Collection of proofs of equations involving trigonometric functions
There are several equivalent ways for defining trigonometric functions , and the proofs of the trigonometric identities between them depend on the chosen definition. The oldest and most elementary definitions are based on the geometry of right triangles and the ratio between their sides. The proofs given in this article use these definitions, and thus apply to non-negative angles not greater than a right angle . For greater and negative angles , see Trigonometric functions .
Other definitions, and therefore other proofs are based on the Taylor series of sine and cosine , or on the differential equation
f
″
+
f
=
0
{\displaystyle f''+f=0}
Elementary trigonometric identities [ edit ] Trigonometric functions specify the relationships between side lengths and interior angles of a right triangle. For example, the sine of angle θ is defined as being the length of the opposite side divided by the length of the hypotenuse. The six trigonometric functions are defined for every real number , except, for some of them, for angles that differ from 0 by a multiple of the right angle (90°). Referring to the diagram at the right, the six trigonometric functions of θ are, for angles smaller than the right angle:
sin
θ
=
o
p
p
o
s
i
t
e
h
y
p
o
t
e
n
u
s
e
=
a
h
{\displaystyle \sin \theta ={\frac {\mathrm {opposite} }{\mathrm {hypotenuse} }}={\frac {a}{h}}}
cos
θ
=
a
d
j
a
c
e
n
t
h
y
p
o
t
e
n
u
s
e
=
b
h
{\displaystyle \cos \theta ={\frac {\mathrm {adjacent} }{\mathrm {hypotenuse} }}={\frac {b}{h}}}
tan
θ
=
o
p
p
o
s
i
t
e
a
d
j
a
c
e
n
t
=
a
b
{\displaystyle \tan \theta ={\frac {\mathrm {opposite} }{\mathrm {adjacent} }}={\frac {a}{b}}}
cot
θ
=
a
d
j
a
c
e
n
t
o
p
p
o
s
i
t
e
=
b
a
{\displaystyle \cot \theta ={\frac {\mathrm {adjacent} }{\mathrm {opposite} }}={\frac {b}{a}}}
sec
θ
=
h
y
p
o
t
e
n
u
s
e
a
d
j
a
c
e
n
t
=
h
b
{\displaystyle \sec \theta ={\frac {\mathrm {hypotenuse} }{\mathrm {adjacent} }}={\frac {h}{b}}}
csc
θ
=
h
y
p
o
t
e
n
u
s
e
o
p
p
o
s
i
t
e
=
h
a
{\displaystyle \csc \theta ={\frac {\mathrm {hypotenuse} }{\mathrm {opposite} }}={\frac {h}{a}}}
In the case of angles smaller than a right angle, the following identities are direct consequences of above definitions through the division identity
a
b
=
(
a
h
)
(
b
h
)
.
{\displaystyle {\frac {a}{b}}={\frac {\left({\frac {a}{h}}\right)}{\left({\frac {b}{h}}\right)}}.}
They remain valid for angles greater than 90° and for negative angles.
tan
θ
=
o
p
p
o
s
i
t
e
a
d
j
a
c
e
n
t
=
(
o
p
p
o
s
i
t
e
h
y
p
o
t
e
n
u
s
e
)
(
a
d
j
a
c
e
n
t
h
y
p
o
t
e
n
u
s
e
)
=
sin
θ
cos
θ
{\displaystyle \tan \theta ={\frac {\mathrm {opposite} }{\mathrm {adjacent} }}={\frac {\left({\frac {\mathrm {opposite} }{\mathrm {hypotenuse} }}\right)}{\left({\frac {\mathrm {adjacent} }{\mathrm {hypotenuse} }}\right)}}={\frac {\sin \theta }{\cos \theta }}}
cot
θ
=
a
d
j
a
c
e
n
t
o
p
p
o
s
i
t
e
=
(
a
d
j
a
c
e
n
t
a
d
j
a
c
e
n
t
)
(
o
p
p
o
s
i
t
e
a
d
j
a
c
e
n
t
)
=
1
tan
θ
=
cos
θ
sin
θ
{\displaystyle \cot \theta ={\frac {\mathrm {adjacent} }{\mathrm {opposite} }}={\frac {\left({\frac {\mathrm {adjacent} }{\mathrm {adjacent} }}\right)}{\left({\frac {\mathrm {opposite} }{\mathrm {adjacent} }}\right)}}={\frac {1}{\tan \theta }}={\frac {\cos \theta }{\sin \theta }}}
sec
θ
=
1
cos
θ
=
h
y
p
o
t
e
n
u
s
e
a
d
j
a
c
e
n
t
{\displaystyle \sec \theta ={\frac {1}{\cos \theta }}={\frac {\mathrm {hypotenuse} }{\mathrm {adjacent} }}}
csc
θ
=
1
sin
θ
=
h
y
p
o
t
e
n
u
s
e
o
p
p
o
s
i
t
e
{\displaystyle \csc \theta ={\frac {1}{\sin \theta }}={\frac {\mathrm {hypotenuse} }{\mathrm {opposite} }}}
tan
θ
=
o
p
p
o
s
i
t
e
a
d
j
a
c
e
n
t
=
(
o
p
p
o
s
i
t
e
×
h
y
p
o
t
e
n
u
s
e
o
p
p
o
s
i
t
e
×
a
d
j
a
c
e
n
t
)
(
a
d
j
a
c
e
n
t
×
h
y
p
o
t
e
n
u
s
e
o
p
p
o
s
i
t
e
×
a
d
j
a
c
e
n
t
)
=
(
h
y
p
o
t
e
n
u
s
e
a
d
j
a
c
e
n
t
)
(
h
y
p
o
t
e
n
u
s
e
o
p
p
o
s
i
t
e
)
=
sec
θ
csc
θ
