In mathematics , trigonometric identities are equalities that involve trigonometric functions that are true for every single value of the occurring variables. These identities are useful whenever expressions involving trigonometric functions need to be simplified. An important application is the integration of non-trigonometric functions: a common trick involves first using the substitution rule with a trigonometric function , and then simplifying the resulting integral with a trigonometric identity.
All of the trigonometric functions of an angle θ can be constructed geometrically in terms of a unit circle centered at O .
Cosines & Sines around the unit circle
To avoid the confusion caused by the ambiguity of sin−1 (x ), the reciprocals and inverses of trigonometric functions are often displayed as in this table. In representing the cosecant function, the longer form 'cosec' is sometimes used in place of 'csc'.
Function
Inverse function
Reciprocal
Inverse reciprocal
sine
sin
arcsine
arcsin
cosecant
csc
arccosecant
arccsc
cosine
cos
arccosine
arccos
secant
sec
arcsecant
arcsec
tangent
tan
arctangent
arctan
cotangent
cot
arccotangent
arccot
Different angular measures can be appropriate in different situations. This table shows some of the more common systems.
Radians is the default angular measure and is the one you use if you use the exponential definitions. All angular measures are unitless.
Degrees
30
45
60
90
120
180
270
360
Radians
π
/
6
{\displaystyle \pi /6}
π
/
4
{\displaystyle \pi /4}
π
/
3
{\displaystyle \pi /3}
π
/
2
{\displaystyle \pi /2}
2
π
/
3
{\displaystyle 2\pi /3}
π
{\displaystyle \pi }
3
π
/
2
{\displaystyle 3\pi /2}
2
π
{\displaystyle 2\pi }
Grads
33 ⅓
50
66 ⅔
100
133 ⅓
200
300
400
Basic relationships [ edit ]
From the two identities above, the following table can be extrapolated. Note however that these conversion equations may not provide the correct sign (+ or −). For example, if sin θ = 1/2, the conversion in the table indicates that
cos
θ
=
1
−
sin
2
θ
=
3
/
2
{\displaystyle \scriptstyle \cos \theta \,=\,{\sqrt {1-\sin ^{2}\theta }}={\sqrt {3}}/2}
, though it is possible that
cos
θ
=
−
3
/
2
{\displaystyle \scriptstyle \cos \theta \,=\,-{\sqrt {3}}/2}
. More information would be needed about which quadrant θ lies in to determine a single, exact answer.
Each trigonometric function in terms of the other five. [ 4]
Function
sin
cos
tan
csc
sec
cot
sin
θ
=
{\displaystyle \sin \theta =}
sin
θ
{\displaystyle \sin \theta \ }
1
−
cos
2
θ
{\displaystyle {\sqrt {1-\cos ^{2}\theta }}}
tan
θ
1
+
tan
2
θ
{\displaystyle {\frac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}}}
1
csc
θ
{\displaystyle {\frac {1}{\csc \theta }}}
sec
2
θ
−
1
sec
θ
{\displaystyle {\frac {\sqrt {\sec ^{2}\theta -1}}{\sec \theta }}}
1
1
+
cot
2
θ
{\displaystyle {\frac {1}{\sqrt {1+\cot ^{2}\theta }}}}
cos
θ
=
{\displaystyle \cos \theta =}
1
−
sin
2
θ
{\displaystyle {\sqrt {1-\sin ^{2}\theta }}}
cos
θ
{\displaystyle \cos \theta \ }
1
1
+
tan
2
θ
{\displaystyle {\frac {1}{\sqrt {1+\tan ^{2}\theta }}}}
csc
2
θ
−
1
csc
θ
{\displaystyle {\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}}
1
sec
θ
{\displaystyle {\frac {1}{\sec \theta }}}
cot
θ
1
+
cot
2
θ
{\displaystyle {\frac {\cot \theta }{\sqrt {1+\cot ^{2}\theta }}}}
tan
θ
=
{\displaystyle \tan \theta =}
sin
θ
1
−
sin
2
θ
{\displaystyle {\frac {\sin \theta }{\sqrt {1-\sin ^{2}\theta }}}}
1
−
cos
2
θ
cos
θ
{\displaystyle {\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}}
tan
θ
{\displaystyle \tan \theta \ }
1
csc
2
θ
−
1
{\displaystyle {\frac {1}{\sqrt {\csc ^{2}\theta -1}}}}
sec
2
θ
−
1
{\displaystyle {\sqrt {\sec ^{2}\theta -1}}}
1
cot
θ
{\displaystyle {\frac {1}{\cot \theta }}}
csc
θ
=
{\displaystyle \csc \theta =}
1
sin
θ
{\displaystyle {1 \over \sin \theta }}
1
1
−
cos
2
θ
{\displaystyle {1 \over {\sqrt {1-\cos ^{2}\theta }}}}
1
+
tan
2
θ
tan
θ
{\displaystyle {{\sqrt {1+\tan ^{2}\theta }} \over \tan \theta }}
csc
θ
{\displaystyle \csc \theta \ }
sec
θ
sec
2
θ
−
1
{\displaystyle {\sec \theta \over {\sqrt {\sec ^{2}\theta -1}}}}
1
+
cot
2
θ
{\displaystyle {\sqrt {1+\cot ^{2}\theta }}}
sec
θ
=
{\displaystyle \sec \theta =}
1
1
−
sin
2
θ
{\displaystyle {1 \over {\sqrt {1-\sin ^{2}\theta }}}}
1
cos
θ
{\displaystyle {1 \over \cos \theta }}
1
+
tan
2
θ
{\displaystyle {\sqrt {1+\tan ^{2}\theta }}}
csc
θ
csc
2
θ
−
1
{\displaystyle {\csc \theta \over {\sqrt {\csc ^{2}\theta -1}}}}
sec
θ
{\displaystyle \sec \theta \ }
1
+
cot
2
θ
cot
θ
{\displaystyle {{\sqrt {1+\cot ^{2}\theta }} \over \cot \theta }}
cot
θ
=
{\displaystyle \cot \theta =}
1
−
sin
2
θ
sin
θ
{\displaystyle {{\sqrt {1-\sin ^{2}\theta }} \over \sin \theta }}
cos
θ
1
−
cos
2
θ
{\displaystyle {\cos \theta \over {\sqrt {1-\cos ^{2}\theta }}}}
1
tan
θ
{\displaystyle {1 \over \tan \theta }}
csc
2
θ
−
1
{\displaystyle {\sqrt {\csc ^{2}\theta -1}}}
1
sec
2
θ
−
1
{\displaystyle {1 \over {\sqrt {\sec ^{2}\theta -1}}}}
cot
θ
{\displaystyle \cot \theta \ }
Historic shorthands [ edit ]
Rarely used today, the versine , coversine , haversine , and exsecant have been defined as below and used in navigation, for example the haversine formula was used to calculate the distance between two points on a sphere.
