Download Inverse Trigonometric Functions

Inverse Trigonometric Functions
The trigonometric functions are not one-to-one. By restricting their domains, we can construct one-to-one functions from
them.
For example,
π π
we have a oneif we restrict the domain of sin x to the interval − ,
2 2
to-one function which has an inverse denoted by arcsin x or sin−1 x .
Similarly if we restrict the domain of cos x to the interval [0, π ] we have a
one-to-one function which has an inverse denoted by arccos x or cos−1 x
If we restrict the domain of tan x to the interval (− π2 , π2 ) we have a oneto-one function which has an inverse denoted by arctan x or tan−1 x .
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The inverses of the other three trigonometric functions are not often
used, but are defined similarly.
We always have the Cancellation Laws, which only hold on the appropriate domains:
sin(sin−1 x) = x, sin−1 (sin x) = x
cos(cos−1 x) = x, cos−1 (cos x) = x
tan(tan−1 x) = x, tan−1 (tan x) = x
sec(sec−1 x) = x, sec−1 (sec x) = x
csc(csc−1 x) = x, csc−1 (csc x) = x
cot(cot−1 x) = x, cot−1 (cot x) = x
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Derivatives of Inverse Trig Functions
By differentiating the first Cancellation Law for each trig function, and
using trigonometric identities we get a differentiation rule for its inverse:
For example:
−1
d sin sin x
d(x)
so:
dx
dx
−1
d sin x
−1
= 1 and therefore
cos sin x
dx
−1
d sin x
1
1
=√
=
dx
1 − x2
cos sin−1 x
r
=
(Remember that cos sin
−1
x =
−1
1 − sin sin
3
x
2
=
√
1 − x 2)
We list the standard differentiation rules for the six inverse trig functions.
The first three should be memorized, and the student should practice
deriving them all from first principles as done above.
−1
d sin x
1
=√
dx
1 − x2
d cos−1 x
1
= −√
dx
1 − x2
1
d tan−1 x
=
dx
1 + x2
d sec−1 x
1
= √
dx
x x2 − 1
d csc−1 x
1
=− √
dx
x x2 − 1
−1
d cot x
1
=−
dx
1 + x2
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Remember that these formulas are only valid when the domains are as
in the definition of the inverse.
The differentiation rules for the inverse trig functions give us a whole new
class of integration formulas, which need not be memorized, because we
will soon get into the technique of trigonometric substitution which can
be used to easily derive these formulas:
Z
dx
√
= sin−1 x + C = − cos−1 x + C
2
1−x
Z
dx
= tan−1 x + C = cot−1 x + C
2
1+x
Z
dx
√
= sec−1 x + C = − csc x + C =
x x2 − 1
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Example:Problem 6.7-66(p.412 of the Brown Stewart
A painting in an art gallery has height h and is hung so that its lower
edge is a distance d above the eye of an observer. How far from the wall
should the observer stand so as to maximize the angle θ subtended at
his eye by the painting?
h
θ
d
α
x
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Solution 1: (Using Inverse Trig Functions)
Variables:
x = observer’s distance from the wall
α = angle between the horizontal and
the bottom of the painting
θ = angle between the top and bottom
of the painting
Relations:
α = arctan
d
x
d+h
x
d+h
d
θ(x) = arctan
− arctan
x
x
θ + α = arctan
We have to find the value of x that will make θ as large as possible, so
we differentiate:
θ 0 (x) =
1+
1
d+h 2
x
d+h
1
d
−
−
2 − 2 =
x2
x
1 + xd
2
2
2
−(d + h)(x + d ) + d x + (d + h)
d
−
+ 2
=−
2
2
2
2
2
x
+
d
x + (d + h)
x + (d + h) (x 2 + d2 )
d+h
−dx 2 − d3 − hx 2 − hd2 + dx 2 + d3 + 2d2 h + dh2
=
2
2
2
2
x + (d + h) (x + d )
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2
=
p
h(d2 + dh − x 2 )
=
0
if
x
=
d(d + h)
2
2
2
2
x + (d + h) (x + d )
Solution 2: (Not Using Inverse Trig Functions)
We use the same variables, but different relations:
Relations:
d
tan α =
x
tan(θ + α) =
d+h
x
Then we have
d
+ tan θ
tan α + tan θ
h+d
d + x tan θ
tan(θ + α) =
= x d
=
or
=
1 − tan α tan θ
x − d tan θ
x
1 − x tan θ
x(d + x tan θ) = (h + d)(x − d tan θ) or
xd + x 2 tan θ = (h + d)x − (h + d)d tan θ or
x 2 tan θ + (h + d)d tan θ = (h + d)x − xd or
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h
i
tan θ x + (h + d)d = xh or
2
tan θ = h
x
x 2 + (h + d)d
We must find the value of x which will make tan θ a maximum. Let
x
, so that tan θ = hf (x)
f (x) = 2
x + (h + d)d
Then
0
f (x) =
x
x 2 + (h + d)d
0
(x 2 + (h + d)d)(x)0 − x(x 2 + (h + d)d)0
=
=
(x 2 + (h + d)d)2
(x 2 + (h + d)d) − x(2x)
(h + d)d − x 2
=
= 0 if
(x 2 + (h + d)d)2
(x 2 + (h + d)d)2
2
p
(h + d)d − x = 0 or x = (h + d)d
This value is known as the geometric mean of d and h + d .
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Example:Problem 6.7-68(p.412 of the Brown Stewart, or 3.6-66(p.234) of
the blue Stewart
A lighthouse is on a small island 3 km away from the nearest point P on a
straight shoreline and its light makes four revolutions per minute. How
fast is the beam of light moving along the shoreline when it is 1 km from
P?
x
θ
3
10
P
Solution 1: (Using Inverse Trig Functions)
Variables:
x = beam’s distance from P
θ = angle between beam of light and
line through the lighthouse and P
Relations:
x
θ = arctan
3
Differentiating, we get
3
1
x0
=
θ =
x 0 , so
2
2
3
9+x
1+ x
0
3
9 + x 2 0 9 + 12 0 10 0 10
80
x =
θ =
θ =
θ =
8π =
π.
3
3
3
3
3
0
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Solution 2: (Not Using Inverse Trig Functions)
We use the relation:
x(t)
, so differentiation gives
3
x 0 (t)
2
0
sec θ(t)θ (t) =
.
3
tan θ(t) =
radians , so
We have θ 0 (t) = 4(2π ) radians
=
8π
min
min
km
x 0 (t) = 3 sec2 θ(t)θ 0 (t) = 24π sec2 θ(t)
.
min
When x = 1, tan θ(t) = 13 , and since sec2 α ≡ tan2 α + 1, we have
sec2 θ(t) = 19 + 1 = 10
9 , so
10 km
80π km
km
x 0 (t) = 24π
=
= 1600π
.
9 min
3 min
hour
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