{\displaystyle \tan \theta ={\frac {\mathrm {opposite} }{\mathrm {adjacent} }}={\frac {\left({\frac {\mathrm {opposite} \times \mathrm {hypotenuse} }{\mathrm {opposite} \times \mathrm {adjacent} }}\right)}{\left({\frac {\mathrm {adjacent} \times \mathrm {hypotenuse} }{\mathrm {opposite} \times \mathrm {adjacent} }}\right)}}={\frac {\left({\frac {\mathrm {hypotenuse} }{\mathrm {adjacent} }}\right)}{\left({\frac {\mathrm {hypotenuse} }{\mathrm {opposite} }}\right)}}={\frac {\sec \theta }{\csc \theta }}}
Or
tan
θ
=
sin
θ
cos
θ
=
(
1
csc
θ
)
(
1
sec
θ
)
=
(
csc
θ
sec
θ
csc
θ
)
(
csc
θ
sec
θ
sec
θ
)
=
sec
θ
csc
θ
{\displaystyle \tan \theta ={\frac {\sin \theta }{\cos \theta }}={\frac {\left({\frac {1}{\csc \theta }}\right)}{\left({\frac {1}{\sec \theta }}\right)}}={\frac {\left({\frac {\csc \theta \sec \theta }{\csc \theta }}\right)}{\left({\frac {\csc \theta \sec \theta }{\sec \theta }}\right)}}={\frac {\sec \theta }{\csc \theta }}}
cot
θ
=
csc
θ
sec
θ
{\displaystyle \cot \theta ={\frac {\csc \theta }{\sec \theta }}}
Complementary angle identities [ edit ] Two angles whose sum is π/2 radians (90 degrees) are complementary . In the diagram, the angles at vertices A and B are complementary, so we can exchange a and b, and change θ to π/2 − θ, obtaining:
sin
(
π
/
2
−
θ
)
=
cos
θ
{\displaystyle \sin \left(\pi /2-\theta \right)=\cos \theta }
cos
(
π
/
2
−
θ
)
=
sin
θ
{\displaystyle \cos \left(\pi /2-\theta \right)=\sin \theta }
tan
(
π
/
2
−
θ
)
=
cot
θ
{\displaystyle \tan \left(\pi /2-\theta \right)=\cot \theta }
cot
(
π
/
2
−
θ
)
=
tan
θ
{\displaystyle \cot \left(\pi /2-\theta \right)=\tan \theta }
sec
(
π
/
2
−
θ
)
=
csc
θ
{\displaystyle \sec \left(\pi /2-\theta \right)=\csc \theta }
csc
(
π
/
2
−
θ
)
=
sec
θ
{\displaystyle \csc \left(\pi /2-\theta \right)=\sec \theta }
Pythagorean identities [ edit ] Identity 1:
sin
2
θ
+
cos
2
θ
=
1
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}
The following two results follow from this and the ratio identities. To obtain the first, divide both sides of
sin
2
θ
+
cos
2
θ
=
1
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}
cos
2
θ
{\displaystyle \cos ^{2}\theta }
sin
2
θ
{\displaystyle \sin ^{2}\theta }
tan
2
θ
+
1
=
sec
2
θ
{\displaystyle \tan ^{2}\theta +1\ =\sec ^{2}\theta }
sec
2
θ
−
tan
2
θ
=
1
{\displaystyle \sec ^{2}\theta -\tan ^{2}\theta =1}
Similarly
1
+
cot
2
θ
=
csc
2
θ
{\displaystyle 1\ +\cot ^{2}\theta =\csc ^{2}\theta }
csc
2
θ
−
cot
2
θ
=
1
{\displaystyle \csc ^{2}\theta -\cot ^{2}\theta =1}
Identity 2:
The following accounts for all three reciprocal functions.
csc
2
θ
+
sec
2
θ
−
cot
2
θ
=
2
+
tan
2
θ
{\displaystyle \csc ^{2}\theta +\sec ^{2}\theta -\cot ^{2}\theta =2\ +\tan ^{2}\theta }
Proof 2:
Refer to the triangle diagram above. Note that
a
2
+
b
2
=
h
2
{\displaystyle a^{2}+b^{2}=h^{2}}
Pythagorean theorem .
csc
2
θ
+
sec
2
θ
=
h
2
a
2
+
h
2
b
2
=
a
2
+
b
2
a
2
+
a
2
+
b
2
b
2
=
2
+
b
2
a
2
+
a
2
b
2
{\displaystyle \csc ^{2}\theta +\sec ^{2}\theta ={\frac {h^{2}}{a^{2}}}+{\frac {h^{2}}{b^{2}}}={\frac {a^{2}+b^{2}}{a^{2}}}+{\frac {a^{2}+b^{2}}{b^{2}}}=2\ +{\frac {b^{2}}{a^{2}}}+{\frac {a^{2}}{b^{2}}}}
Substituting with appropriate functions -
2
+
b
2
a
2
+
a
2
b
2
=
2
+
tan
2
θ
+
cot
2
θ
{\displaystyle 2\ +{\frac {b^{2}}{a^{2}}}+{\frac {a^{2}}{b^{2}}}=2\ +\tan ^{2}\theta +\cot ^{2}\theta }
Rearranging gives:
csc
2
θ
+
sec
2
θ
−
cot
2
θ
=
2
+
tan
2
θ
{\displaystyle \csc ^{2}\theta +\sec ^{2}\theta -\cot ^{2}\theta =2\ +\tan ^{2}\theta }
Angle sum identities [ edit ] Illustration of the sum formula. Draw a horizontal line (the x -axis); mark an origin O. Draw a line from O at an angle
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
x -axis is
α
+
β
{\displaystyle \alpha +\beta }
Place P on the line defined by
α
+
β
{\displaystyle \alpha +\beta }
Let PQ be a line perpendicular to line OQ defined by angle
α
{\displaystyle \alpha }
∴
{\displaystyle \therefore }
Let QA be a perpendicular from point A on the x -axis to Q and PB be a perpendicular from point B on the x -axis to P.
∴
{\displaystyle \therefore }
Draw R on PB so that QR is parallel to the x -axis.