Name(s)
Abbreviation(s)
Value [ 5]
versed sineversine
versin
θ
{\displaystyle {\textrm {versin}}\,\theta }
vers
θ
{\displaystyle {\textrm {vers}}\,\theta }
1
−
cos
θ
{\displaystyle 1-\cos \theta \,}
coversed sine coversine
coversin
θ
{\displaystyle {\textrm {coversin}}\,\theta }
cover
θ
{\displaystyle {\textrm {cover}}\,\theta }
1
−
sin
θ
{\displaystyle 1-\sin \theta \,}
haversed sine haversine
haversin
θ
{\displaystyle {\textrm {haversin}}\,\theta }
hav
θ
{\displaystyle {\textrm {hav}}\,\theta }
1
2
versin
θ
{\displaystyle {\tfrac {1}{2}}{\textrm {versin}}\theta \,}
hacoversed sine hacoversine cohaversine havercosine
hacoversin
θ
{\displaystyle {\textrm {hacoversin}}\,\theta }
hacov
θ
{\displaystyle {\textrm {hacov}}\,\theta }
1
2
coversin
θ
{\displaystyle {\tfrac {1}{2}}{\textrm {coversin}}\theta \,}
exsecant
exsec
θ
{\displaystyle {\textrm {exsec}}\,\theta \,}
sec
θ
−
1
{\displaystyle \sec \theta -1\,}
excosecant
excsc
θ
{\displaystyle {\textrm {excsc}}\,\theta \,}
csc
θ
−
1
{\displaystyle \csc \theta -1\,}
Symmetry, shifts, and periodicity[ edit ]
By examining the unit circle, the following properties of the trigonometric functions can be established.
When the trigonometric functions are reflected from certain values of
θ
{\displaystyle \theta }
, The result is often one of the other trigonometric functions. This leads to the following identities:
Reflected in
θ
=
0
{\displaystyle \theta =0}
[ 6]
Reflected in
θ
=
π
/
2
{\displaystyle \theta =\pi /2}
(co-function identities)[ 7]
Reflected in
θ
=
π
{\displaystyle \theta =\pi }
sin
(
0
−
θ
)
=
−
sin
θ
cos
(
0
−
θ
)
=
+
cos
θ
tan
(
0
−
θ
)
=
−
tan
θ
csc
(
0
−
θ
)
=
−
csc
θ
sec
(
0
−
θ
)
=
+
sec
θ
cot
(
0
−
θ
)
=
−
cot
θ
{\displaystyle {\begin{aligned}\sin(0-\theta )&=-\sin \theta \\\cos(0-\theta )&=+\cos \theta \\\tan(0-\theta )&=-\tan \theta \\\csc(0-\theta )&=-\csc \theta \\\sec(0-\theta )&=+\sec \theta \\\cot(0-\theta )&=-\cot \theta \end{aligned}}}
sin
(
π
2
−
θ
)
=
+
cos
θ
cos
(
π
2
−
θ
)
=
+
sin
θ
tan
(
π
2
−
θ
)
=
+
cot
θ
csc
(
π
2
−
θ
)
=
+
sec
θ
sec
(
π
2
−
θ
)
=
+
csc
θ
cot
(
π
2
−
θ
)
=
+
tan
θ
{\displaystyle {\begin{aligned}\sin({\tfrac {\pi }{2}}-\theta )&=+\cos \theta \\\cos({\tfrac {\pi }{2}}-\theta )&=+\sin \theta \\\tan({\tfrac {\pi }{2}}-\theta )&=+\cot \theta \\\csc({\tfrac {\pi }{2}}-\theta )&=+\sec \theta \\\sec({\tfrac {\pi }{2}}-\theta )&=+\csc \theta \\\cot({\tfrac {\pi }{2}}-\theta )&=+\tan \theta \end{aligned}}}
sin
(
π
−
θ
)
=
+
sin
θ
cos
(
π
−
θ
)
=
−
cos
θ
tan
(
π
−
θ
)
=
−
tan
θ
csc
(
π
−
θ
)
=
+
csc
θ
sec
(
π
−
θ
)
=
−
sec
θ
cot
(
π
−
θ
)
=
−
cot
θ
{\displaystyle {\begin{aligned}\sin(\pi -\theta )&=+\sin \theta \\\cos(\pi -\theta )&=-\cos \theta \\\tan(\pi -\theta )&=-\tan \theta \\\csc(\pi -\theta )&=+\csc \theta \\\sec(\pi -\theta )&=-\sec \theta \\\cot(\pi -\theta )&=-\cot \theta \\\end{aligned}}}
Shifts and periodicity [ edit ]
By shifting the function round by certain angles, it is often possible to find different trigonometric functions that express the result more simply. Some examples of this are shown by shifting functions round by π/2, π and 2π radians. Because the periods of these functions are either π or 2π, there are cases where the new function is exactly the same as the old function without the shift.
Shift by π/2
Shift by π Period for tan and cot[ 8]
Shift by 2π Period for sin, cos, csc and sec[ 9]
sin
(
θ
+
π
2
)
=
+
cos
θ
cos
(
θ
+
π
2
)
=
−
sin
θ
tan
(
θ
+
π
2
)
=
−
cot
θ
csc
(
θ
+
π
2
)
=
+
sec
θ
sec
(
θ
+
π
2
)
=
−
csc
θ
cot
(
θ
+
π
2
)
=
−
tan
θ
{\displaystyle {\begin{aligned}\sin(\theta +{\tfrac {\pi }{2}})&=+\cos \theta \\\cos(\theta +{\tfrac {\pi }{2}})&=-\sin \theta \\\tan(\theta +{\tfrac {\pi }{2}})&=-\cot \theta \\\csc(\theta +{\tfrac {\pi }{2}})&=+\sec \theta \\\sec(\theta +{\tfrac {\pi }{2}})&=-\csc \theta \\\cot(\theta +{\tfrac {\pi }{2}})&=-\tan \theta \end{aligned}}}
sin
(
θ
+
π
)
=
−
sin
θ
cos
(
θ
+
π
)
=
−
cos
θ
tan
(
θ
+
π
)
=
+
tan
θ
csc
(
θ
+
π
)
=
−
csc
θ
sec
(
θ
+
π
)
=
−
sec
θ
cot
(
θ
+
π
)
=
+
cot
θ
{\displaystyle {\begin{aligned}\sin(\theta +\pi )&=-\sin \theta \\\cos(\theta +\pi )&=-\cos \theta \\\tan(\theta +\pi )&=+\tan \theta \\\csc(\theta +\pi )&=-\csc \theta \\\sec(\theta +\pi )&=-\sec \theta \\\cot(\theta +\pi )&=+\cot \theta \\\end{aligned}}}
sin
(
θ
+
2
π
)
=
+
sin
θ
cos
(
θ
+
2
π
)
=
+
cos
θ
tan
(
θ
+
2
π
)
=
+
tan
θ
csc
(
θ
+
2
π
)
=
+
csc
θ
sec
(
θ
+
2
π
)
=
+
sec
θ
cot
(
θ
+
2
π
)
=
+
cot
θ
{\displaystyle {\begin{aligned}\sin(\theta +2\pi )&=+\sin \theta \\\cos(\theta +2\pi )&=+\cos \theta \\\tan(\theta +2\pi )&=+\tan \theta \\\csc(\theta +2\pi )&=+\csc \theta \\\sec(\theta +2\pi )&=+\sec \theta \\\cot(\theta +2\pi )&=+\cot \theta \end{aligned}}}
Angle sum and difference identities [ edit ]
These are also known as the addition and subtraction theorems or formulæ .
