Inverse trig functions 11/21/2011 Remember: f 1 ( x ) is the - PDF document

Inverse trig functions 11/21/2011 Remember: f � 1 ( x ) is the inverse function of f ( x ) if f � 1 ( y ) = x . y = f ( x ) implies For inverse functions to the trigonometric functions, there are two notations: f � 1 ( x ) f ( x ) sin � 1 ( x ) = arcsin( x ) sin( x ) cos � 1 ( x ) = arccos( x ) cos( x ) tan � 1 ( x ) = arctan( x ) tan( x ) sec � 1 ( x ) = arcsec( x ) sec( x ) csc � 1 ( x ) = arccsc( x ) csc( x ) cot � 1 ( x ) = arccot( x ) cot( x )

In general: arc ( - ) takes in a ratio and spits out an angle: ! # θ " cos( θ ) = a / c so arccos( a / c ) = θ sin( θ ) = b / c so arcsin( b / c ) = θ tan( θ ) = b / a so arctan( b / a ) = θ There are lots of points we know on these functions... Examples: 1. Since sin( π / 2) = 1, we have arcsin(1) = π / 2 2. Since cos( π / 2) = 0, we have arccos(0) = π / 2 3. arccos(1) = √ 4. arcsin( 2 / 2) = 5. arctan(1) =

Domain/range Domain/range y = sin( x ) y = sin( x ) y = arcsin( x ) Domain: − 1 ≤ x ≤ 1 Domain/range Domain/range y = arcsin( x ) y = arcsin( x ) ! /2 ! /2 - - - - "! /2 "! /2 Domain: − 1 ≤ x ≤ 1 Domain: − 1 ≤ x ≤ 1 Range: − π / 2 ≤ y ≤ π / 2

Domain/range Domain/range y = cos( x ) y = cos( x ) y = arccos( x ) Domain: − 1 ≤ x ≤ 1 Domain/range Domain/range y = arccos( x ) y = arccos( x ) ! - ! - 0 0 Domain: − 1 ≤ x ≤ 1 Domain: − 1 ≤ x ≤ 1 Range: 0 ≤ y ≤ π

Domain/range Domain/range y = tan( x ) y = tan( x ) y = arctan( x ) Domain: −∞ ≤ x ≤ ∞ Domain/range Domain/range y = arctan( x ) y = arctan( x ) ! /2 ! /2 - - - - "! /2 "! /2 Domain: −∞ ≤ x ≤ ∞ Domain: −∞ ≤ x ≤ ∞ Range: − π / 2 < y < π / 2

Domain/range Domain/range y = sec( x ) y = sec( x ) y = arcsec( x ) Domain: x ≤ − 1 and 1 ≤ x Domain/range Domain/range y = arcsec( x ) y = arcsec( x ) ! - ! - 0 0 Domain: x ≤ − 1 and 1 ≤ x Domain: x ≤ − 1 and 1 ≤ x Range: 0 ≤ y ≤ π

Domain/range Domain/range y = csc( x ) y = csc( x ) y = arccsc( x ) Domain: x ≤ − 1 and 1 ≤ x Domain/range Domain/range y = arccsc( x ) y = arccsc( x ) ! /2 ! /2 - - - - "! /2 "! /2 Domain: x ≤ − 1 and 1 ≤ x Domain: x ≤ − 1 and 1 ≤ x Range: − π / 2 ≤ y ≤ π / 2

Domain/range Domain/range y = cot( x ) y = cot( x ) y = arccot( x ) Domain: −∞ ≤ x ≤ ∞ Domain/range Domain/range y = arccot( x ) y = arccot( x ) ! - ! - 0 0 Domain: −∞ ≤ x ≤ ∞ Domain: −∞ ≤ x ≤ ∞ Range: 0 < y < π

Derivatives Use implicit di ff erentiation (just like ln( x )). Q. Let y = arcsin( x ). What is dy dx ? If y = arcsin( x ) then x = sin( y ). d Take dx of both sides of x = sin( y ): LHS: d RHS: d dx sin( y ) = cos( y ) dy dx = cos(arcsin( x )) dy dx x = 1 dx So 1 dy dx = cos(arcsin( x )) . Simplifying cos(arcsin( x )) Call arcsin( x ) = θ . sin( θ ) = x 1 ! θ √ 1"#"$ ² ! p 1 − x 2 So cos( arcsin( x )) = d 1 1 So dx arcsin( x ) = cos(arcsin( x )) = 1 − x 2 . √

d Calculate dx arctan( x ). 1. Rewrite y = arctan( x ) as x = tan( y ). 2. Use implicit di ff erentiation and solve for dy dx . 3. Your answer will have sec(arctan( x )) in it. Simplify this expression using ! arctan(x) 1 !

Recall: In general, if y = f � 1 ( x ), then x = f ( y ). So 1 = f 0 ( y ) dy f � 1 ( x ) dx = f 0 � � , and so d 1 dx f � 1 ( x ) = f 0 ( f ( x )) f 0 ( x ) f ( x ) f 0 ( x ) f ( x ) 1 arctan( x ) − sin(arccos( x )) cos( x ) − sin( x ) 1 arcsec( x ) sec( x ) sec( x ) tan( x ) sec(arcsec( x )) tan(arcsec( x )) 1 arccsc( x ) csc( x ) − csc( x ) cot( x ) − csc(arccsc( x )) cot(arccsc( x )) 1 − csc 2 ( x ) arccot( x ) cot( x ) − � 2 � csc(arccot( x )) To simplify, use the triangles 1 ! arccos(x) arcsec(x) ! 1 ! ! ! 1 1 arccsc(x) arccot(x) ! ! !

More examples 1 d Since dx arctan( x ) = 1+ x 2 , we know d 1. dx arctan(ln( x )) = Z 1 2. 1 + x 2 dx = 1 Z 3. (1 + x ) √ x dx =