Now angle
R
P
Q
=
α
{\displaystyle RPQ=\alpha }
O
Q
A
=
π
2
−
α
{\displaystyle OQA={\frac {\pi }{2}}-\alpha }
R
Q
O
=
α
,
R
Q
P
=
π
2
−
α
{\displaystyle RQO=\alpha ,RQP={\frac {\pi }{2}}-\alpha }
R
P
Q
=
α
{\displaystyle RPQ=\alpha }
R
P
Q
=
π
2
−
R
Q
P
=
π
2
−
(
π
2
−
R
Q
O
)
=
R
Q
O
=
α
{\displaystyle RPQ={\tfrac {\pi }{2}}-RQP={\tfrac {\pi }{2}}-({\tfrac {\pi }{2}}-RQO)=RQO=\alpha }
O
P
=
1
{\displaystyle OP=1}
P
Q
=
sin
β
{\displaystyle PQ=\sin \beta }
O
Q
=
cos
β
{\displaystyle OQ=\cos \beta }
A
Q
O
Q
=
sin
α
{\displaystyle {\frac {AQ}{OQ}}=\sin \alpha }
A
Q
=
sin
α
cos
β
{\displaystyle AQ=\sin \alpha \cos \beta }
P
R
P
Q
=
cos
α
{\displaystyle {\frac {PR}{PQ}}=\cos \alpha }
P
R
=
cos
α
sin
β
{\displaystyle PR=\cos \alpha \sin \beta }
sin
(
α
+
β
)
=
P
B
=
R
B
+
P
R
=
A
Q
+
P
R
=
sin
α
cos
β
+
cos
α
sin
β
{\displaystyle \sin(\alpha +\beta )=PB=RB+PR=AQ+PR=\sin \alpha \cos \beta +\cos \alpha \sin \beta }
By substituting
−
β
{\displaystyle -\beta }
β
{\displaystyle \beta }
reflection identities of even and odd functions , we also get:
sin
(
α
−
β
)
=
sin
α
cos
(
−
β
)
+
cos
α
sin
(
−
β
)
{\displaystyle \sin(\alpha -\beta )=\sin \alpha \cos(-\beta )+\cos \alpha \sin(-\beta )}
sin
(
α
−
β
)
=
sin
α
cos
β
−
cos
α
sin
β
{\displaystyle \sin(\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta }
Using the figure above,
O
P
=
1
{\displaystyle OP=1}
P
Q
=
sin
β
{\displaystyle PQ=\sin \beta }
O
Q
=
cos
β
{\displaystyle OQ=\cos \beta }
O
A
O
Q
=
cos
α
{\displaystyle {\frac {OA}{OQ}}=\cos \alpha }
O
A
=
cos
α
cos
β
{\displaystyle OA=\cos \alpha \cos \beta }
R
Q
P
Q
=
sin
α
{\displaystyle {\frac {RQ}{PQ}}=\sin \alpha }
R
Q
=
sin
α
sin
β
{\displaystyle RQ=\sin \alpha \sin \beta }
cos
(
α
+
β
)
=
O
B
=
O
A
−
B
A
=
O
A
−
R
Q
=
cos
α
cos
β
−
sin
α
sin
β
{\displaystyle \cos(\alpha +\beta )=OB=OA-BA=OA-RQ=\cos \alpha \cos \beta \ -\sin \alpha \sin \beta }
By substituting
−
β
{\displaystyle -\beta }
β
{\displaystyle \beta }
reflection identities of even and odd functions , we also get:
cos
(
α
−
β
)
=
cos
α
cos
(
−
β
)
−
sin
α
sin
(
−
β
)
,
{\displaystyle \cos(\alpha -\beta )=\cos \alpha \cos(-\beta )-\sin \alpha \sin(-\beta ),}
cos
(
α
−
β
)
=
cos
α
cos
β
+
sin
α
sin
β
{\displaystyle \cos(\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta }
Also, using the complementary angle formulae ,
cos
(
α
+
β
)
=
sin
(
π
/
2
−
(
α
+
β
)
)
=
sin
(
(
π
/
2
−
α
)
−
β
)
=
sin
(
π
/
2
−
α
)
cos
β
−
cos
(
π
/
2
−
α
)
sin
β
=
cos
α
cos
β
−
sin
α
sin
β
{\displaystyle {\begin{aligned}\cos(\alpha +\beta )&=\sin \left(\pi /2-(\alpha +\beta )\right)\\&=\sin \left((\pi /2-\alpha )-\beta \right)\\&=\sin \left(\pi /2-\alpha \right)\cos \beta -\cos \left(\pi /2-\alpha \right)\sin \beta \\&=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\\end{aligned}}}
Tangent and cotangent [ edit ] From the sine and cosine formulae, we get
tan
(
α
+
β
)
=
sin
(
α
+
β
)
cos
(
α
+
β
)
=
sin
α
cos
β
+
cos
α
sin
β
cos
α
cos
β
−
sin
α
sin
β
{\displaystyle \tan(\alpha +\beta )={\frac {\sin(\alpha +\beta )}{\cos(\alpha +\beta )}}={\frac {\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta -\sin \alpha \sin \beta }}}
Dividing both numerator and denominator by
cos
α
cos
β
{\displaystyle \cos \alpha \cos \beta }
tan
(
α
+
β
)
=
tan
α
+
tan
β
1
−
tan
α
tan
β
{\displaystyle \tan(\alpha +\beta )={\frac {\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }}}
Subtracting
β
{\displaystyle \beta }
α
{\displaystyle \alpha }
tan
(
−
β
)
=
−
tan
β
{\displaystyle \tan(-\beta )=-\tan \beta }
tan
(
α
−
β
)
=
tan
α
+
tan
(
−
β
)
1
−
tan
α
tan
(
−
β
)
=
tan
α
−
tan
β
1
+
tan
α
tan
β
{\displaystyle \tan(\alpha -\beta )={\frac {\tan \alpha +\tan(-\beta )}{1-\tan \alpha \tan(-\beta )}}={\frac {\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }}}
Similarly, from the sine and cosine formulae, we get
cot
(
α
+
β
)
=
cos
(
α
+
β
)
sin
(
α
+
β
)
=
cos
α
cos
β
−
sin
α
sin
β
sin
α
cos
β
+
cos
α
sin
β
{\displaystyle \cot(\alpha +\beta )={\frac {\cos(\alpha +\beta )}{\sin(\alpha +\beta )}}={\frac {\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\sin \alpha \cos \beta +\cos \alpha \sin \beta }}}
Then by dividing both numerator and denominator by
sin
α
sin
β
{\displaystyle \sin \alpha \sin \beta }
cot
(
α
+
β
)
=
cot
α
cot
β
−
1
cot
α
+
cot
β
{\displaystyle \cot(\alpha +\beta )={\frac {\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }}}
Or, using
cot
θ
=
1
tan
θ
{\displaystyle \cot \theta ={\frac {1}{\tan \theta }}}
cot
(
α
+
β
)
=
1
−
tan
α
tan
β
tan
α
+
tan
β
=
1
tan
α
tan
β
−
1
1
tan