The quickest way to prove these is Euler's formula .
The sum and difference formulas for sine and cosine can be written in matrix form, thus:
[
cos
α
−
sin
α
sin
α
cos
α
]
[
cos
β
−
sin
β
sin
β
cos
β
]
=
[
cos
(
α
+
β
)
−
sin
(
α
+
β
)
sin
(
α
+
β
)
cos
(
α
+
β
)
]
.
{\displaystyle \left[{\begin{matrix}\cos \alpha &-\sin \alpha \\\sin \alpha &\cos \alpha \end{matrix}}\right]\left[{\begin{matrix}\cos \beta &-\sin \beta \\\sin \beta &\cos \beta \end{matrix}}\right]=\left[{\begin{matrix}\cos(\alpha +\beta )&-\sin(\alpha +\beta )\\\sin(\alpha +\beta )&\cos(\alpha +\beta )\end{matrix}}\right].}
Sines and cosines of sums of infinitely many terms [ edit ]
sin
(
∑
i
=
1
∞
θ
i
)
=
∑
o
d
d
k
≥
1
(
−
1
)
(
k
−
1
)
/
2
∑
A
⊆
{
1
,
2
,
3
,
…
}
|
A
|
=
k
(
∏
i
∈
A
sin
θ
i
∏
i
∉
A
cos
θ
i
)
{\displaystyle \sin \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{\mathrm {odd} \ k\geq 1}(-1)^{(k-1)/2}\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}
cos
(
∑
i
=
1
∞
θ
i
)
=
∑
e
v
e
n
k
≥
0
(
−
1
)
k
/
2
∑
A
⊆
{
1
,
2
,
3
,
…
}
|
A
|
=
k
(
∏
i
∈
A
sin
θ
i
∏
i
∉
A
cos
θ
i
)
{\displaystyle \cos \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{\mathrm {even} \ k\geq 0}~(-1)^{k/2}~~\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}
In these two identities an asymmetry appears that is not seen in the case of sums of finitely many terms: in each product, there are only finitely many sine factors and cofinitely many cosine factors.
If only finitely many of the terms θi are nonzero, then only finitely many of the terms on the right side will be nonzero because sine factors will vanish, and in each term, all but finitely many of the cosine factors will be unity.
Tangents of sums of finitely many terms [ edit ]
Let x i = tan(θi ), for i = 1, ..., n . Let e k be the k th-degree elementary symmetric polynomial in the variables x i , i = 1, ..., n , k = 0, ..., n . Then
tan
(
θ
1
+
⋯
+
θ
n
)
=
e
1
−
e
3
+
e
5
−
⋯
e
0
−
e
2
+
e
4
−
⋯
,
{\displaystyle \tan(\theta _{1}+\cdots +\theta _{n})={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }},}
the number of terms depending on n .
For example:
tan
(
θ
1
+
θ
2
+
θ
3
)
=
e
1
−
e
3
e
0
−
e
2
=
(
x
1
+
x
2
+
x
3
)
−
(
x
1
x
2
x
3
)
1
−
(
x
1
x
2
+
x
1
x
3
+
x
2
x
3
)
,
tan
(
θ
1
+
θ
2
+
θ
3
+
θ
4
)
=
e
1
−
e
3
e
0
−
e
2
+
e
4
=
(
x
1
+
x
2
+
x
3
+
x
4
)
−
(
x
1
x
2
x
3
+
x
1
x
2
x
4
+
x
1
x
3
x
4
+
x
2
x
3
x
4
)
1
−
(
x
1
x
2
+
x
1
x
3
+
x
1
x
4
+
x
2
x
3
+
x
2
x
4
+
x
3
x
4
)
+
(
x
1
x
2
x
3
x
4
)
,
{\displaystyle {\begin{aligned}\tan(\theta _{1}+\theta _{2}+\theta _{3})&{}={\frac {e_{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\\\\tan(\theta _{1}+\theta _{2}+\theta _{3}+\theta _{4})&{}={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\\\&{}={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}}
and so on. The general case can be proved by mathematical induction .
Tn is the n th Chebyshev polynomial
cos
n
θ
=
T
n
(
cos
θ
)
{\displaystyle \cos n\theta =T_{n}(\cos \theta )\,}
[ 14]
S n is the n th spread polynomial
sin
2
n
θ
=
S
n
(
sin
2
θ
)
{\displaystyle \sin ^{2}n\theta =S_{n}(\sin ^{2}\theta )\,}
de Moivre's formula ,
i
{\displaystyle i}
is the Imaginary unit
cos
n
θ
+
i
sin
n
θ
=
(
cos
(
θ
)
+
i
sin
(
θ
)
)
n
{\displaystyle \cos n\theta +i\sin n\theta =(\cos(\theta )+i\sin(\theta ))^{n}\,}
[ 15]
1
+
2
cos
(
x
)
+
2
cos
(
2
x
)
+
2
cos
(
3
x
)
+
⋯
+
2
cos
(
n
x
)
=
sin
(
(
n
+
1
2
)
x
)
sin
(
x
/
2
)
.
{\displaystyle 1+2\cos(x)+2\cos(2x)+2\cos(3x)+\cdots +2\cos(nx)={\frac {\sin \left(\left(n+{\frac {1}{2}}\right)x\right)}{\sin(x/2)}}.}
(This function of x is the Dirichlet kernel .)
