10/13/2010
f(x) = sin(x)
y = sin x is graphed below. The restricted
portion is highlighted.
3.5 Inverse Trig Functions.
• The inverse sine function is denoted by sin-1
or arcsin. y = sinx is equivalent to the
equation x = arcsin y and x = sin-1 y.
6
4
2
-10
-5
• In other words, arcsin y is the angle whose
sine is y.
5
-π/2
10
π/2
-2
-4
• Since the sine function is not one-to-one, its
domain must be restricted so that it has an
inverse that is a function.
-6
-8
•
y = cos x is graphed below. The restricted
portion is highlighted. We use this
restricted portion to graph y = cos-1 x..
The graph at the left shows
π
π
y = sin x when − ≤ x ≤
2
2
f(x) = cos(x)
4
2
f(x) = sin-1(x)
(0,1)
4
π/2 ≈ 1.571
2
-1
-5
1
Since the restricted domain of sine is
[-π/2, π/2] and its range is [-1,1], the
domain of arcsine is [-1,1] and its
range is [-π/2, π/2]
-5
((π/2,, 0))
π
0
5
(π,-1)
-2
5
-4
-π/2 ≈ -1.571
-2
-4
Interchange the x- and y- values of the ordered
pairs to graph the inverse function.
y
π
(-1, π)
y = cos-1x or y = arccosx means
x = cos y,for y in [0, π]
y = tan x is graphed below. The restricted
portion is highlighted in green.
f(x) = tan(x)
4
(π/2, 0)
2
(π/4, 1)
(0, 0)
x
-1
o
(1, 0)
-5
5
-π/2
π/2
5
( -π/4, -1)
-2
-4
Restricted domain
(-π/2, π/2)
1
10/13/2010
y = tan-1x or y = arctanx means x= tany, for y in
(-π/2, π/2)
y
DERIVATIVE OF INVERSE SINE FUNCTION
y = sin-1 x. Then
π/2
x = sin y and –π/2≤ y≤ π/2
Differentiate x = siny implicitly with respect to x.
d
d
(x)= siny
dx
dx
(1, π/4)
x
(0, 0)
( 1 -π/4)
(-1,
π/4)
1 = cosy
-π/2
dy
dx
1
dy
=
cos y dx
1
1-sin2 y
SUMMARY OF INVERSE FUNCTIONS
Notation
Restriction
Inverse
Cosine
cos-11 x = y
−
π
2
2
2
1
≤
x
≤
1
r
o
f
x
=
x
1
s
o
c
s
o
c
1
s
o
c
1
≤
x
≤
1
r
o
f
x
=
x
1
n
i
s
︵
π
≤
x
≤
0
r
o
f
x
=
x
s
o
c
π 2
≤
x
≤
π 2
r
o
f
x
=
x
n
i
s
︶
(
︶
and since x = siny,
then x2 = sin2y
︶
)
Ex. 1 Evaluate arctan (-1).
arctan (-1) means the angle in (-π/2, π/2) whose
tangent is -1.
Note: the angle must be in quad 1 or 4.
tangent is negative in quad 4.
Ans. arctan (-1) = -π/4
Note: the angle must be in quad 1 or 4.
sine is negative in quad 4.
Ex. 5 Evaluate tan ⎛⎜ sin−1 ⎛⎜ − 4 ⎞⎟ ⎞⎟
5
⎝
⎝
Note: the angle must be in quads 1 or 2
the side adjacent to θ is
3π ⎞
⎟
4 ⎠
θ
5
3π
We cannott assume that
W
th t the
th answer is
i
just
j t because
b
4
3π
arcsin and sin are inverse functions of each other.
lies
−
4 3
⎞⎞
⎟⎟
⎠⎠
=
⎛
⎜
⎝
1
⎛
⎜
⎝
4 5
-
n
i
s
n
a
t
.
s
n
A
π 4
⎛ 2⎞
arcsin⎜⎜ 2 ⎟⎟ =
⎝
⎠
4
52 − 42 = 3
the tangent of θ is 4/3.
