Inverse trigonometric functions (Sect. 7.6) Review ...
Inverse trigonometric functions (Sect. 7.6)
Today: Derivatives and integrals.
Review: Definitions and properties. Derivatives. Integrals.
Last class: Definitions and properties.
Domains restrictions and inverse trigs. Evaluating inverse trigs at simple values. Few identities for inverse trigs.
Review: Definitions and properties
Remark: On certain domains the trigonometric functions are
invertible.
y
y = sin(x)
y
1
1
y = cos(x)
y y = tan(x)
- / 2
/2 x
0
-1 -1
y
y = csc(x)
y
/ 2
x
y = sec(x)
- / 2
/2 x
y y = cot(x)
1
- / 2
0
-1
/2 x
1
0
/ 2
-1
x
0
/ 2
x
Review: Definitions and properties
Remark: The graph of the inverse function is a reflection of the
original function graph about the y = x axis.
y y = arcsin(x)
/ 2
y y = arccos(x)
y y = arctan(x)
/ 2
-1 - / 2
1x
y y = arccsc(x)
/ 2
-1 0 1
x
- / 2
/ 2
-1
0
1x
y y = arcsec(x)
/ 2
-1 0 1
x
x
- / 2
y y = arccot(x)
/ 2
0
x
Review: Definitions and properties
Theorem
For all x [-1, 1] the following identities hold,
arccos(x) + arccos(-x) = ,
arccos(x) + arcsin(x) = .
2
Proof:
y
1
arccos(-x)
y
1
arcsin(x)
arccos(x)
-
-x = cos(-)
x = cos ()
x
x = sin(/2-) /2 -
x = cos ()
arccos(x)
x
Review: Definitions and properties
Theorem
For all x [-1, 1] the following identities hold,
arcsin(-x) = - arcsin(x), arctan(-x) = - arctan(x), arccsc(-x) = -arccsc(x).
Proof:
y y = arcsin(x)
/ 2
y y = arctan(x)
/ 2
y y = arccsc(x)
/ 2
-1 - / 2
1x
- / 2
x
-1 0 1
x
- / 2
Inverse trigonometric functions (Sect. 7.6)
Today: Derivatives and integrals.
Review: Definitions and properties. Derivatives. Integrals.
Derivatives of inverse trigonometric functions
Remark: Derivatives inverse functions can be computed with
f -1 (x) =
1 .
f f -1(x)
Theorem
1
The derivative of arcsin is given by arcsin (x) =
.
1 - x2
Proof: For x [-1, 1] holds
1
1
arcsin (x) =
=
sin arcsin(x) cos arcsin(x)
For x [-1, 1] we get arcsin(x) = y , , and the cosine is
22 positive in that interval, then cos(y ) = + 1 - sin2(y ), hence
1
1
arcsin (x) =
arcsin (x) =
.
1 - sin2 arcsin(x)
1 - x2
Derivatives of inverse trigonometric functions
Theorem
The derivative of inverse trigonometric functions are:
1
arcsin (x) =
,
1 - x2
1 arctan (x) = 1 + x2 ,
1
arcsec (x) =
,
|x| x2 - 1
1
arccos (x) = -
,
1 - x2
1 arccot (x) = - 1 + x2 ,
1
arccsc (x) = -
,
|x| x2 - 1
|x| 1, x R,
|x| 1.
Proof: arctan (x) =
tan
1 arctan(x )
,
cos2(y ) + sin2(y )
tan (y ) =
cos2(y )
tan (y ) = 1 + tan2(y ), y = arctan(x),
1 arctan (x) = 1 + x2 .
Derivatives of inverse trigonometric functions
Proof: arcsec (x) =
1 , for |x| 1.
sec arcsec(x)
Then y = arcsec(x) satisfies y [0, ] - {/2}. Recall,
1 sec (y ) =
cos(y )
sin(y ) = cos2(y ) ,
sin(y ) = + 1 - cos2(y ),
1 - cos2(y )
1
1 - cos2(y )
sec (y ) =
cos2(y )
= | cos(y )|
, | cos(y )|
1 sec (y ) =
| cos(y )|
1 cos2(y ) - 1 = | sec(y )|
sec2(y ) - 1.
1
We conclude: arcsec (x) =
.
|x| x2 - 1
Derivatives of inverse trigonometric functions
Example
Compute the derivative of y (x) = arcsec(3x + 7).
1
Solution: Recall the main formula: arcsec (u) =
.
|u| u2 - 1
3
Then, chain rule implies, y (x) =
.
|3x + 7| (3x + 7)2 - 1
Example
Compute the derivative of y (x) = arctan(4 ln(x)).
1 Solution: Recall the main formula: arctan (u) = 1 + u2 . Therefore, chain rule implies,
1
4
4
y (x) = 1 + 4 ln(x) 2 x
y
= x
1 + 16 ln2(x)
.
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