These can be shown by using either the sum and difference identities or the multiple-angle formulae.
Double-angle formulae[ 16] [ 17]
sin
2
θ
=
2
sin
θ
cos
θ
=
2
tan
θ
1
+
tan
2
θ
{\displaystyle {\begin{aligned}\sin 2\theta &=2\sin \theta \cos \theta \ \\&={\frac {2\tan \theta }{1+\tan ^{2}\theta }}\end{aligned}}}
cos
2
θ
=
cos
2
θ
−
sin
2
θ
=
2
cos
2
θ
−
1
=
1
−
2
sin
2
θ
=
1
−
tan
2
θ
1
+
tan
2
θ
{\displaystyle {\begin{aligned}\cos 2\theta &=\cos ^{2}\theta -\sin ^{2}\theta \\&=2\cos ^{2}\theta -1\\&=1-2\sin ^{2}\theta \\&={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}\end{aligned}}}
tan
2
θ
=
2
tan
θ
1
−
tan
2
θ
{\displaystyle \tan 2\theta ={\frac {2\tan \theta }{1-\tan ^{2}\theta }}\,}
cot
2
θ
=
cot
2
θ
−
1
2
cot
θ
{\displaystyle \cot 2\theta ={\frac {\cot ^{2}\theta -1}{2\cot \theta }}\,}
Triple-angle formulae[ 18] [ 14]
sin
3
θ
=
3
sin
θ
−
4
sin
3
θ
{\displaystyle \sin 3\theta =3\sin \theta -4\sin ^{3}\theta \,}
cos
3
θ
=
4
cos
3
θ
−
3
cos
θ
{\displaystyle \cos 3\theta =4\cos ^{3}\theta -3\cos \theta \,}
tan
3
θ
=
3
tan
θ
−
tan
3
θ
1
−
3
tan
2
θ
{\displaystyle \tan 3\theta ={\frac {3\tan \theta -\tan ^{3}\theta }{1-3\tan ^{2}\theta }}}
cot
3
θ
=
3
cot
θ
−
cot
3
θ
1
−
3
cot
2
θ
{\displaystyle \cot 3\theta ={\frac {3\cot \theta -\cot ^{3}\theta }{1-3\cot ^{2}\theta }}}
Half-angle formulae[ 19] [ 20]
sin
θ
2
=
±
1
−
cos
θ
2
{\displaystyle \sin {\tfrac {\theta }{2}}=\pm \,{\sqrt {\frac {1-\cos \theta }{2}}}}
cos
θ
2
=
±
1
+
cos
θ
2
{\displaystyle \cos {\tfrac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{2}}}}
tan
θ
2
=
csc
θ
−
cot
θ
=
±
1
−
cos
θ
1
+
cos
θ
=
sin
θ
1
+
cos
θ
=
1
−
cos
θ
sin
θ
{\displaystyle {\begin{aligned}\tan {\tfrac {\theta }{2}}&=\csc \theta -\cot \theta \\&=\pm \,{\sqrt {1-\cos \theta \over 1+\cos \theta }}\\&={\frac {\sin \theta }{1+\cos \theta }}\\&={\frac {1-\cos \theta }{\sin \theta }}\end{aligned}}}
cot
θ
2
=
csc
θ
+
cot
θ
=
±
1
+
cos
θ
1
−
cos
θ
=
sin
θ
1
−
cos
θ
=
1
+
cos
θ
sin
θ
{\displaystyle {\begin{aligned}\cot {\tfrac {\theta }{2}}&=\csc \theta +\cot \theta \\&=\pm \,{\sqrt {1+\cos \theta \over 1-\cos \theta }}\\&={\frac {\sin \theta }{1-\cos \theta }}\\&={\frac {1+\cos \theta }{\sin \theta }}\end{aligned}}}
See also Tangent half-angle formula .
Sine, cosine, and tangent of multiple angles[ edit ]
sin
n
θ
=
∑
k
=
0
n
(
n
k
)
cos
k
θ
sin
n
−
k
θ
sin
(
1
2
(
n
−
k
)
π
)
{\displaystyle \sin n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\sin \left({\frac {1}{2}}(n-k)\pi \right)}
cos
n
θ
=
∑
k
=
0
n
(
n
k
)
cos
k
θ
sin
n
−
k
θ
cos
(
1
2
(
n
−
k
)
π
)
{\displaystyle \cos n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\cos \left({\frac {1}{2}}(n-k)\pi \right)}
tan nθ can be written in terms of tan θ using the recurrence relation:
tan
(
n
+
1
)
θ
=
tan
n
θ
+
tan
θ
1
−
tan
n
θ
tan
θ
.
{\displaystyle \tan \,(n{+}1)\theta ={\frac {\tan n\theta +\tan \theta }{1-\tan n\theta \,\tan \theta }}.}
cot nθ can be written in terms of cot θ using the recurrence relation:
cot
(
n
+
1
)
θ
=
cot
n
θ
cot
θ
−
1
cot
n
θ
+
cot
θ
.
{\displaystyle \cot \,(n{+}1)\theta ={\frac {\cot n\theta \,\cot \theta -1}{\cot n\theta +\cot \theta }}.}
Tangent of an average [ edit ]
tan
(
α
+
β
2
)
=
sin
α
+
sin
β
cos
α
+
cos
β
=
−
cos
α
−
cos
β
sin
α
−
sin
β
{\displaystyle \tan \left({\frac {\alpha +\beta }{2}}\right)={\frac {\sin \alpha +\sin \beta }{\cos \alpha +\cos \beta }}=-\,{\frac {\cos \alpha -\cos \beta }{\sin \alpha -\sin \beta }}}
Setting either α or β to 0 gives the usual tangent half-angle formulas.
Euler's infinite product[ edit ]
cos
(
θ
2
)
⋅
cos
(
θ
4
)
⋅
cos
(
θ
8
)
⋯
=
∏
n
=
1
∞
cos
(
θ
2
n
)
=
sin
(
θ
)
θ
.
{\displaystyle \cos \left({\theta \over 2}\right)\cdot \cos \left({\theta \over 4}\right)\cdot \cos \left({\theta \over 8}\right)\cdots =\prod _{n=1}^{\infty }\cos \left({\theta \over 2^{n}}\right)={\sin(\theta ) \over \theta }.}
Obtained by solving the second and third versions of the cosine double-angle formula.