4
4
outside the range of arcsin, which is [-π/2, π/2]. What is
3π
the angle in [-π/2, π/2] that has the same sine as sin ?
⎛
3π
2⎞
=
sin
⎟
⎜⎜
4
2 ⎟⎠
⎝
⎠⎠
The angle θ is in quad 4. Draw the triangle.
Ans. cos-1 (-1/2) = 2π/3
⎝
dy
1
=
dx
1-x 2
Ans. sin-1 (-1/2) = -π/6
Ex. 3 Evaluate cos-1 (-1/2)
Ex. 4 Evaluate arcsin ⎛⎜ sin
then
sin-1 (-1/2) means the angle in [-π/2, π/2] whose sine is -1/2.
n
i
s
︵
1
︵
dy
dx
Ex. 2 Evaluate sin-1 (-1/2).
2
⇔ tan y = x 2
π
< y<
x
+
1
π
n
i
s
tan-1x = y
π
⇔ cos y = x 0 ≤ y ≤ π
−
Inverse
Tangent
≤ y ≤
1
<
x
<
1
-
sin-1 x = y
1
<
x
<
1
-
⇔ y = sin x
Inverse
Sine
Derivative
2
2
x1
x
1 - 1 1
1
-
1. Functi
on
=
note: sin2 y +cos2 y =1
→ cosy = 1-sin2 y
( Ans )
2
10/13/2010
Notation
Restriction
sin-1 x = y
tan-1x = y
cot-1 x = y
< y<
π
2
2
•
d 2
( x − 1) =
dx
1
1− x + 2x 2 − 1
4
• 2x
Ans.
1
=y
1 − ( x − 1)
2
4
Inverse
cotangent
2
1
y' =
=
Inverse
secant
sec-1x
π
1
-
csc-1x = y
2
0≤ y ≤π
−
Inverse
Cosecant
π
≤ y ≤
Ex. 6 Find the derivative.
y = sin-1 (x2 – 1)
x
x 2
2 x
2
Inverse
Tangent
2
2
2
x
21 x
2
1 +
1 x
x
1 +
1
x 1 -x
cos-1 x = y
π
1
<
x
<
1
-
−
Inverse
Cosine
Derivative
1
<
x
<
1
-
Inverse
Sine
2
x
2
x 1 1
1 1 -
1. Function
Find the derivative:
y = xln(arctanx)
Find the derivative:
4
f(x) = x⋅ln(tan-1(x))
3
y = xcos-1x − 1 − x 2
2
1
-4
-2
2
4
-1
6
⎛
y ' = (1)cos-1x + x ⎜ ⎝
-2
⎞ 1
−1/ 2
2
⎟ - 2 (1 − x ) i( −2 x )
1-x ⎠
1
2
-3
Find the derivative:
Find the derivative:
y = csc ( x + 1)
-1
y = cos-1(e 2 x )
1
2
x
2
+
x 4
2 x
-
︶
︵
i2 x
x
4
( x + 1) ( x 2 + 1) 2 − 1
2
1
+
2
x
=
'
y
︵
2
x
2 e
e
2 1
=
'
y
y'=−
x
1
s
o
c
=
'
y
︶
x
n
a
t
c
r
a
2
︶
︵
⎛ x ⎞
⎛ x ⎞
= cos-11x - ⎜
⎟ +⎜
⎟
2
2
⎝ 1-x ⎠
⎝ 1-x ⎠
x
=︵
x
+
1
+
x
n
a
t
c
r
a
n
l
1
⎛
⎞
⎜ 1+x 2 ⎟
y' = (1)ln(arctanx) + x ⎜
⎟
⎜ arctanx ⎟
⎝
⎠
︶
3
10/13/2010
Find the derivative:
y = arctan(arcsin x )
y'=
y' =
1
1 + (arcsin x )2
i
1
1− ( x )
2
i
1 −1/ 2
x
2
1
1 + (arcsin x )2
4