Sign up Log in
Inverse Trigonometric Identities
Quiz
Inverse Trigonometric Identities
Relevant For...
Geometry>
Sum and Difference Trigonometric Formulas
Omkar Kulkarni, Pranjal Jain, Jimin Khim, and 1 other contributed
Before reading this, make sure you are familiar with inverse trigonometric functions.
The following inverse trigonometric identities give an angle in different ratios. Before the more complicated identities come some seemingly obvious ones. Be observant of the conditions the identities call for.
−
1
(
−
x
)
=
−
−
1
x
,
∣
x
∣
≤
1
−
1
(
−
x
)
=
π
−
−
1
x
,
∣
x
∣
≤
1
−
1
(
−
x
)
=
−
−
1
x
,
x
∈
R
−
1
(
−
x
)
=
π
−
−
1
x
,
x
∈
R
−
1
x
=
−
1
(
1
x
)
,
∣
x
∣
≥
1
−
1
x
=
−
1
(
1
x
)
,
∣
x
∣
≥
1
−
1
x
=
−
1
(
1
x
)
,
x
>
0
−
1
x
=
π
+
−
1
(
1
x
)
,
x
<
0
−
1
x
+
−
1
x
=
π
2
,
∣
x
∣
≤
1
−
1
x
+
−
1
x
=
π
2
,
∣
x
∣
≥
1
sin
−1
(−x)
cos
−1
(−x)
tan
−1
(−x)
cot
−1
(−x)
csc
−1
x
sec
−1
x
cot
−1
x
cot
−1
x
sin
−1
x+cos
−1
x
csc
−1
x+sec
−1
x
=−sin
−1
x,
=π−cos
−1
x,
=−tan
−1
x,
=π−cot
−1
x,
=sin
−1
(
x
1
),
=cos
−1
(
x
1
),
=tan
−1
(
x
1
),
=π+tan
−1
(
x
1
),
=
2
π
,
=
2
π
,
∣x∣≤1
∣x∣≤1
x∈R
x∈R
∣x∣≥1
∣x∣≥1
x>0
x<0
∣x∣≤1
∣x∣≥1
Now for the more complicated identities. These come handy very often, and can easily be derived using the basic trigonometric identities.
−
1
x
=
−
1
(
1
−
x
2
)
,
x
≥
0
−
1
x
=
−
1
(
1
−
x
2
)
,
x
≥
0
−
1
x
=
π
−
−
1
(
1
−
x
2
)
,
x
<
0
−
1
x
+
−
1
y
=
−
1
(
x
+
y
1
−
x
y
)
,
x
y
<
1
−
1
x
+
−
1
y
=
π
+
−
1
(
x
+
y
1
−
x
y
)
,
x
y
>
1
−
1
x
−
−
1
y
=
−
1
(
x
−
y
1
+
x
y
)
−
1
x
=
−
1
(
x
1
−
x
2
)
,
x
∈
(
0
,
1
)
−
1
x
=
−
1
(
1
−
x
2
x
)
,
x
∈
(
0
,
1
)
−
1
x
=
−
1
(
x
x
2
+
1
)
,
x
>
0
−
1
x
=
−
1
(
1
x
2
+
1
)
,
x
>
0
sin
−1
x
cos
−1
x
cos
−1
x
tan
−1
x+tan
−1
y
tan
−1
x+tan
−1
y
tan
−1
x−tan
−1
y
sin
−1
x
cos
−1
x
tan
−1
x
tan
−1
x
=cos
−1
(
1−x
2
),
=sin
−1
(
1−x
2
),
=π−sin
−1
(
1−x
2
),
=tan
−1
(
1−xy
x+y
),
=π+tan
−1
(
1−xy
x+y
),
=tan
−1
(
1+xy
x−y
)
=tan
−1
(
1−x
2
x
),
=tan
−1
(
x
1−x
2
),
=sin
−1
(
x
2
+1
x
),
=cos
−1
(
x
2
+1
1
),
x≥0
x≥0
x<0
xy<1
xy>1
x∈(0,1)
x∈(0,1)
x>0
x>0
Find the value of
x
x for which
sin
(
−
1
(
1
+
x
)
)
=
cos
(
−
1
(
x
)
)
.
sin(cot
−1
(1+x))=cos(tan
−1
(x)).
We have
−
1
(
1
+
x
)
=
−
1
(
1
1
+
(
1
+
x
)
2
)
−
1
x
=
−
1
(
1
x
2
−
1
)
.
cot
−1
(1+x)
tan
−1
x
=sin
−1
(
1+(1+x)
2
1
)
=cos
−1
(
x
2
−1
1
).
Therefore, we have
sin
(
−
1
(
1
+
x
)
)
=
cos
(
−
1
(
x
)
)
sin
(
−
1
(
1
1
+
(
1
+
x
)
2
)
)
=
cos
(
−
1
(
1
x
2
−
1
)
)
1
1
+
(
1
+
x
)
2
=
1
x
2
+
1
x
2
+
1
=
(
x
+
1
)
2
+
1
x
2
+
2
x
+
1
=
x
2
x
=
−
1
2
.
□
sin(cot
−1
(1+x))
sin(sin
−1
(
1+(1+x)
2
1
))
1+(1+x)
2
1
x
2
+1
x
2
+2x+1
x
=cos(tan
−1
(x))
=cos(cos
−1
(
x
2
−1
1
))
=
x
2
+1
1
=(x+1)
2
+1
=x
2
=−
2
1
.
□
Cite as: Inverse Trigonometric Identities. Brilliant.org. Retrieved 22:45, February 27, 2020, from https://brilliant.org/wiki/inverse-trigonometric-identities/
Get more Brilliant. Sign upSign up Log in
Inverse Trigonometric Identities
Quiz
Inverse Trigonometric Identities
Relevant For...