Sine
Cosine
Other
sin
2
θ
=
1
−
cos
2
θ
2
{\displaystyle \sin ^{2}\theta ={\frac {1-\cos 2\theta }{2}}}
cos
2
θ
=
1
+
cos
2
θ
2
{\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}}
sin
2
θ
cos
2
θ
=
1
−
cos
4
θ
8
{\displaystyle \sin ^{2}\theta \cos ^{2}\theta ={\frac {1-\cos 4\theta }{8}}}
sin
3
θ
=
3
sin
θ
−
sin
3
θ
4
{\displaystyle \sin ^{3}\theta ={\frac {3\sin \theta -\sin 3\theta }{4}}}
cos
3
θ
=
3
cos
θ
+
cos
3
θ
4
{\displaystyle \cos ^{3}\theta ={\frac {3\cos \theta +\cos 3\theta }{4}}}
sin
3
θ
cos
3
θ
=
3
sin
2
θ
−
sin
6
θ
32
{\displaystyle \sin ^{3}\theta \cos ^{3}\theta ={\frac {3\sin 2\theta -\sin 6\theta }{32}}}
sin
4
θ
=
3
−
4
cos
2
θ
+
cos
4
θ
8
{\displaystyle \sin ^{4}\theta ={\frac {3-4\cos 2\theta +\cos 4\theta }{8}}}
cos
4
θ
=
3
+
4
cos
2
θ
+
cos
4
θ
8
{\displaystyle \cos ^{4}\theta ={\frac {3+4\cos 2\theta +\cos 4\theta }{8}}}
sin
4
θ
cos
4
θ
=
3
−
4
cos
4
θ
+
cos
8
θ
128
{\displaystyle \sin ^{4}\theta \cos ^{4}\theta ={\frac {3-4\cos 4\theta +\cos 8\theta }{128}}}
sin
5
θ
=
10
sin
θ
−
5
sin
3
θ
+
sin
5
θ
16
{\displaystyle \sin ^{5}\theta ={\frac {10\sin \theta -5\sin 3\theta +\sin 5\theta }{16}}}
cos
5
θ
=
10
cos
θ
+
5
cos
3
θ
+
cos
5
θ
16
{\displaystyle \cos ^{5}\theta ={\frac {10\cos \theta +5\cos 3\theta +\cos 5\theta }{16}}}
sin
5
θ
cos
5
θ
=
10
sin
2
θ
−
5
sin
6
θ
+
sin
10
θ
512
{\displaystyle \sin ^{5}\theta \cos ^{5}\theta ={\frac {10\sin 2\theta -5\sin 6\theta +\sin 10\theta }{512}}}
and in general terms of powers of sin θ or cos θ the following is true, and can be deduced using De Moivre's formula , Euler's formula and binomial expansion .
Cosine
Sine
if
n
is odd
{\displaystyle {\mbox{if }}n{\mbox{ is odd}}}
cos
n
θ
=
2
2
n
∑
k
=
0
n
−
1
2
(
n
k
)
cos
(
n
−
2
k
)
θ
{\displaystyle \cos ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}{\binom {n}{k}}\cos {(n-2k)\theta }}
sin
n
θ
=
2
2
n
∑
k
=
0
n
−
1
2
(
−
1
)
(
n
−
1
2
−
k
)
(
n
k
)
sin
(
n
−
2
k
)
θ
{\displaystyle \sin ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}(-1)^{({\frac {n-1}{2}}-k)}{\binom {n}{k}}\sin {(n-2k)\theta }}
if
n
is even
{\displaystyle {\mbox{if }}n{\mbox{ is even}}}
cos
n
θ
=
1
2
n
(
n
n
2
)
+
2
2
n
∑
k
=
0
n
2
−
1
(
n
k
)
cos
(
n
−
2
k
)
θ
{\displaystyle \cos ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}{\binom {n}{k}}\cos {(n-2k)\theta }}
sin
n
θ
=
1
2
n
(
n
n
2
)
+
2
2
n
∑
k
=
0
n
2
−
1
(
−
1
)
(
n
2
−
k
)
(
n
k
)
cos
(
n
−
2
k
)
θ
{\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{({\frac {n}{2}}-k)}{\binom {n}{k}}\cos {(n-2k)\theta }}
Product-to-sum and sum-to-product identities [ edit ]
The product-to-sum identities can be proven by expanding their right-hand sides using the angle addition theorems . See beat frequency for an application of the sum-to-product formulas.
Product-to-sum[ 21]
cos
θ
cos
φ
=
cos
(
θ
−
φ
)
+
cos
(
θ
+
φ
)
2
{\displaystyle \cos \theta \cos \varphi ={\cos(\theta -\varphi )+\cos(\theta +\varphi ) \over 2}}
sin
θ
sin
φ
=
cos
(
θ
−
φ
)
−
cos
(
θ
+
φ
)
2
{\displaystyle \sin \theta \sin \varphi ={\cos(\theta -\varphi )-\cos(\theta +\varphi ) \over 2}}
sin
θ
cos
φ
=
sin
(
θ
+
φ
)
+
sin
(
θ
−
φ
)
2
{\displaystyle \sin \theta \cos \varphi ={\sin(\theta +\varphi )+\sin(\theta -\varphi ) \over 2}}
cos
θ
sin
φ
=
sin
(
θ
+
φ
)
−
sin
(
θ
−
φ
)
2
{\displaystyle \cos \theta \sin \varphi ={\sin(\theta +\varphi )-\sin(\theta -\varphi ) \over 2}}
Sum-to-product[ 22]
sin
θ
+
sin
φ
=
2
sin
(
θ
+
φ
2
)
cos
(
θ
−
φ
2
)
{\displaystyle \sin \theta +\sin \varphi =2\sin \left({\frac {\theta +\varphi }{2}}\right)\cos \left({\frac {\theta -\varphi }{2}}\right)}
cos
θ
+
cos
φ
=
2
cos
(
θ
+
φ
2
)
cos
(
θ
−
φ
2
)
{\displaystyle \cos \theta +\cos \varphi =2\cos \left({\frac {\theta +\varphi }{2}}\right)\cos \left({\frac {\theta -\varphi }{2}}\right)}
cos
θ
−
cos
φ
=
−
2
sin
(
θ
+
φ
2
)
sin
(
θ
−
φ
2
)
{\displaystyle \cos \theta -\cos \varphi =-2\sin \left({\theta +\varphi \over 2}\right)\sin \left({\theta -\varphi \over 2}\right)}
sin
θ
−
sin
φ
=
2
cos
(
θ
+
φ
2
)
sin
(
θ
−
φ
2
)
{\displaystyle \sin \theta -\sin \varphi =2\cos \left({\theta +\varphi \over 2}\right)\sin \left({\theta -\varphi \over 2}\right)\;}
If x , y , and z are the three angles of any triangle, or in other words
if
x
+
y
+
z
=
π
=
half circle,
{\displaystyle {\mbox{if }}x+y+z=\pi ={\mbox{half circle,}}\,}
then
tan
(
x
)
+
tan
(
y
)
+
tan
(
z
)
=
tan
(
x
)
tan
(
y
)
tan
(
z
)
.