α
+
1
tan
β
=
cot
α
cot
β
−
1
cot
α
+
cot
β
{\displaystyle \cot(\alpha +\beta )={\frac {1-\tan \alpha \tan \beta }{\tan \alpha +\tan \beta }}={\frac {{\frac {1}{\tan \alpha \tan \beta }}-1}{{\frac {1}{\tan \alpha }}+{\frac {1}{\tan \beta }}}}={\frac {\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }}}
Using
cot
(
−
β
)
=
−
cot
β
{\displaystyle \cot(-\beta )=-\cot \beta }
cot
(
α
−
β
)
=
cot
α
cot
(
−
β
)
−
1
cot
α
+
cot
(
−
β
)
=
cot
α
cot
β
+
1
cot
β
−
cot
α
{\displaystyle \cot(\alpha -\beta )={\frac {\cot \alpha \cot(-\beta )-1}{\cot \alpha +\cot(-\beta )}}={\frac {\cot \alpha \cot \beta +1}{\cot \beta -\cot \alpha }}}
Double-angle identities [ edit ] From the angle sum identities, we get
sin
(
2
θ
)
=
2
sin
θ
cos
θ
{\displaystyle \sin(2\theta )=2\sin \theta \cos \theta }
and
cos
(
2
θ
)
=
cos
2
θ
−
sin
2
θ
{\displaystyle \cos(2\theta )=\cos ^{2}\theta -\sin ^{2}\theta }
The Pythagorean identities give the two alternative forms for the latter of these:
cos
(
2
θ
)
=
2
cos
2
θ
−
1
{\displaystyle \cos(2\theta )=2\cos ^{2}\theta -1}
cos
(
2
θ
)
=
1
−
2
sin
2
θ
{\displaystyle \cos(2\theta )=1-2\sin ^{2}\theta }
The angle sum identities also give
tan
(
2
θ
)
=
2
tan
θ
1
−
tan
2
θ
=
2
cot
θ
−
tan
θ
{\displaystyle \tan(2\theta )={\frac {2\tan \theta }{1-\tan ^{2}\theta }}={\frac {2}{\cot \theta -\tan \theta }}}
cot
(
2
θ
)
=
cot
2
θ
−
1
2
cot
θ
=
cot
θ
−
tan
θ
2
{\displaystyle \cot(2\theta )={\frac {\cot ^{2}\theta -1}{2\cot \theta }}={\frac {\cot \theta -\tan \theta }{2}}}
It can also be proved using Euler's formula
e
i
φ
=
cos
φ
+
i
sin
φ
{\displaystyle e^{i\varphi }=\cos \varphi +i\sin \varphi }
Squaring both sides yields
e
i
2
φ
=
(
cos
φ
+
i
sin
φ
)
2
{\displaystyle e^{i2\varphi }=(\cos \varphi +i\sin \varphi )^{2}}
But replacing the angle with its doubled version, which achieves the same result in the left side of the equation, yields
e
i
2
φ
=
cos
2
φ
+
i
sin
2
φ
{\displaystyle e^{i2\varphi }=\cos 2\varphi +i\sin 2\varphi }
It follows that
(
cos
φ
+
i
sin
φ
)
2
=
cos
2
φ
+
i
sin
2
φ
{\displaystyle (\cos \varphi +i\sin \varphi )^{2}=\cos 2\varphi +i\sin 2\varphi }
Expanding the square and simplifying on the left hand side of the equation gives
i
(
2
sin
φ
cos
φ
)
+
cos
2
φ
−
sin
2
φ
=
cos
2
φ
+
i
sin
2
φ
{\displaystyle i(2\sin \varphi \cos \varphi )+\cos ^{2}\varphi -\sin ^{2}\varphi \ =\cos 2\varphi +i\sin 2\varphi }
Because the imaginary and real parts have to be the same, we are left with the original identities
cos
2
φ
−
sin
2
φ
=
cos
2
φ
{\displaystyle \cos ^{2}\varphi -\sin ^{2}\varphi \ =\cos 2\varphi }
and also
2
sin
φ
cos
φ
=
sin
2
φ
{\displaystyle 2\sin \varphi \cos \varphi =\sin 2\varphi }
Half-angle identities [ edit ] The two identities giving the alternative forms for cos 2θ lead to the following equations:
cos
θ
2
=
±
1
+
cos
θ
2
,
{\displaystyle \cos {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{2}}},}
sin
θ
2
=
±
1
−
cos
θ
2
.
{\displaystyle \sin {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1-\cos \theta }{2}}}.}
The sign of the square root needs to be chosen properly—note that if 2π is added to θ, the quantities inside the square roots are unchanged, but the left-hand-sides of the equations change sign. Therefore, the correct sign to use depends on the value of θ.
For the tan function, the equation is:
tan
θ
2
=
±
1
−
cos
θ
1
+
cos
θ
.
{\displaystyle \tan {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}.}
Then multiplying the numerator and denominator inside the square root by (1 + cos θ) and using Pythagorean identities leads to:
tan
θ
2
=
sin
θ
1
+
cos
θ
.
{\displaystyle \tan {\frac {\theta }{2}}={\frac {\sin \theta }{1+\cos \theta }}.}
Also, if the numerator and denominator are both multiplied by (1 - cos θ), the result is:
tan
θ
2
=
1
−
cos
θ
sin
θ
.
{\displaystyle \tan {\frac {\theta }{2}}={\frac {1-\cos \theta }{\sin \theta }}.}
This also gives:
tan
θ
2
=
csc
θ
−
cot
θ
.
{\displaystyle \tan {\frac {\theta }{2}}=\csc \theta -\cot \theta .}
Similar manipulations for the cot function give:
cot
θ
2
=
±
1
+
cos
θ
1
−
cos
θ
=
1
+
cos
θ
sin
θ
=
sin
θ
1
−
cos
θ
=
csc
θ
+
cot
θ
.
{\displaystyle \cot {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{1-\cos \theta }}}={\frac {1+\cos \theta }{\sin \theta }}={\frac {\sin \theta }{1-\cos \theta }}=\csc \theta +\cot \theta .}
[ edit ] If
ψ
+
θ
+
ϕ
=
π
=
{\displaystyle \psi +\theta +\phi =\pi =}
ψ
{\displaystyle \psi }
θ
{\displaystyle \theta }
ϕ
{\displaystyle \phi }
tan
(
ψ
)
+
tan
(
θ
)
+
tan
(
ϕ
)
=
tan
(
ψ
)
tan
(
θ
)
tan
(
ϕ
)
.
