Download MA 2231 Lecture 15 - Inverse Trig Functions Wednesday, March 1

MA 2231 Lecture 15 - Inverse Trig Functions
Wednesday, March 1, 2017
Objectives: Discuss, derive, introduce derivatives of inverse trig functions.
Inverse Trig Functions
For any function or operation we have, we would like to have a function or operation that does the opposite
(that undoes it), like squares and square roots, addition and subtraction, exponentials and logs, etc.
Functions are always defined so that each input has a unique output. This is a critical feature that make
functions such a central object in the study of mathematics. As a result, most functions can’t have an exact
inverse function. It’s generally sufficient, however, to define our inverse functions over a restricted range,
and mathematicians have mostly settled on standard ranges for each case.
As an example, the sine function is periodic. Its inverse function, the inverse sine function is written
arcsin(x) or sin−1 (x) (Remember: A −1 in this context indicates inverse function, and any other number is
an exponent.). We want
(1)
sin(arcsin(x)) = x and arcsin(sin(x)) = x,
but to make this happen, we would need, for example, arcsin(0) = 0 and arcsin(0) = π and arcsin(0) = 2π
and etc. This would make the expression arcsin(0) not really anything in particular. It’s better to just pick
one of these values, and go with it.
Therefore, the inverse sine function is defined to satisfy
sin(arcsin(x)) = x and arcsin(sin(x)) = x, − π2 ≤ arcsin(x) ≤
(2)
π
2
.
All the other inverse trig functions are defined similarly over the ranges given below.
inverse
inverse
inverse
inverse
inverse
inverse
sine
cosine
tangent
cotangent
secant
cosecant
arcsin(x) = sin−1 (x)
arccos(x) = cos−1 (x)
arctan(x) = tan−1 (x)
arccot(x) = cot−1 (x)
arcsec(x) = sec −1 (x)
arccsc(x) = csc −1 (x)
− π2 ≤ arcsin(x) ≤
0 ≤ arccos(x) ≤ π
− π2 ≤ x ≤ π2
0 ≤ arctan(x) ≤ π
0 ≤ arcsec(x) ≤ π
− π2 ≤ arccsc(x) ≤
π
2
π
2
We won’t be terribly concerned with these ranges, because we’ll use the inverse trig functions more for their
derivative patterns than for trigonometry. You should know the ranges exist, and check them when it’s time
to put specific numbers in.
Derivatives of Inverse Trig Functions
With the derivative of any inverse function, the first thing you should think of is the trick we used to get
the derivative of ln(x).
Let’s try it on the inverse sine. We’ll differentiate sin(arcsin(x)) two ways. The easy way:
[ sin(arcsin(x)) ]0 = [ x ]0 = 1.
(3)
Then do it again using the chain rule:
(4)
0
0
[ sin(arcsin(x)) ] = cos(arcsin(x)) · [ arcsin(x) ] .
These two things are the same, and so we have an equation, and solving it gives us
1
0
(5)
[ arcsin(x) ] =
.
cos(arcsin(x))
1
MA 2231 Lecture 15 - Inverse Trig Functions
2
We can use some trigonometry to get this derivative in a more useful form. The trick is to think of arcsin(x)
as an angle in a right triangle. We also know that
opposite
x
(6)
sin(arcsin(x)) =
=x= .
hypotenuse
1
In other words, one right triangle that has arcsin(x) as an angle has opposite side x and hypotenuse 1. The
Pythagorean Theorem then tells us that
adjacent2 + x2 = 12 ,
(7)
and so
(8)
adjacent =
p
1 − x2 .
Finally,
adjacent
cos(arcsin(x)) =
=
hypotenuse
(9)
√
1 − x2 p
= 1 − x2 ,
1
and so
0
(10)
[ arcsin(x) ] =
1
1
.
= √
cos(arcsin(x))
1 − x2
The other inverse trig functions work out the same way.
0
[ arcsin(x) ] = √
1
1 − x2
0
[ arccos(x) ] = √
−1
1 − x2
[ arctan(x) ] =
1
1 + x2
[ arccot(x) ] =
−1
1 + x2
[ arcsec(x) ]0 =
1
√
|x| x2 − 1
[ arccsc(x) ]0 =
−1
√
|x| x2 − 1
0
0
Note that these functions are came up in relation to the trig functions, but their derivatives are algebraic (we
also got x1 from ln(x)). Distances are often computed using the Pythagorean Theorem, and so expressions
√
like 1 − x2 come up relatively often in any situation where position or motion are involved. Applications
in mathematics, especially those that involve calculus in some way, most often involve doing derivatives
backwards, and so these are important formulas. There are some combinations that are missing, and the
hyperbolic functions covered next fill some of those in.
Quiz 15
1.
The expression arctan(x) can represent the angle in some right triangle, and
(11)
tan(arctan(x)) = x =
x
opposite
=
,
1
adjacent
a.
Draw a right triangle with arccos(x) as one of the non-right angles, and label the hypotenuse and adjacent
sides with x and 1, as appropriate.
b.
Use the Pythagorean Theorem to find the length of the remaining side (the opposite).
c.
Find sin(arccos(x)) using
(12)
d.
sin(arctan(x)) =
Find cos(arccos(x)).
opposite
.
hypotenuse
MA 2231 Lecture 15 - Inverse Trig Functions
e.
Find tan(arccos(x)).
f.
Find cot(arccos(x)).
g.
Find sec(arccos(x)).
h.
Find csc(arccos(x)).
3
Homework 15
The expression arctan(x) can represent the angle in some right triangle, and
x
opposite
(13)
tan(arctan(x)) = x = =
,
1
adjacent
1.
a.
Draw a right triangle with arctan(x) as one of the non-right angles, and label the opposite and adjacent
sides with x and 1, as appropriate.
b.
Use the Pythagorean Theorem to find the length of the remaining side (the hypotenuse).
c.
Find sin(arctan(x)) using
(14)
sin(arctan(x)) =
opposite
.
hypotenuse
d.
Find cos(arctan(x)).
e.
Find tan(arctan(x)).
f.
Find cot(arctan(x)).
g.
Find sec(arctan(x)).
h.
Find csc(arctan(x)).
2.
Derive the derivative of arctan(x). Use the same tricks as with the inverse sine and Problem 1.
Answers: 1b)
√
x2 + 1. 1c)
√ x
x2 +1
1g)
√
x2 + 1.
2) Using the chain rule, you would get
(15)
1 = sec2 (arctan(x)) · [ arctan(x) ] .
0