{\displaystyle {\mbox{then }}\tan(x)+\tan(y)+\tan(z)=\tan(x)\tan(y)\tan(z).\,}
(If any of x , y , z is a right angle, one should take both sides to be ∞. This is neither +∞ nor −∞; for present purposes it makes sense to add just one point at infinity to the real line , that is approached by tan(θ) as tan(θ) either increases through positive values or decreases through negative values. This is a one-point compactification of the real line.)
If
x
+
y
+
z
=
π
=
half circle,
{\displaystyle {\mbox{If }}x+y+z=\pi ={\mbox{half circle,}}\,}
then
sin
(
2
x
)
+
sin
(
2
y
)
+
sin
(
2
z
)
=
4
sin
(
x
)
sin
(
y
)
sin
(
z
)
.
{\displaystyle {\mbox{then }}\sin(2x)+\sin(2y)+\sin(2z)=4\sin(x)\sin(y)\sin(z).\,}
If
w
+
x
+
y
+
z
=
π
=
half circle,
{\displaystyle {\mbox{If }}w+x+y+z=\pi ={\mbox{half circle,}}\,}
then
sin
(
w
+
x
)
sin
(
x
+
y
)
=
sin
(
x
+
y
)
sin
(
y
+
z
)
=
sin
(
y
+
z
)
sin
(
z
+
w
)
=
sin
(
z
+
w
)
sin
(
w
+
x
)
=
sin
(
w
)
sin
(
y
)
+
sin
(
x
)
sin
(
z
)
.
{\displaystyle {\begin{aligned}{\mbox{then }}&\sin(w+x)\sin(x+y)\\&{}=\sin(x+y)\sin(y+z)\\&{}=\sin(y+z)\sin(z+w)\\&{}=\sin(z+w)\sin(w+x)=\sin(w)\sin(y)+\sin(x)\sin(z).\end{aligned}}}
(The first three equalities are trivial; the fourth is the substance of this identity.) Essentially this is Ptolemy's theorem adapted to the language of trigonometry.
Linear combinations [ edit ]
For some purposes it is important to know that any linear combination of sine waves of the same period but different phase shifts is also a sine wave with the same period, but a different phase shift. In the case of a linear combination of a sine and cosine wave (which is just a sine wave with a phase shift of π/2), we have
a
sin
x
+
b
cos
x
=
a
2
+
b
2
⋅
sin
(
x
+
φ
)
{\displaystyle a\sin x+b\cos x={\sqrt {a^{2}+b^{2}}}\cdot \sin(x+\varphi )\,}
where
φ
=
arcsin
(
b
a
2
+
b
2
)
+
{
0
if
a
≥
0
,
π
if
a
<
0
,
{\displaystyle \varphi =\arcsin \left({\frac {b}{\sqrt {a^{2}+b^{2}}}}\right)+{\begin{cases}0&{\text{if }}a\geq 0,\\\pi &{\text{if }}a<0,\end{cases}}}
or equivalently
φ
=
arctan
(
b
a
)
+
{
0
if
a
≥
0
,
π
if
a
<
0.
{\displaystyle \varphi =\arctan \left({\frac {b}{a}}\right)+{\begin{cases}0&{\text{if }}a\geq 0,\\\pi &{\text{if }}a<0.\end{cases}}}
More generally, for an arbitrary phase shift, we have
a
sin
x
+
b
sin
(
x
+
α
)
=
c
sin
(
x
+
β
)
{\displaystyle a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\,}
where
c
=
a
2
+
b
2
+
2
a
b
cos
α
,
{\displaystyle c={\sqrt {a^{2}+b^{2}+2ab\cos \alpha }},\,}
and
β
=
arctan
(
b
sin
α
a
+
b
cos
α
)
.
{\displaystyle \beta =\arctan \left({\frac {b\sin \alpha }{a+b\cos \alpha }}\right).}
Other sums of trigonometric functions [ edit ]
Sum of sines and cosines with arguments in arithmetic progression:
sin
φ
+
sin
(
φ
+
α
)
+
sin
(
φ
+
2
α
)
+
⋯
+
sin
(
φ
+
n
α
)
=
sin
(
(
n
+
1
)
α
2
)
⋅
sin
(
φ
+
n
α
2
)
sin
α
2
.
{\displaystyle \sin {\varphi }+\sin {(\varphi +\alpha )}+\sin {(\varphi +2\alpha )}+\cdots +\sin {(\varphi +n\alpha )}={\frac {\sin {\left({\frac {(n+1)\alpha }{2}}\right)}\cdot \sin {(\varphi +{\frac {n\alpha }{2}})}}{\sin {\frac {\alpha }{2}}}}.}
cos
φ
+
cos
(
φ
+
α
)
+
cos
(
φ
+
2
α
)
+
⋯
+
cos
(
φ
+
n
α
)
=
sin
(
(
n
+
1
)
α
2
)
⋅
cos
(
φ
+
n
α
2
)
sin
α
2
.
{\displaystyle \cos {\varphi }+\cos {(\varphi +\alpha )}+\cos {(\varphi +2\alpha )}+\cdots +\cos {(\varphi +n\alpha )}={\frac {\sin {\left({\frac {(n+1)\alpha }{2}}\right)}\cdot \cos {(\varphi +{\frac {n\alpha }{2}})}}{\sin {\frac {\alpha }{2}}}}.}
For any a and b :
a
cos
(
x
)
+
b
sin
(
x
)
=
a
2
+
b
2
cos
(
x
−
arctan
(
b
,
a
)
)
{\displaystyle a\cos(x)+b\sin(x)={\sqrt {a^{2}+b^{2}}}\cos(x-\arctan(b,a))\;}
where arctan(y , x ) is the generalization of arctan(y /x ) which covers the entire circular range (see also the account of this same identity in "symmetry, periodicity, and shifts" above for this generalization of arctan).
tan
(
x
)
+
sec
(
x
)
=
tan
(
x
2
+
π
4
)
.
{\displaystyle \tan(x)+\sec(x)=\tan \left({x \over 2}+{\pi \over 4}\right).}
The above identity is sometimes convenient to know when thinking about the Gudermannian function , which relates the circular and hyperbolic trigonometric functions without resorting to complex numbers .