{\displaystyle \tan(\psi )+\tan(\theta )+\tan(\phi )=\tan(\psi )\tan(\theta )\tan(\phi ).}
Proof:[ 1]
ψ
=
π
−
θ
−
ϕ
tan
(
ψ
)
=
tan
(
π
−
θ
−
ϕ
)
=
−
tan
(
θ
+
ϕ
)
=
−
tan
θ
−
tan
ϕ
1
−
tan
θ
tan
ϕ
=
tan
θ
+
tan
ϕ
tan
θ
tan
ϕ
−
1
(
tan
θ
tan
ϕ
−
1
)
tan
ψ
=
tan
θ
+
tan
ϕ
tan
ψ
tan
θ
tan
ϕ
−
tan
ψ
=
tan
θ
+
tan
ϕ
tan
ψ
tan
θ
tan
ϕ
=
tan
ψ
+
tan
θ
+
tan
ϕ
{\displaystyle {\begin{aligned}\psi &=\pi -\theta -\phi \\\tan(\psi )&=\tan(\pi -\theta -\phi )\\&=-\tan(\theta +\phi )\\&={\frac {-\tan \theta -\tan \phi }{1-\tan \theta \tan \phi }}\\&={\frac {\tan \theta +\tan \phi }{\tan \theta \tan \phi -1}}\\(\tan \theta \tan \phi -1)\tan \psi &=\tan \theta +\tan \phi \\\tan \psi \tan \theta \tan \phi -\tan \psi &=\tan \theta +\tan \phi \\\tan \psi \tan \theta \tan \phi &=\tan \psi +\tan \theta +\tan \phi \\\end{aligned}}}
[ edit ] If
ψ
+
θ
+
ϕ
=
π
2
=
{\displaystyle \psi +\theta +\phi ={\tfrac {\pi }{2}}=}
cot
(
ψ
)
+
cot
(
θ
)
+
cot
(
ϕ
)
=
cot
(
ψ
)
cot
(
θ
)
cot
(
ϕ
)
{\displaystyle \cot(\psi )+\cot(\theta )+\cot(\phi )=\cot(\psi )\cot(\theta )\cot(\phi )}
Proof:
Replace each of
ψ
{\displaystyle \psi }
θ
{\displaystyle \theta }
ϕ
{\displaystyle \phi }
Given
ψ
+
θ
+
ϕ
=
π
2
{\displaystyle \psi +\theta +\phi ={\tfrac {\pi }{2}}}
∴
(
π
2
−
ψ
)
+
(
π
2
−
θ
)
+
(
π
2
−
ϕ
)
=
3
π
2
−
(
ψ
+
θ
+
ϕ
)
=
3
π
2
−
π
2
=
π
{\displaystyle \therefore ({\tfrac {\pi }{2}}-\psi )+({\tfrac {\pi }{2}}-\theta )+({\tfrac {\pi }{2}}-\phi )={\tfrac {3\pi }{2}}-(\psi +\theta +\phi )={\tfrac {3\pi }{2}}-{\tfrac {\pi }{2}}=\pi }
so the result follows from the triple tangent identity.
Sum to product identities [ edit ]
sin
θ
±
sin
ϕ
=
2
sin
(
θ
±
ϕ
2
)
cos
(
θ
∓
ϕ
2
)
{\displaystyle \sin \theta \pm \sin \phi =2\sin \left({\frac {\theta \pm \phi }{2}}\right)\cos \left({\frac {\theta \mp \phi }{2}}\right)}
cos
θ
+
cos
ϕ
=
2
cos
(
θ
+
ϕ
2
)
cos
(
θ
−
ϕ
2
)
{\displaystyle \cos \theta +\cos \phi =2\cos \left({\frac {\theta +\phi }{2}}\right)\cos \left({\frac {\theta -\phi }{2}}\right)}
cos
θ
−
cos
ϕ
=
−
2
sin
(
θ
+
ϕ
2
)
sin
(
θ
−
ϕ
2
)
{\displaystyle \cos \theta -\cos \phi =-2\sin \left({\frac {\theta +\phi }{2}}\right)\sin \left({\frac {\theta -\phi }{2}}\right)}
Proof of sine identities [ edit ] First, start with the sum-angle identities:
sin
(
α
+
β
)
=
sin
α
cos
β
+
cos
α
sin
β
{\displaystyle \sin(\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta }
sin
(
α
−
β
)
=
sin
α
cos
β
−
cos
α
sin
β
{\displaystyle \sin(\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta }
By adding these together,
sin
(
α
+
β
)
+
sin
(
α
−
β
)
=
sin
α
cos
β
+
cos
α
sin
β
+
sin
α
cos
β
−
cos
α
sin
β
=
2
sin
α
cos
β
{\displaystyle \sin(\alpha +\beta )+\sin(\alpha -\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta +\sin \alpha \cos \beta -\cos \alpha \sin \beta =2\sin \alpha \cos \beta }
Similarly, by subtracting the two sum-angle identities,
sin
(
α
+
β
)
−
sin
(
α
−
β
)
=
sin
α
cos
β
+
cos
α
sin
β
−
sin
α
cos
β
+
cos
α
sin
β
=
2
cos
α
sin
β
{\displaystyle \sin(\alpha +\beta )-\sin(\alpha -\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta -\sin \alpha \cos \beta +\cos \alpha \sin \beta =2\cos \alpha \sin \beta }
Let
α
+
β
=
θ
{\displaystyle \alpha +\beta =\theta }
α
−
β
=
ϕ
{\displaystyle \alpha -\beta =\phi }
∴
α
=
θ
+
ϕ
2
{\displaystyle \therefore \alpha ={\frac {\theta +\phi }{2}}}
β
=
θ
−
ϕ
2
{\displaystyle \beta ={\frac {\theta -\phi }{2}}}
Substitute
θ
{\displaystyle \theta }
ϕ
{\displaystyle \phi }
sin
θ
+
sin
ϕ
=
2
sin
(
θ
+
ϕ
2
)
cos
(
θ
−
ϕ
2
)
{\displaystyle \sin \theta +\sin \phi =2\sin \left({\frac {\theta +\phi }{2}}\right)\cos \left({\frac {\theta -\phi }{2}}\right)}
