Download Inverse Trig Functions: For a function to have an inverse, it must be

Inverse Trig Functions:
For a function to have an inverse, it must be one-to-one – that is, it must pass the Horizontal Line test.
The graph of the sine function does not pass this test.
If we restrict our domain to
h π πi
1. On the interval − , ,
2 2
h π πi
2. On the interval − , ,
2 2
h π πi
3. On the interval − , ,
2 2
−
π
π
≤ x ≤ , we have:
2
2
y = sin x is increasing.
y = sin x takes on its full range of values, −1 ≤ sin x ≤ 1.
y = sin x is one-to-one.
h π πi
Hence, on the restricted domain − , , y = sin x has a unique inverse called the inverse sine function,
2 2
denoted by
y = arcsin x
or
y = sin−1 x.
Note: The notation sin−1 x is consistent with the notation f −1 (x); here,
sin−1 x 6=
1
.
sin x
The inverse sine function is defined by
y = arcsin x
if and only if
sin y = x,
h π πi
π
π
where −1 ≤ x ≤ 1 and − ≤ y ≤ . The domain of y = arcsin x is [−1, 1], and the range in − , .
2
2
2 2
Ex: If possible, find the exact value of
√ !
− 2
arcsin
sin−1 (−1)
arcsin 3
2
√
√ !
π
2
π
π
2
π
Since sin −
=−
for − ≤ y ≤ , we have arcsin −
=− .
4
2
2
2
2
4
π
π
π
π
Since sin −
= −1 for − ≤ y ≤ , we have sin−1 (−1) = − .
2
2
2
2
Since −1 ≤ sin x ≤ 1 for all real numbers, arcsin 3 has no solution.
To define the inverse for the cosine function, we restrict our domain to 0 ≤ x ≤ π.
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Definitions of the Inverse Trigonometric Functions:
Function
y = arcsin x = sin x if and only if sin y = x
Domain
−1 ≤ x ≤ 1
y = arccos x = cos−1 x if and only if cos y = x
y = arctan x = tan−1 x if and only if tan y = x
−1 ≤ x ≤ 1
−∞ ≤ x ≤ ∞
−1
Range
π
π
− ≤y≤
2
2
0≤y≤π
π
π
− <y<
2
2
To obtain these graphs, you can either follow the process on page 196 by filling out a table, plotting
those points, and then filling in the curve, or you can reflect the graph of your function (with its restricted
domain) across the!line y = x.
√
√
3
3
5π
5π
5π
Ex: arccos −
=
since cos
=−
and 0 ≤
≤ π.
2
6
6
2
6
√
√ π
π
π
π
π
3 = since tan
= 3 and − ≤ ≤ .
Ex: tan−1
3
3
2
3
2
Inverse Property of Trigonometric Functions
π
π
If −1 ≤ x ≤ 1 and − ≤ y ≤ , then
2
2
sin (arcsin x) = x
and
arcsin (sin y) = y.
and
arccos (cos y) = y.
and
arctan (tan y) = y.
If −1 ≤ x ≤ 1 and 0 ≤ y ≤ π, then
cos (arccos x) = x
If x is a real number and −
π
π
≤ y ≤ , then
2
2
tan (arctan x) = x
Ex: If possible, find the exact value.
cos
−1
3π
cos −
2
1
sin arcsin
2
3π
3π
π
Note that −
is not in the interval [0, π], but the coterminal angle −
+ 2π =
is in the interval.
2
2
2
So, we have
π π
3π
−1
cos
cos −
= cos−1 cos
= .
2
2
2
1
Since −1 ≤ ≤ 1, we have
2
1
1
sin arcsin
= .
2
2
Ex: Find the exact values of
1
tan arccos
3
2
sin arctan −
.
3
2
1
1
Pic. Let u = arccos
so that cos u = . We draw a triangle in the first quadrant since the angle is
3
3
positive. We have hyp = 3 and adj = 1, and
q
q
√
2
2
2
2
2
2
2
(adj) + (opp) = (hyp)
⇒ opp = (hyp) − (adj) = (3) − (1) = 2 2.
So,
√
√
1
opp
2 2
tan arccos
= tan (u) =
=
=2 2
3
adj
1
2
2
Pic. Let v = arctan −
so that tan v = − . We draw our triangle in the fourth quadrant since tan u
3
3
is negative. We have opp = 2 and adj = 3. Thus,
q
p
√
2
2
hyp = (adj) + (opp) = 22 + 32 = 13,
and
2
sin arctan −
3
= sin (v) =
2
opp
= −√
hyp
13
Note that its negative since we are in Quadrant IV.
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Ex: Write sec sin−1 7x , with 0 ≤ x ≤ (to avoid complex numbers), as an algebraic expression.
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Let u = sin−1 7x so that sin u = 7x, where −1 ≤ 7x ≤ 1. We have opp = 7x and hyp = 1. We draw our
7x
opp
=
. Thus,
triangle in Quadrant I with u an acute angle since sin u =
hyp
1
q
q
p
2
2
2
2
2
2
2
(adj) + (opp) = (hyp)
⇒ adj = (hyp) − (opp) = (1) − (7x) = 1 − 49x2 .
Thus,
1
hyp
=√
sec sin−1 7x = sec (u) =
.
adj
1 − 49x2
The remaining inverse trigonometric functions are not used as frequently and are summarized here.
The choice of intervals for y in the definitions of csc−1 and sec−1 is not universally agreed upon. For
instance, some authors use y ∈ [0, π/2) ∪ (π/2, π] in the definition of sec−1 .
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