If x , y , and z are the three angles of any triangle, i.e. if x + y + z = π, then
cot
(
x
)
cot
(
y
)
+
cot
(
y
)
cot
(
z
)
+
cot
(
z
)
cot
(
x
)
=
1.
{\displaystyle \cot(x)\cot(y)+\cot(y)\cot(z)+\cot(z)\cot(x)=1.\,}
If ƒ (x ) is given by the linear fractional transformation
f
(
x
)
=
(
cos
α
)
x
−
sin
α
(
sin
α
)
x
+
cos
α
,
{\displaystyle f(x)={\frac {(\cos \alpha )x-\sin \alpha }{(\sin \alpha )x+\cos \alpha }},}
and similarly
g
(
x
)
=
(
cos
β
)
x
−
sin
β
(
sin
β
)
x
+
cos
β
,
{\displaystyle g(x)={\frac {(\cos \beta )x-\sin \beta }{(\sin \beta )x+\cos \beta }},}
then
f
(
g
(
x
)
)
=
g
(
f
(
x
)
)
=
(
cos
(
α
+
β
)
)
x
−
sin
(
α
+
β
)
(
sin
(
α
+
β
)
)
x
+
cos
(
α
+
β
)
.
{\displaystyle f(g(x))=g(f(x))={\frac {(\cos(\alpha +\beta ))x-\sin(\alpha +\beta )}{(\sin(\alpha +\beta ))x+\cos(\alpha +\beta )}}.}
More tersely stated, if for all α we let ƒ α be what we called ƒ above, then
f
α
∘
f
β
=
f
α
+
β
.
{\displaystyle f_{\alpha }\circ f_{\beta }=f_{\alpha +\beta }.\,}
If x is the slope of a line, then ƒ (x ) is the slope of its rotation through an angle of −α .
Inverse trigonometric functions [ edit ]
arcsin
(
x
)
+
arccos
(
x
)
=
π
/
2
{\displaystyle \arcsin(x)+\arccos(x)=\pi /2\;}
arctan
(
x
)
+
arccot
(
x
)
=
π
/
2.
{\displaystyle \arctan(x)+\operatorname {arccot}(x)=\pi /2.\;}
arctan
(
x
)
+
arctan
(
1
/
x
)
=
{
π
/
2
,
if
x
>
0
−
π
/
2
,
if
x
<
0
{\displaystyle \arctan(x)+\arctan(1/x)=\left\{{\begin{matrix}\pi /2,&{\mbox{if }}x>0\\-\pi /2,&{\mbox{if }}x<0\end{matrix}}\right.}
Compositions of trig and inverse trig functions [ edit ]
sin
[
arccos
(
x
)
]
=
1
−
x
2
{\displaystyle \sin[\arccos(x)]={\sqrt {1-x^{2}}}\,}
tan
[
arcsin
(
x
)
]
=
x
1
−
x
2
{\displaystyle \tan[\arcsin(x)]={\frac {x}{\sqrt {1-x^{2}}}}}
sin
[
arctan
(
x
)
]
=
x
1
+
x
2
{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {1+x^{2}}}}}
tan
[
arccos
(
x
)
]
=
1
−
x
2
x
{\displaystyle \tan[\arccos(x)]={\frac {\sqrt {1-x^{2}}}{x}}}
cos
[
arctan
(
x
)
]
=
1
1
+
x
2
{\displaystyle \cos[\arctan(x)]={\frac {1}{\sqrt {1+x^{2}}}}}
cot
[
arcsin
(
x
)
]
=
1
−
x
2
x
{\displaystyle \cot[\arcsin(x)]={\frac {\sqrt {1-x^{2}}}{x}}}
cos
[
arcsin
(
x
)
]
=
1
−
x
2
{\displaystyle \cos[\arcsin(x)]={\sqrt {1-x^{2}}}\,}
cot
[
arccos
(
x
)
]
=
x
1
−
x
2
{\displaystyle \cot[\arccos(x)]={\frac {x}{\sqrt {1-x^{2}}}}}
Relation to the complex exponential function [ edit ]
e
i
x
=
cos
(
x
)
+
i
sin
(
x
)
{\displaystyle e^{ix}=\cos(x)+i\sin(x)\,}
[ 23] (Euler's formula ),
e
−
i
x
=
cos
(
−
x
)
+
i
sin
(
−
x
)
=
cos
(
x
)
−
i
sin
(
x
)
{\displaystyle e^{-ix}=\cos(-x)+i\sin(-x)=\cos(x)-i\sin(x)\,}
e
i
π
=
−
1
{\displaystyle e^{i\pi }=-1\,}
cos
(
x
)
=
e
i
x
+
e
−
i
x
2
{\displaystyle \cos(x)={\frac {e^{ix}+e^{-ix}}{2}}\;}
[ 24]
sin
(
x
)
=
e
i
x
−
e
−
i
x
2
i
{\displaystyle \sin(x)={\frac {e^{ix}-e^{-ix}}{2i}}\;}
[ 25]
and hence the corollary:
tan
(
x
)
=
e
i
x
−
e
−
i
x
i
(
e
i
x
+
e
−
i
x
)
=
sin
(
x
)
cos
(
x
)
{\displaystyle \tan(x)={\frac {e^{ix}-e^{-ix}}{i({e^{ix}+e^{-ix}})}}\;={\frac {\sin(x)}{\cos(x)}}}
where
i
2
=
−
1
{\displaystyle i^{2}=-1}
.
For applications to special functions , the following infinite product formulæ for trigonometric functions are useful:[ 26] [ 27]
Identities without variables [ edit ]
The curious identity
cos
20
∘
⋅
cos
40
∘
⋅
cos
80
∘
=
1
8
{\displaystyle \cos 20^{\circ }\cdot \cos 40^{\circ }\cdot \cos 80^{\circ }={\frac {1}{8}}}
is a special case of an identity that contains one variable:
∏
j
=
0
k
−
1
cos
(
2
j
x
)
=
sin
(
2
k
x
)
2
k
sin
(
x
)
.
{\displaystyle \prod _{j=0}^{k-1}\cos(2^{j}x)={\frac {\sin(2^{k}x)}{2^{k}\sin(x)}}.}
A similar-looking identity is
cos
π
7
cos
2
π
7
cos
3
π
7
=
1
8
,
{\displaystyle \cos {\frac {\pi }{7}}\cos {\frac {2\pi }{7}}\cos {\frac {3\pi }{7}}={\frac {1}{8}},}
and in addition
sin
20
∘
⋅
sin
40
∘
⋅
sin
80
∘
=
3
8
.