sin
θ
−
sin
ϕ
=
2
cos
(
θ
+
ϕ
2
)
sin
(
θ
−
ϕ
2
)
=
2
sin
(
θ
−
ϕ
2
)
cos
(
θ
+
ϕ
2
)
{\displaystyle \sin \theta -\sin \phi =2\cos \left({\frac {\theta +\phi }{2}}\right)\sin \left({\frac {\theta -\phi }{2}}\right)=2\sin \left({\frac {\theta -\phi }{2}}\right)\cos \left({\frac {\theta +\phi }{2}}\right)}
Therefore,
sin
θ
±
sin
ϕ
=
2
sin
(
θ
±
ϕ
2
)
cos
(
θ
∓
ϕ
2
)
{\displaystyle \sin \theta \pm \sin \phi =2\sin \left({\frac {\theta \pm \phi }{2}}\right)\cos \left({\frac {\theta \mp \phi }{2}}\right)}
Proof of cosine identities [ edit ] Similarly for cosine, start with the sum-angle identities:
cos
(
α
+
β
)
=
cos
α
cos
β
−
sin
α
sin
β
{\displaystyle \cos(\alpha +\beta )=\cos \alpha \cos \beta \ -\sin \alpha \sin \beta }
cos
(
α
−
β
)
=
cos
α
cos
β
+
sin
α
sin
β
{\displaystyle \cos(\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta }
Again, by adding and subtracting
cos
(
α
+
β
)
+
cos
(
α
−
β
)
=
cos
α
cos
β
−
sin
α
sin
β
+
cos
α
cos
β
+
sin
α
sin
β
=
2
cos
α
cos
β
{\displaystyle \cos(\alpha +\beta )+\cos(\alpha -\beta )=\cos \alpha \cos \beta \ -\sin \alpha \sin \beta +\cos \alpha \cos \beta +\sin \alpha \sin \beta =2\cos \alpha \cos \beta }
cos
(
α
+
β
)
−
cos
(
α
−
β
)
=
cos
α
cos
β
−
sin
α
sin
β
−
cos
α
cos
β
−
sin
α
sin
β
=
−
2
sin
α
sin
β
{\displaystyle \cos(\alpha +\beta )-\cos(\alpha -\beta )=\cos \alpha \cos \beta \ -\sin \alpha \sin \beta -\cos \alpha \cos \beta -\sin \alpha \sin \beta =-2\sin \alpha \sin \beta }
Substitute
θ
{\displaystyle \theta }
ϕ
{\displaystyle \phi }
cos
θ
+
cos
ϕ
=
2
cos
(
θ
+
ϕ
2
)
cos
(
θ
−
ϕ
2
)
{\displaystyle \cos \theta +\cos \phi =2\cos \left({\frac {\theta +\phi }{2}}\right)\cos \left({\frac {\theta -\phi }{2}}\right)}
cos
θ
−
cos
ϕ
=
−
2
sin
(
θ
+
ϕ
2
)
sin
(
θ
−
ϕ
2
)
{\displaystyle \cos \theta -\cos \phi =-2\sin \left({\frac {\theta +\phi }{2}}\right)\sin \left({\frac {\theta -\phi }{2}}\right)}
Illustration of the sine and tangent inequalities. The figure at the right shows a sector of a circle with radius 1. The sector is θ /(2π )θ /2θ < π /2
O
A
=
O
D
=
1
{\displaystyle OA=OD=1}
A
B
=
sin
θ
{\displaystyle AB=\sin \theta }
C
D
=
tan
θ
{\displaystyle CD=\tan \theta }
The area of triangle OAD AB /2sin(θ )/2 . The area of triangle OCD CD /2tan(θ )/2 .
Since triangle OAD OCD
sin
θ
<
θ
<
tan
θ
.
{\displaystyle \sin \theta <\theta <\tan \theta .}
This geometric argument relies on definitions of arc length and
area , which act as assumptions, so it is rather a condition imposed in construction of trigonometric functions than
a provable property.[ 2] θ > π /2θ > 1sin θ ≤ 1 (because of the Pythagorean identity), so sin θ < θ . So we have
sin
θ
θ
<
1
i
f
0
<
θ
.
{\displaystyle {\frac {\sin \theta }{\theta }}<1\ \ \ \mathrm {if} \ \ \ 0<\theta .}
For negative values of θ
sin
θ
θ
=
sin
(
−
θ
)
−
θ
<
1.
{\displaystyle {\frac {\sin \theta }{\theta }}={\frac {\sin(-\theta )}{-\theta }}<1.}
Hence
sin
θ
θ
<
1
if
θ
≠
0
,
{\displaystyle {\frac {\sin \theta }{\theta }}<1\quad {\text{if }}\quad \theta \neq 0,}
and
tan
θ
θ
>
1
if
0
<
θ
<
π
2
.
{\displaystyle {\frac {\tan \theta }{\theta }}>1\quad {\text{if }}\quad 0<\theta <{\frac {\pi }{2}}.}
Identities involving calculus [ edit ]
lim
θ
→
0
sin
θ
=
0
{\displaystyle \lim _{\theta \to 0}{\sin \theta }=0}
lim
θ
→
0
cos
θ
=
1
{\displaystyle \lim _{\theta \to 0}{\cos \theta }=1}
Sine and angle ratio identity [ edit ]
lim
θ
→
0
sin
θ
θ
=
1
{\displaystyle \lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}=1}
In other words, the function sine is differentiable at 0, and its derivative is 1.