Geometry>
Sum and Difference Trigonometric Formulas
Omkar Kulkarni, Pranjal Jain, Jimin Khim, and 1 other contributed
Before reading this, make sure you are familiar with inverse trigonometric functions.
The following inverse trigonometric identities give an angle in different ratios. Before the more complicated identities come some seemingly obvious ones. Be observant of the conditions the identities call for.
−
1
(
−
x
)
=
−
−
1
x
,
∣
x
∣
≤
1
−
1
(
−
x
)
=
π
−
−
1
x
,
∣
x
∣
≤
1
−
1
(
−
x
)
=
−
−
1
x
,
x
∈
R
−
1
(
−
x
)
=
π
−
−
1
x
,
x
∈
R
−
1
x
=
−
1
(
1
x
)
,
∣
x
∣
≥
1
−
1
x
=
−
1
(
1
x
)
,
∣
x
∣
≥
1
−
1
x
=
−
1
(
1
x
)
,
x
>
0
−
1
x
=
π
+
−
1
(
1
x
)
,
x
<
0
−
1
x
+
−
1
x
=
π
2
,
∣
x
∣
≤
1
−
1
x
+
−
1
x
=
π
2
,
∣
x
∣
≥
1
sin
−1
(−x)
cos
−1
(−x)
tan
−1
(−x)
cot
−1
(−x)
csc
−1
x
sec
−1
x
cot
−1
x
cot
−1
x
sin
−1
x+cos
−1
x
csc
−1
x+sec
−1
x
=−sin
−1
x,
=π−cos
−1
x,
=−tan
−1
x,
=π−cot
−1
x,
=sin
−1
(
x
1
),
=cos
−1
(
x
1
),
=tan
−1
(
x
1
),
=π+tan
−1
(
x
1
),
=
2
π
,
=
2
π
,
∣x∣≤1
∣x∣≤1
x∈R
x∈R
∣x∣≥1
∣x∣≥1
x>0
x<0
∣x∣≤1
∣x∣≥1
Now for the more complicated identities. These come handy very often, and can easily be derived using the basic trigonometric identities.
−
1
x
=
−
1
(
1
−
x
2
)
,
x
≥
0
−
1
x
=
−
1
(
1
−
x
2
)
,
x
≥
0
−
1
x
=
π
−
−
1
(
1
−
x
2
)
,
x
<
0
−
1
x
+
−
1
y
=
−
1
(
x
+
y
1
−
x
y
)
,
x
y
<
1
−
1
x
+
−
1
y
=
π
+
−
1
(
x
+
y
1
−
x
y
)
,
x
y
>
1
−
1
x
−
−
1
y
=
−
1
(
x
−
y
1
+
x
y
)
−
1
x
=
−
1
(
x
1
−
x
2
)
,
x
∈
(
0
,
1
)
−
1
x
=
−
1
(
1
−
x
2
x
)
,
x
∈
(
0
,
1
)
−
1
x
=
−
1
(
x
x
2
+
1
)
,
x
>
0
−
1
x
=
−
1
(
1
x
2
+
1
)
,
x
>
0
sin
−1
x
cos
−1
x
cos
−1
x
tan
−1
x+tan
−1
y
tan
−1
x+tan
−1
y
tan
−1
x−tan
−1
y
sin
−1
x
cos
−1
x
tan
−1
x
tan
−1
x
=cos
−1
(
1−x
2
),
=sin
−1
(
1−x
2
),
=π−sin
−1
(
1−x
2
),
=tan
−1
(
1−xy
x+y
),
=π+tan
−1
(
1−xy
x+y
),
=tan
−1
(
1+xy
x−y
)
=tan
−1
(
1−x
2
x
),
=tan
−1
(
x
1−x
2
),
=sin
−1
(
x
2
+1
x
),
=cos
−1
(
x
2
+1
1
),
x≥0
x≥0
x<0
xy<1
xy>1
x∈(0,1)
x∈(0,1)
x>0
x>0
Find the value of
x
x for which
sin
(
−
1
(
1
+
x
)
)
=
cos
(
−
1
(
x
)
)
.
sin(cot
−1
(1+x))=cos(tan
−1
(x)).
We have
−
1
(
1
+
x
)
=
−
1
(
1
1
+
(
1
+
x
)
2
)
−
1
x
=
−
1
(
1
x
2
−
1
)
.
cot
−1
(1+x)
tan
−1
x
=sin
−1
(
1+(1+x)
2
1
)
=cos
−1
(
x
2
−1
1
).
Therefore, we have
sin
(
−
1
(
1
+
x
)
)
=
cos
(
−
1
(
x
)
)
sin
(
−
1
(
1
1
+
(
1
+
x
)
2
)
)
=
cos
(
−
1
(
1
x
2
−
1
)
)
1
1
+
(
1
+
x
)
2
=
1
x
2
+
1
x
2
+
1
=
(
x
+
1
)
2
+
1
x
2
+
2
x
+
1
=
x
2
x
=
−
1
2
.
□
sin(cot
−1
(1+x))
sin(sin
−1
(
1+(1+x)
2
1
))
1+(1+x)
2
1
x
2
+1
x
2
+2x+1
x
=cos(tan
−1
(x))
=cos(cos
−1
(
x
2
−1
1
))
=
x
2
+1
1
=(x+1)
2
+1
=x
2
=−
2
1
.
□
Cite as: Inverse Trigonometric Identities. Brilliant.org. Retrieved 22:45, February 27, 2020, from https://brilliant.org/wiki/inverse-trigonometric-identities/
Get more Brilliant. Sign up
