3/5/2012
Math 103 – Rimmer
3.9Inverse Trig.
Functions
f ( x ) = arcsin x
or f ( x ) = sin −1 x
sin −1 x ≠
1
sin x
arcsin ( sin x ) = x
sin ( arcsin x ) = x
Math 103 – Rimmer
3.9Inverse Trig.
Functions
 −π π 
Definition : y = arcsin x is the number in 
,
for which sin y = x
 2 2 
 3 π
Find arcsin 
 =
3
 2 
3
when plugged into the sine function.
2
π
π
3

Find the angle  b / w − and  on the unit circle that goes with a point with a y − value of
.
2
2
2

Find the angle that gives
 −1  −π
Find arcsin   =
6
 2 
π
π
−1

Find the angle  b / w − and  on the unit circle that goes with a point with a y − value of
.
2
2
2

Find arcsin ( −1) =
−π
2
π
π

Find the angle  b / w − and  on the unit circle that goes with a point with a y − value of − 1.
2
2

1
3/5/2012
What is the derivative of y = arcsin x?
from before:
f ( x ) = sin x
1
′
( f −1 ) ( x ) = f ′ f −1 ( x )
f ′ ( x ) = cos x
(
)
1
( f )′ ( x ) = cos ( arcsin
x)
1
−1
( f )′ ( x ) =
′ x =
( )
(f )
( )( )
this is better written as
1 − sin ( arcsin x )
2
1
−1
−1
f −1 ( x ) = arcsin x
f −1 ′ x = ?
since cos x = 1 − sin 2 x , we have
−1
( f )′ ( x ) =
Math 103 – Rimmer
3.9Inverse Trig.
Functions
remember sin ( arcsin x ) = x
1 − sin ( arcsin x ) 
2
y = arcsin x
1
y′ =
1 − x2
1
1 − x2
Math 103 – Rimmer
3.9Inverse Trig.
Functions
Let y = arcsin ( 3x ) .
 3
Find y ′ 
 .
 6 
y′ =
1
1 − ( 3x )
2
⋅ ( 3)
3⋅
3
=
6
3
2
2
 3
3
3
=
y ′ 
 =
1
3
 6 
1−
4
4
 3 3
y ′ 
 =
 6  1
2
 3
3

 = 4
 2 
1
1
=
2
4
 3
y ′ 
 = 6
 6 
2
3/5/2012
Math 103 – Rimmer
3.9Inverse Trig.
Functions
f ( x ) = arccos x or f ( x ) = cos −1 x
cos −1 x ≠
1
cos x
arccos ( cos x ) = x
cos ( arccos x ) = x
Math 103 – Rimmer
3.9Inverse Trig.
Functions
Definition : y = arccos x is the number in [ 0, π ] for which cos y = x
 −1  2π
Find arccos   =
3
 2 
Find the angle ( b / w 0 and π ) on the unit circle that goes with a point with a x − value of
−1
.
2
 3 π
Find arccos 
 =
6
 2 
Find the angle ( b / w 0 and π ) on the unit circle that goes with a point with a x − value of
3
.
2
Find arccos ( −1) = π
Find the angle ( b / w 0 and π ) on the unit circle that goes with a point with a x − value of − 1.
3
3/5/2012
What is the derivative of y = arccos x?
from before:
f ( x ) = cos x
1
′
f −1 ( x ) =
f ′ ( x ) = − sin x
f ′ f −1 ( x )
( )
(
)
1
( f )′ ( x ) = − sin ( arccos
x)
1
−1
f −1 ( x ) = arccos x
f −1 ′ x = ?
( )( )
since sin x = 1 − cos 2 x , we have
−1
( f )′ ( x ) =
Math 103 – Rimmer
3.9Inverse Trig.
Functions
− 1 − cos ( arccos x )
this is better written as
2
( f )′ ( x ) =
−1
−1
remember cos ( arccos x ) = x
1 −  cos ( arccos x ) 
2
y = arccos x
( f )′ ( x ) =
−1
−1
y′ =
1 − x2
Fall 2010
y′ =
 −1

⋅  x −3/ 2 
2

 1  
1− 

 x
⇒ y′ ( 4 ) =
−1
2
 −1
⋅
1  2 ⋅ 43/ 2
1− 
4
−1
( )
−1
1 − x2
g=
1
= x −1/ 2
x
g′ =
−1 −3/2
x
2
2
 1 
1
since 
 =
x
 x
y′ =
Math 103 – Rimmer
3.9Inverse Trig.
Functions
 −1 
⋅  3/2 
1  2x 
1−
x
−1

3/ 2
−1  −1 
 =
⋅
since 43/ 2 =  2 2  = 8


3  2 ⋅8 

4
=
1
−1  −1 
⋅  =
3  16  8 3
2
y′ ( 4 ) =
1
8 3
4
3/5/2012
f ( x ) = arctan x or f ( x ) = tan x
−1
lim arctan x =
x→− ∞
Math 103 – Rimmer
3.9Inverse Trig.
Functions
−π
2
lim arctan x =
x →∞
1
tan −1 x ≠
tan x
π
2
arctan ( tan x ) = x
tan ( arctan x ) = x
Math 103 – Rimmer
3.9Inverse Trig.
Functions
 −π π 
Definition : y = arctan x is the number in 
,
for which tan y = x
 2 2 
Find arctan ( −1) =
−π
4
tan ( π2 ) = undef
tan ( π3 ) = 3
tan ( π4 ) = 1
Find arctan
( 3 ) = π3
1
3
tan ( π6 ) =
tan ( 0 ) = 0
 −1  −π
Find arctan 
=
6
 3
tan ( −6π ) =
−1
3
tan ( −4π ) = −1
tan ( −3π ) = − 3
tan ( −2π ) = undef
5
3/5/2012
What is the derivative of y = arctan x?
from before:
f ( x ) = tan x
1
′
f −1 ( x ) =
f ′ ( x ) = sec 2 x
f ′ f −1 ( x )
( )
(
)
1
( f )′ ( x ) = sec ( arctan
x)
Math 103 – Rimmer
3.9Inverse Trig.
Functions
f −1 ( x ) = arctan x
f −1 ′ x = ?
( )( )
since sec2 x = 1 + tan 2 x, we have
−1
2
( f )′ ( x ) = 1 + tan (1arctan x )
−1
this is better written as
2
( f )′ ( x ) =
1
−1
1 +  tan ( arctan x ) 
remember tan ( arctan x ) = x
2
y = arctan x
′ x =
( )
(f )
−1
1
1 + x2
Why is the word arc used?
y′ =
1
1 + x2
Math 103 – Rimmer
3.9Inverse Trig.
Functions
arclentgh s = rθ
On the unit circle r = 1,so s = θ
6