Proof: From the previous inequalities, we have, for small angles
sin
θ
<
θ
<
tan
θ
{\displaystyle \sin \theta <\theta <\tan \theta }
Therefore,
sin
θ
θ
<
1
<
tan
θ
θ
{\displaystyle {\frac {\sin \theta }{\theta }}<1<{\frac {\tan \theta }{\theta }}}
Consider the right-hand inequality. Since
tan
θ
=
sin
θ
cos
θ
{\displaystyle \tan \theta ={\frac {\sin \theta }{\cos \theta }}}
∴
1
<
sin
θ
θ
cos
θ
{\displaystyle \therefore 1<{\frac {\sin \theta }{\theta \cos \theta }}}
Multiply through by
cos
θ
{\displaystyle \cos \theta }
cos
θ
<
sin
θ
θ
{\displaystyle \cos \theta <{\frac {\sin \theta }{\theta }}}
Combining with the left-hand inequality:
cos
θ
<
sin
θ
θ
<
1
{\displaystyle \cos \theta <{\frac {\sin \theta }{\theta }}<1}
Taking
cos
θ
{\displaystyle \cos \theta }
θ
→
0
{\displaystyle \theta \to 0}
lim
θ
→
0
cos
θ
=
1
{\displaystyle \lim _{\theta \to 0}{\cos \theta }=1}
Therefore,
lim
θ
→
0
sin
θ
θ
=
1
{\displaystyle \lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}=1}
Cosine and angle ratio identity [ edit ]
lim
θ
→
0
1
−
cos
θ
θ
=
0
{\displaystyle \lim _{\theta \to 0}{\frac {1-\cos \theta }{\theta }}=0}
Proof:
1
−
cos
θ
θ
=
1
−
cos
2
θ
θ
(
1
+
cos
θ
)
=
sin
2
θ
θ
(
1
+
cos
θ
)
=
(
sin
θ
θ
)
×
sin
θ
×
(
1
1
+
cos
θ
)
{\displaystyle {\begin{aligned}{\frac {1-\cos \theta }{\theta }}&={\frac {1-\cos ^{2}\theta }{\theta (1+\cos \theta )}}\\&={\frac {\sin ^{2}\theta }{\theta (1+\cos \theta )}}\\&=\left({\frac {\sin \theta }{\theta }}\right)\times \sin \theta \times \left({\frac {1}{1+\cos \theta }}\right)\\\end{aligned}}}
The limits of those three quantities are 1, 0, and 1/2, so the resultant limit is zero.
Cosine and square of angle ratio identity [ edit ]
lim
θ
→
0
1
−
cos
θ
θ
2
=
1
2
{\displaystyle \lim _{\theta \to 0}{\frac {1-\cos \theta }{\theta ^{2}}}={\frac {1}{2}}}
Proof:
As in the preceding proof,
1
−
cos
θ
θ
2
=
sin
θ
θ
×
sin
θ
θ
×
1
1
+
cos
θ
.
{\displaystyle {\frac {1-\cos \theta }{\theta ^{2}}}={\frac {\sin \theta }{\theta }}\times {\frac {\sin \theta }{\theta }}\times {\frac {1}{1+\cos \theta }}.}
The limits of those three quantities are 1, 1, and 1/2, so the resultant limit is 1/2.
Proof of compositions of trig and inverse trig functions [ edit ] All these functions follow from the Pythagorean trigonometric identity. We can prove for instance the function
sin
[
arctan
(
x
)
]
=
x
1
+
x
2
{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {1+x^{2}}}}}
Proof:
We start from
sin
2
θ
+
cos
2
θ
=
1
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1}
Then we divide this equation (I) by
cos
2
θ
{\displaystyle \cos ^{2}\theta }
cos
2
θ
=
1
tan
2
θ
+
1
{\displaystyle \cos ^{2}\theta ={\frac {1}{\tan ^{2}\theta +1}}}
1
−
sin
2
θ
=
1
tan
2
θ
+
1
{\displaystyle 1-\sin ^{2}\theta ={\frac {1}{\tan ^{2}\theta +1}}}
Then use the substitution
θ
=
arctan
(
x
)
{\displaystyle \theta =\arctan(x)}
1
−
sin
2
[
arctan
(
x
)
]
=
1
tan
2
[
arctan
(
x
)
]
+
1
{\displaystyle 1-\sin ^{2}[\arctan(x)]={\frac {1}{\tan ^{2}[\arctan(x)]+1}}}
sin
2
[
arctan
(
x
)
]
=
tan
2
[
arctan
(
x
)
]
tan
2
[
arctan
(
x
)
]
+
1
{\displaystyle \sin ^{2}[\arctan(x)]={\frac {\tan ^{2}[\arctan(x)]}{\tan ^{2}[\arctan(x)]+1}}}
Then we use the identity
tan
[
arctan
(
x
)
]
≡
x
{\displaystyle \tan[\arctan(x)]\equiv x}
sin
[
arctan
(
x
)
]
=
x
x
2
+
1
{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {x^{2}+1}}}}
And initial Pythagorean trigonometric identity proofed...
sin
2
θ
{\displaystyle \sin ^{2}\theta }
sin
2
θ
=
1
1
1
+
1
tan
2
θ
{\displaystyle \sin ^{2}\theta ={\frac {\frac {1}{1}}{1+{\frac {1}{\tan ^{2}\theta }}}}}
sin
2
θ
=
tan
2
θ
tan
2
θ
+
1
{\displaystyle \sin ^{2}\theta ={\frac {\tan ^{2}\theta }{\tan ^{2}\theta +1}}}
Then use the substitution
θ
=
arctan
(
x
)
{\displaystyle \theta =\arctan(x)}
sin
2
[
arctan
(
x
)
]
=
tan
2
[
arctan
(
x
)
]
tan
2
[
arctan
(
x
)
]
+
1
{\displaystyle \sin ^{2}[\arctan(x)]={\frac {\tan ^{2}[\arctan(x)]}{\tan ^{2}[\arctan(x)]+1}}}
Then we use the identity
tan
[
arctan
(
x
)
]
≡
x
{\displaystyle \tan[\arctan(x)]\equiv x}
sin
[
arctan
(
x
)
]
=
x
x
2
+
1
{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {x^{2}+1}}}}
And initial Pythagorean trigonometric identity proofed...
