Download Section 3.5 Inverse Trigonometric Functions

Section 3.5 Inverse Trigonometric Functions
f(x) = sin(x)
By inspection, does this function have an inverse?
No, it fails the horizontal line test. (not one-to-one)
Hm...but we have seen things like sin1 x how??
Domain Limitations
[
]
,
, it passes the
2
2
horizontal line test and is locally one-to-one.
For sin(x), if x What does sin1 x look like?
1
So, sin x = y sin y = x if Domain sin1 x = [1, 1 ]
1
Note: sin 1
sin x
y
2
2
[
1
Range sin x = ,
2 2
]
sin
Example:
3 = 23
sin1
3
3
=
2
Must be between , 2
2
Unit Circle
Notation: Sometimes instead of sin1 x = arcsin x
sin
Example:
2
3
=
3
2
sin1
2
3
3
2
Must be between , 2
2
Instead, always sin1
3
=
2
3
Just like other inverse functions: f f 1 x = x
So:
sin sin1 x = x
And:
sin1 sin x = x
Example: Because sin x is continuous and differentiable so is sin1 x . How do we find its
derivative?
Implicitly:
y = sin1 x
The derivative will be
sin y = x
Hence we know
y
2
2
dy
dx
Work on easier function:
d sin y = d x
dx
dx
dy
cos y
=1
dx
dy
1
=
dx
cos y
Here's the tricky part: we know cos y 0 since y
2
Recall:
sin2 y cos 2 y = 1
cos2 y = 1 sin2 y
Take the positive one: cos y = 1sin 2 y
So:
dy
=
dx
1
1sin2 y
But, what is sin y? x (from above)
So:
And
dy
=
dx
1
1 x2
d
[ sin1 x ] =
dx
1
1 x2
provided 1 x 1
Example: Work cos1 x on your own or through this example.
Limit the domain to
So,
y = cos1 x
x [0, ] (now the function locally is one-to-one)
cos y = x
0 y
cos1 x
cos x
1. Domain cos1 x : [-1, 1]
2. Range cos1 x : [0, ]
1
Note: cos x 1
cos x
Find derivative:
y = cos1 x x = cos y
Let
dy
d
cos1 x
=
dx
dx
Start with:
x = cos y
d
d
x =
cos y dx
dx
dy
1 = sin y
dx
dy
1
=
dx
sin y
Tricky part, sign of sine when input 0 y sin y 0
sin2 y cos 2 y = 1
sin2 y = 1cos2 y
sin y = 1cos2 y
sin y = 1 x2
So:
1
dy
=
provided 1 x 1
dx
1 x 2
d
[ cos1 x ] = 1 2
dx
1 x
Formally:
Example: Evaluate cos1
2
3
=
(must be between [0, ])
2
4
Example: Work tan1 x on your own or through this example.
Limit the domain
x ,
2 2
Domain: ,
2 2
Range: , So,
y = tan1 x
tan y = x
Domain tan1 x : , Range tan1 x : ,
2 2
y
2
2
tan1 x
tan x
Find derivative
y = tan1 x x = tan y
dy
d
tan1 x =
Let
dx
dx
d
d
x =
tan y dx
dx
1 = sec2 y
dy
dx
1
dy
=
dx
sec2 y
1 tan2 y = sec2 y
dy
1
=
dx
1 tan2 y
dy
1
=
dx
1 x2
So:
d
[ tan1 x ] = 1 2
dx
1 x
Example: Find an exact value for:
1.
1
=?
2
How to think through:
sin1
1
and
(i.e. quadrant 4 or 1) has sin x = ?
2
2
2
1 1
family sin
=
6
2
6
What angle between Quadrant 1,
2.
sin1 3
=?
2
Where is sin x = 3.
cos1
3 ? Quadrant 4, family sin1 3 = 2
3
2
2
=?
2
What angle between 0 and (i.e. quadrant 1 or 2) has cos x =
family 4
Quadrant 1,
4.
2
2
=
2
4
1
=?
2
1
? quadrant 2,
2
family 3
cos1 1
2
=
2
3
tan1 1 = ?
What angle between and
(i.e. quadrant 4 or 1) has tan x = 1 ?
2
2
family 4
Quadrant 4,
6.
cos1
2 ?
cos1 Where is cos x = 5.
3
tan1 1 = tan1 3 = ?
3
2
tan x = 3 =
1
2
sin x
cos x
Example: Simplify the expression:
cos tan1 x Let
y = tan1 x
Then tan y = x
y
2
2
So, we are trying to find cos y
4
Relationships known:
So
sec2 y = 1 tan 2 y
sec2 y = 1 x 2
sec y = 1 x2
1
1
cos y =
=
sec y
1 x 2
cos tan1 x = cos y =
Hence:
1
1 x 2
Another way (pictorially):
tan y =
opposite
x
= =x
adjacent
1
So cos y =
1
1 x2
Example: Simplify the expression:
sin tan1 x 1
From picture (above) sin tan x =
Example: Find derivatives:
y = tan1 x
Recall:
y' =
d
tan1 x = 1 2
dx
1 x
d
tan1 x dx
[ ]
1
d
=
tan1 x 2
dx
Apply Chain Rule
x
1 x 2
=
1 2
2
d
x2
dx
1 1
=
x
1 x 2 1 x
=
1
1
1
2
1
2 x 1 x
y = sin1 2x1
Recall:
y' =
d
sin1 x =
dx
1
1 x 2
d
sin1 2x 1 dx
Apply Chain Rule
=
1
d
2x 1
dx
12x1
2
=
12x 12
2
H x = 1 x 2 arctan x
d
d
tan1 x tan1 x dx
1 x2 dx
H ' x = 1 x 2 Product Rule
1
= 1 x tan1 x 2x 2
1 x
= 1 2x tan1 x
= 12x arctan x
2
Example: Find derivative and domain of g x = cos1 32x
Domain: cos1 x :
x [1, 1]
So Domain: cos1 32x 32x [1, ] 1
1 32x 1
4 2x 2
2x1
Domain: x [ 1, 2]
Range:
[0, ]
Derivative:
g ' x =
1
d
3 2x
dx
133x 12
=
133x 2
2
=
133x 2
2
132x2 0
Domain:
32x 2 1
32x 1
2x 2
x 1 x [ 1, 2 ]