following triangle. 1 1− x 2 θ Then tan(θ) = x 1− x 2 x Problem. Find the derivative of cos−1(2 x +.1) Solution. d cos −1(2 x +1) =− dx 1 1− (2 x +1) d (2x +1) =− 2 dx 2 1−(2 x +1)2 Problem. Find the derivative of tan−1( x .) Solution. d tan −1( x ) = 1 d ( x)= 1 1 = 1 dx (1+ x) 2 x 2 x (1+ x) (1+ ( x ) 2 ) dx Problem. Find the derivative of sec−1(ex) . Solution. d sec−1(e x ) = dx |e x | d ex = 1 x= 1 e dx 2 2 x −1 e x e2x −1 x e e −1 1 ( ) Problem. Find the derivative of ln sin −1( x) . Solution. ( d ln sin −1( x) dx ) 1 d sin −1(x) = 1 1 = − 1 dx sin −1( x) 1− x2 sin (x) Problem. Find the derivative of Solution. d tan −1( x) = dx tan −1( x). 1 d tan −1( x) = 1 1 2 dx − 1 − 1 1 + x 2 tan (x) 2 tan ( x) Problem. Find the derivative of sin−1( x) +.cos−1( x) Solution. Why? d sin −1(x) + cos−1( x) = dx + −1 = 0 1− x 2 1− x 2 1 Because sin−1( x) + cos−1( x) = π 2 3 − 1 Problem. Find the derivative of x2 sin ( x) . ( d x2 sin −1( x) dx ) 3 3 2 − 1 2 − 1 = 2 x sin (x) + x 3 sin ( x) d sin −1(x) dx 3 2 1 − 1 2 − 1 = 2x sin ( x) + x 3 sin ( x) 1− x2 2 − 1 sin (x) 3 − 1 = 2 x sin ( x) +3x2 1− x 2 Solution. ( ) ( Problem. Find the derivative of sin tan −1.(x) Solution. ( ) d sin tan −1(x) dx = cos tan −1( x) d tan −1( x) dx = cos tan −1( x) 1 1+ x2 ) Problem. Find dy by implicit differentiation, if dx . x3 + x tan −1( y) = e y Solution. 3x2 + tan−1( y) + x dy = e y dy dx 1+ y 2 dx so x dy − e y dy =−3x2 − tan−1( y) dx 1+ y 2 dx x dy y − e =−3x 2 − tan −1( y ) 2 1 + y dx dy = −3x2 − tan −1( y) dx x y −e 1+ y 2