[
arctan
(
x
)
]
=
[
arcsin
(
x
x
2
+
1
)
]
{\displaystyle [\arctan(x)]=[\arcsin({\frac {x}{\sqrt {x^{2}+1}}})]}
y
=
x
x
2
+
1
{\displaystyle y={\frac {x}{\sqrt {x^{2}+1}}}}
y
2
=
x
2
x
2
+
1
{\displaystyle y^{2}={\frac {x^{2}}{x^{2}+1}}}
x
=
y
1
−
y
2
{\displaystyle x={\frac {y}{\sqrt {1-y^{2}}}}}
x
2
=
y
2
1
−
y
2
{\displaystyle x^{2}={\frac {y^{2}}{1-y^{2}}}}
y
2
=
y
2
(
1
−
y
2
)
y
2
(
1
−
y
2
)
+
1
{\displaystyle y^{2}={\frac {\frac {y^{2}}{(1-y^{2})}}{{\frac {y^{2}}{(1-y^{2})}}+1}}}
y
2
=
y
2
(
1
−
y
2
)
1
(
1
−
y
2
)
{\displaystyle y^{2}={\frac {\frac {y^{2}}{(1-y^{2})}}{\frac {1}{(1-y^{2})}}}}
So it's true:
y
2
=
y
2
{\displaystyle y^{2}=y^{2}}
x
=
y
1
−
y
2
{\displaystyle x={\frac {y}{\sqrt {1-y^{2}}}}}
[
arctan
(
x
)
]
=
[
arcsin
(
x
x
2
+
1
)
]
=
[
arcsin
(
y
)
]
=
[
arctan
(
y
1
−
y
2
)
]
{\displaystyle [\arctan(x)]=[\arcsin({\frac {x}{\sqrt {x^{2}+1}}})]=[\arcsin(y)]=[\arctan({\frac {y}{\sqrt {1-y^{2}}}})]}
[
arcsin
(
x
)
]
=
[
arctan
(
x
1
−
x
2
)
]
{\displaystyle [\arcsin(x)]=[\arctan({\frac {x}{\sqrt {1-x^{2}}}})]}
[
arccos
(
x
)
]
{\displaystyle [\arccos(x)]}
cos
[
arccos
(
x
)
]
=
x
{\displaystyle \cos[\arccos(x)]=x}
cos
(
π
2
−
(
π
2
−
[
arccos
(
x
)
]
)
)
=
x
{\displaystyle \cos({\frac {\pi }{2}}-({\frac {\pi }{2}}-[\arccos(x)]))=x}
sin
(
π
2
−
[
arccos
(
x
)
]
)
=
x
{\displaystyle \sin({\frac {\pi }{2}}-[\arccos(x)])=x}
π
2
−
[
arccos
(
x
)
]
=
[
arcsin
(
x
)
]
{\displaystyle {\frac {\pi }{2}}-[\arccos(x)]=[\arcsin(x)]}
[
arccos
(
x
)
]
=
π
2
−
[
arcsin
(
x
)
]
{\displaystyle [\arccos(x)]={\frac {\pi }{2}}-[\arcsin(x)]}
[
arcsin
(
x
)
]
{\displaystyle [\arcsin(x)]}
[
arccos
(
x
)
]
=
π
2
−
[
arctan
(
x
1
−
x
2
)
]
{\displaystyle [\arccos(x)]={\frac {\pi }{2}}-[\arctan({\frac {x}{\sqrt {1-x^{2}}}})]}
[
arccos
(
x
)
]
=
π
2
−
[
arccot
(
1
−
x
2
x
)
]
{\displaystyle [\arccos(x)]={\frac {\pi }{2}}-[\operatorname {arccot}({\frac {\sqrt {1-x^{2}}}{x}})]}
And finally we have [arccos] expressed through [arctan]...
[
arccos
(
x
)
]
=
[
arctan
(
1
−
x
2
x
)
]
{\displaystyle [\arccos(x)]=[\arctan({\frac {\sqrt {1-x^{2}}}{x}})]}
![{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {1+x^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d3ed8cedb4c73b01b65314a15152647c9000086)
![{\displaystyle 1-\sin ^{2}[\arctan(x)]={\frac {1}{\tan ^{2}[\arctan(x)]+1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0bb45d076e27715801c0437e364d4b1f45dfb1a)
![{\displaystyle \sin ^{2}[\arctan(x)]={\frac {\tan ^{2}[\arctan(x)]}{\tan ^{2}[\arctan(x)]+1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31457873262613ebf792de2499d564919e4d18c9)
![{\displaystyle \tan[\arctan(x)]\equiv x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49ce3dc146ff12eb2efb89a4388afcb984b7da8e)
![{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {x^{2}+1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c5222bb21dc2d342d1363c855694bae8b967914)
![{\displaystyle [\arctan(x)]=[\arcsin({\frac {x}{\sqrt {x^{2}+1}}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dbc17437ca395ab8ec9e04f2d21096288df5b9a2)
![{\displaystyle [\arctan(x)]=[\arcsin({\frac {x}{\sqrt {x^{2}+1}}})]=[\arcsin(y)]=[\arctan({\frac {y}{\sqrt {1-y^{2}}}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fbaafc98c5960a737537039cff02fdd72be65da)
![{\displaystyle [\arcsin(x)]=[\arctan({\frac {x}{\sqrt {1-x^{2}}}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/552a4ab628e13245dc277ab57029427f50ee8ebd)
![{\displaystyle [\arccos(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f61004b1d4ad7ff39f9a7ee50976ec672d2c198)
![{\displaystyle \cos[\arccos(x)]=x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9d5691af1ed2de3248e08dbaf90a24a80c1c816)
![{\displaystyle \cos({\frac {\pi }{2}}-({\frac {\pi }{2}}-[\arccos(x)]))=x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7562bfeec9737b56e7affe3409239315302693c3)
![{\displaystyle \sin({\frac {\pi }{2}}-[\arccos(x)])=x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ad20b817a17336b1fc1af8c339a3d7e4cf111f6)
![{\displaystyle {\frac {\pi }{2}}-[\arccos(x)]=[\arcsin(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20f54833eb7dbe561e4cb31e51453ebd14a5cc51)
![{\displaystyle [\arccos(x)]={\frac {\pi }{2}}-[\arcsin(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfc6a970c74fe335d2b0684f23027e6ca255a9cc)
![{\displaystyle [\arcsin(x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef6f2837ad8f82968478d6c5e93863549f833207)
![{\displaystyle [\arccos(x)]={\frac {\pi }{2}}-[\arctan({\frac {x}{\sqrt {1-x^{2}}}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6675d621ebaf7c4c4b10adbee9a1fd4b6f1d6eb)
![{\displaystyle [\arccos(x)]={\frac {\pi }{2}}-[\operatorname {arccot}({\frac {\sqrt {1-x^{2}}}{x}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4fe5f411d564cd54abb12c188ca5885150f1949)
![{\displaystyle [\arccos(x)]=[\arctan({\frac {\sqrt {1-x^{2}}}{x}})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/13428b7bd426702f18227deb2d8da5f116c16db4)