Inverse Trigonometric Functions and Their Derivatives None of the - PowerPoint PPT Presentation

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 ) ≤ ≤ π 0 x π ≠ x 2 sec − 1 ( x ) = arcsec( x )

Evaluating inverse trigonometric functions − = 1 sin ( ) The equation is equivalent with the equation y x = Thus sin − 1 ( x ) should be thought of as the angle sin( ). x y 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 π 2 − . 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 . π 1. Let y = sin − 1 ( − 1). Then sin( y ) = − 1, so =− y 2 y π 2. Let y = cos − 1 (1/2). Then cos( y ) = 1/2, so = 3 We can also find identities involving the inverse trigonometric functions, by using ordinary trigonometric identities. 2 Example . Show that cos(sin − 1 ( x )) = − 1 x 1 x The result is obvious − 1 sin ( ) x from the diagram: 2 − 1 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. π 5 1 For example, if x = then sin( x ) = . But the angle in 4 2 π π π π 5 − ≤ ≤ the range having that sine is not . x 4 4 2 2 All such expressions ( inversetrigfn ( trigfn ) = ) must be treated with care.

Derivatives of Inverse Trigonometric Functions −  = Theorem . 1 1 d sin ( ) x     dx 2 − 1 x − = = 1 Proof . Let so sin( ) sin ( ) . Then y x y x dy = cos( ) 1 y dx 1 1 1 dy = = = cos( ) dx y 2 2 − − 1 sin ( ) 1 y x

In a similar way we can show that − −  = 1 1 d cos ( ) x     dx 2 − 1 x − Theorem .  = 1 1 d tan ( ) x     2 dx + 1 x − = = 1 Proof . Let so tan( ) tan ( ) . Then y x y x dy = 2 sec ( ) 1 y dx 1 1 1 dy = = = 2 2 2 + + dx sec ( ) 1 tan ( ) 1 y y x

−  = Theorem . 1 1 d sec ( ) x     dx 2 − | | 1 x x − = = 1 Proof . Let so sec( ) sec ( ) . Then y x y x dy = sec( )tan( ) 1 y y dx 1 cos( ) dy y = = cos( ) y x 2 1 x − sec( )tan( ) sin( ) dx y y y y 2 cos ( ) 1 1 y x = = = 1 2 sin( ) y 2 2 − − x 1 | | 1 x x x

We have now shown the following rules for differentiation. Basic Rule Generalized Rule −  = − 1 1  = d du 1 1 d sin ( ) u 1. sin ( ) x         dx 2 dx − dx 2 − 1 u 1 x −   = −  = 1 1 d du 1 1 d tan ( ) 2. tan ( ) u x     2   2   dx + dx + dx 1 u 1 x −  = −  = 1 1 d du 1 1 d sec ( ) u 3. sec ( ) x         dx 2 dx dx 2 − − | | 1 u u | | 1 x x The other three inverse trig functions have derivatives that are the negatives of their respective cofunctions.

Miscellaneous Problems   1 − θ = 1 cos Problem . Suppose that . Find the exact values   2   θ θ θ θ θ sin( ),tan( ),cot( ),sec( ),csc( ). of Solution . Construct the following triangle. 1 2 θ so that cos( ) = 3 2 θ 1

The other functions can be read directly from that triangle. 3 θ = sin( ) 2 2 3 1 θ = θ = θ tan( ) 3 cot( ) 3 1 2 θ = θ = sec( ) 2 csc( ) 3

    − Problem . Compute . 3 1 −   sec sin     4   Solution . Construct the following triangle. 7 θ − 3 4 Then sec( θ ) = 4 7

−  = 1 tan cos ( ) ? Problem . Complete the identity . x     Solution . Construct the following triangle. 1 2 − 1 x 2 − 1 x Then tan( θ ) = θ x x

− + 1 cos (2 1) Problem . Find the derivative of . x Solution . −   1 1 2 d d + =− + =− cos (2 1) (2 1) x x     dx 2 dx 2 − + − + 1 (2 1) 1 (2 1) x x − 1 tan ( ) Problem . Find the derivative of . x Solution . −  = 1 1 1 1 1 d d = = tan ( ) ( ) x x   + +   2 + (1 )2 dx dx x 2 (1 ) x x x (1 ( ) ) x

− 1 x Problem . Find the derivative of . sec ( ) e Solution . −     1 1 1 1 d x d x = = = x sec ( ) e e e     ( )     dx 2 dx 2 2 x x x − − 1 1 x x e e e − | | 1 e e −   1 Problem . Find the derivative of . ln sin ( ) x     Solution . ( ) − −     1 1 1 1 1 d d = = ln sin ( ) sin ( ) x x     − −   1 1   dx dx 2 − sin ( ) sin ( ) 1 x x x

− 1 Problem . Find the derivative of . tan ( ) x Solution .   − −   1 1 1 1 1 d d = = tan ( ) tan ( ) x x     − −   2   + dx 1 dx 1 1 x 2 tan ( ) 2 tan ( ) x x − − + 1 1 Problem . Find the derivative of . sin ( ) cos ( ) x x Solution . − − −   1 1 1 1 d + = + = sin ( ) cos ( ) 0 x x     dx 2 2 − − 1 1 x x π − − + = 1 1 Why? Because sin ( ) cos ( ) x x 2

3 −   1 2 sin Problem . Find the derivative of . ( ) x x     ( )   3 Solution . − 2 1 d   sin ( ) x x   dx   3 2 − − −         1 2 1 1 d = + 2 sin ( ) 3 sin ( ) sin ( ) x x x x x                 dx 3 2 − −       1 2 1 1 = + 2 sin ( ) 3 sin ( ) x x x x             2 − 1 x ( ) 2 − 1 sin ( ) 3 x −   1 = + 2 2 sin ( ) 3 x x x     2 − 1 x

( ) − 1 Problem . Find the derivative of . sin tan ( ) x Solution . ( ) −   1 d sin tan ( ) x     dx − −     1 1 d = cos tan ( ) tan ( ) x x         dx −   1 1 = cos tan ( ) x   2   + 1 x

dy Problem . Find by implicit differentiation, if dx . − + = y 3 1 tan ( ) x x y e − x dy dy + + = y Solution . 2 1 3 tan ( ) x y e 2 + dx dx 1 y − x dy dy − y =− − 2 1 so 3 tan ( ) e x y 2 + dx dx 1 y   x − y dy − =− − 2 1 3 tan ( )  e  x y 2 + dx   1 y − 2 1 − − 3 tan ( ) dy x y =   dx x y −   e 2   +   1 y