Simplify cos (sin −1 x ) and tan(sin −1 x ).
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sin
is not one-to-one
does not have an . inverse
y x
y
*** Our restricted domains will be the largest interval
(containing q
sin with re
uadrant I) in which is strictly monotoni
stricted domain in ,2 2
c.
x
f
x
,12
, 12
1,
2
1,2
1
12
2
x
y
siny x 1
1
y
x2
2
11
y
x
2
2
sin with
restricted domain
y x1siny x
(with the appropriate restricted domaiSine
is one-to-one and does have an in
n)
verse.
3
2
2 3
2
2
Restricted : /2 2
1
s
/
i
: 1
n x
D x x
R y y
1sin
: / 1 1
: /2 2
D x x
R y y
x
1sin is the unique in , such that sin2 2
y x y x
1,2
1,2
11
y
x
2
2
1siny x
1. . sin 1 sin 12 2
i e
,12
, 12
1
1
y
x2
2
sin with
restricted domain
y x
Solve for xOur restricted domain of sine
x
sin x
0 2 3 4 6 6 4 3 2
1 3
2
2
2
1
2 0
1
2
2
2
3
21
x
1sin x
1 3
2
2
2
1
2 0
1
2
2
2
3
21
2
3
4
6
0
6
4
3
2
,12
, 12
1,
2
1,2
1
12
2
x
ysiny x 1
1
y
x2
2
11
y
x
2
2
sin with
restricted domain
y x1siny x
1sin sin ?4
1 5sin
5Expla siin why n .
4 4
1 5in
4s sin
4
4
1 2
2sin
4
If our initial is not in the restricted domain of sine, we must find an
(within the restricted domain) for which sine has an equivalent output.
1cos is the unique in 0, such that cosy x y x
To compute the derivatives of the inverse trigonometric functions we simply need to simplify composite expressions, such as cos(sin−1 x) and tan(sec−1 x), by referring to the appropriate right triangle.
, 1
0,1 1,
1,0
1
12
2
x
ycosy x 1
1
y
x2
11
y
xcos with
restricted domain
y x1cosy x
cosine of "the whose sine is "xSimplify cos(sin−1 x) and tan(sin−1 x).
-1 2c 1os sin cosx x
-1
2tan sin
1tanx
x
x
2 2 1b x
1
2
1
1
1 1si
sin sin
1 1
coscos s 1in
n
dx
f x x f x g x x
xdx x
1
1
2
1 1cos cos
1 1
sinsin
1cos
1cos
f x x f x g x x
dx
dx x x
THEOREM 1 Derivatives of Arcsine and Arccosine
Derivative of an i1
'v rse'
n e g xf g x
CV
1 1
2 2
1 1sin , cos
1 1
d dx x
dx dxx x
I can't use the same ,
but I can use the same right :-)
Derivatives of Arcsine and Arccosine 1g x f x
QED
Right 's THUse and to find thM 2 . e..
1
2
1sin
1
dx
dx x
If f (x) = arcsin(x2)
1 ' ?
2f
1 2
4
2 1 1sin '
4 4
2 11 116
15
15
1
151
156
d xx f
dx x
or for McNeal...
1tan is the unique in , such that tan2 2
y x y x
1cot is the unique in 0, such that coty x y x
1sec is the unique in [0, ) ( , ] such that sec2 2
y x y x
secy x
x
y y
1secy x
x
1csc is the unique in [ ,0) (0, ] such that csc2 2
y x y x
2
2
1
1
2
2
x
111csc x cscf
y
cscy x
x
y
1cscy xx
THEOREM 2 Derivatives of Inverse Trigonometric Functions
1 12 2
1 1
2 2
1 1tan , cot
1 11 1
sec , csc1 1
d dx x
dx x dx xd d
x xdx dxx x x x
122
3
9 6
1 3tan
1 1 23 1
3
dx
dx x x x
Day 2
THEOREM 2 Derivatives of Inverse Trigonometric Functions
1 12 2
1 1
2 2
1 1tan , cot
1 11 1
sec , csc1 1
d dx x
dx x dx xd d
x xdx dxx x x x
1
20
0
20 0
1csc 1
1 1 1
1 1 21
1
3
x
x
x xx
ede
dx e e
e
e e
Sorry McNeal...
The formulas for the derivatives of the inverse trigonometric functions yield the following integration formulas.
Integral Formulas
1
2
12
1
2
sin1
tan1
sec1
dxx C
xdx
x Cxdx
x Cx x
In this list, we omit the integral formulas corresponding to the derivatives of cos−1 x, cot−1 x, and csc−1 x
... are nothing more than simple transformations
of the antiderivatives shown above.
1
2
1
2
1sin
1
sin1
dx
dx xdt
t Ct
We can use these formulas to express the inverse trigonometric functions as definite integrals. For example, because sin−1 0 = 0, we have:
1
20
sin for 1 11
x dtx x
t
Area model, in terms of .x
0C
11
0tan
40
4x
1
20
?1
dx
x
221
22
1
2
1sec 2 s
12 sec
11
ec 2
2
duu
u u
2 2u x du dx
1
2sec
1
dxx C
x x
Using Substitution1
21/ 2
?4 1
dx
x x
Because of our bounds,
the is not necessary.
4 4
3 3u x du dx
222 16 4
9 16 9 1 3 19 3
x xx
0 0
2 21 1
01
1
31443 1 1
1 1sin 0
4 4 2 8
du du
u u
u
Using Substitution
1
2sin
1
dxx C
x
0
23/ 4
?9 16
dx
x
21
dx
x
Suggested Problems 2 Days
Day 1: 5,7,17,21,27,29,33,37,41,45
Day 2: 59-107 EOO (Use integration techniques discussed
thus far in the class)
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