{\displaystyle \sin 20^{\circ }\cdot \sin 40^{\circ }\cdot \sin 80^{\circ }={\frac {\sqrt {3}}{8}}.}
The following is perhaps not as readily generalized to an identity containing variables (but see explanation below):
cos
24
∘
+
cos
48
∘
+
cos
96
∘
+
cos
168
∘
=
1
2
.
{\displaystyle \cos 24^{\circ }+\cos 48^{\circ }+\cos 96^{\circ }+\cos 168^{\circ }={\frac {1}{2}}.}
Degree measure ceases to be more felicitous than radian measure when we consider this identity with 21 in the denominators:
cos
(
2
π
21
)
+
cos
(
2
⋅
2
π
21
)
+
cos
(
4
⋅
2
π
21
)
{\displaystyle \cos \left({\frac {2\pi }{21}}\right)\,+\,\cos \left(2\cdot {\frac {2\pi }{21}}\right)\,+\,\cos \left(4\cdot {\frac {2\pi }{21}}\right)}
+
cos
(
5
⋅
2
π
21
)
+
cos
(
8
⋅
2
π
21
)
+
cos
(
10
⋅
2
π
21
)
=
1
2
.
{\displaystyle \,+\,\cos \left(5\cdot {\frac {2\pi }{21}}\right)\,+\,\cos \left(8\cdot {\frac {2\pi }{21}}\right)\,+\,\cos \left(10\cdot {\frac {2\pi }{21}}\right)={\frac {1}{2}}.}
The factors 1, 2, 4, 5, 8, 10 may start to make the pattern clear: they are those integers less than 21/2 that are relatively prime to (or have no prime factors in common with) 21. The last several examples are corollaries of a basic fact about the irreducible cyclotomic polynomials : the cosines are the real parts of the zeroes of those polynomials; the sum of the zeroes is the Möbius function evaluated at (in the very last case above) 21; only half of the zeroes are present above. The two identities preceding this last one arise in the same fashion with 21 replaced by 10 and 15, respectively.
An efficient way to compute π is based on the following identity without variables, due to Machin :
π
4
=
4
arctan
1
5
−
arctan
1
239
{\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}}
or, alternatively, by using Euler 's formula:
π
4
=
5
arctan
1
7
+
2
arctan
3
79
.
{\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}.}
A useful mnemonic for certain values of sines and cosines [ edit ]
For certain simple angles, the sines and cosines take the form
n
/
2
{\displaystyle \scriptstyle {\sqrt {n}}/2}
for 0 ≤ n ≤ 4, which makes them easy to remember.
sin
0
=
sin
0
∘
=
0
/
2
=
cos
90
∘
=
cos
(
π
2
)
=
0
/
2
sin
(
π
6
)
=
sin
30
∘
=
1
/
2
=
cos
60
∘
=
cos
(
π
3
)
=
1
/
2
sin
(
π
4
)
=
sin
45
∘
=
2
/
2
=
cos
45
∘
=
cos
(
π
4
)
=
2
/
2
sin
(
π
3
)
=
sin
60
∘
=
3
/
2
=
cos
30
∘
=
cos
(
π
6
)
=
3
/
2
sin
(
π
2
)
=
sin
90
∘
=
4
/
2
=
cos
0
∘
=
cos
0
=
4
/
2
{\displaystyle {\begin{matrix}\sin 0&=&\sin 0^{\circ }&=&{\sqrt {0}}/2&=&\cos 90^{\circ }&=&\cos \left({\frac {\pi }{2}}\right)&={\sqrt {0}}/2\\\\\sin \left({\frac {\pi }{6}}\right)&=&\sin 30^{\circ }&=&{\sqrt {1}}/2&=&\cos 60^{\circ }&=&\cos \left({\frac {\pi }{3}}\right)&={\sqrt {1}}/2\\\\\sin \left({\frac {\pi }{4}}\right)&=&\sin 45^{\circ }&=&{\sqrt {2}}/2&=&\cos 45^{\circ }&=&\cos \left({\frac {\pi }{4}}\right)&={\sqrt {2}}/2\\\\\sin \left({\frac {\pi }{3}}\right)&=&\sin 60^{\circ }&=&{\sqrt {3}}/2&=&\cos 30^{\circ }&=&\cos \left({\frac {\pi }{6}}\right)&={\sqrt {3}}/2\\\\\sin \left({\frac {\pi }{2}}\right)&=&\sin 90^{\circ }&=&{\sqrt {4}}/2&=&\cos 0^{\circ }&=&\cos 0&={\sqrt {4}}/2\end{matrix}}}
Other interesting values [ edit ]
sin
π
7
=
7
6
−
7
189
∑
j
=
0
∞
(
3
j
+
1
)
!
189
j
j
!
(
2
j
+
2
)
!
{\displaystyle \sin {\frac {\pi }{7}}={\frac {\sqrt {7}}{6}}-{\frac {\sqrt {7}}{189}}\sum _{j=0}^{\infty }{\frac {(3j+1)!}{189^{j}j!\,(2j+2)!}}\!}
sin
π
18
=
1
6
∑
j
=
0
∞
(
3
j
)
!
27
j
j
!
(
2
j
+
1
)
!
{\displaystyle \sin {\frac {\pi }{18}}={\frac {1}{6}}\sum _{j=0}^{\infty }{\frac {(3j)!}{27^{j}j!\,(2j+1)!}}\!}
With the golden ratio φ:
cos
(
π
5
)
=
cos
36
∘
=
5
+
1
4
=
φ
/
2
{\displaystyle \cos \left({\frac {\pi }{5}}\right)=\cos 36^{\circ }={{\sqrt {5}}+1 \over 4}=\varphi /2}
sin
(
π
10
)
=
sin
18
∘
=
5
−
1
4
=
φ
−
1
2
=
1
2
φ
{\displaystyle \sin \left({\frac {\pi }{10}}\right)=\sin 18^{\circ }={{\sqrt {5}}-1 \over 4}={\varphi -1 \over 2}={1 \over 2\varphi }}
Also see exact trigonometric constants .
In calculus the relations stated below require angles to be measured in radians ; the relations would become more complicated if angles were measured in another unit such as degrees. If the trigonometric functions are defined in terms of geometry, their derivatives can be found by verifying two limits. The first is:
lim
x
→
0
sin
x
x
=
1
,
{\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1,}
verified using the unit circle and squeeze theorem . It may be tempting to propose to use L'Hôpital's rule to establish this limit. However, if one uses this limit in order to prove that the derivative of the sine is the cosine, and then uses the fact that the derivative of the sine is the cosine in applying L'Hôpital's rule, one is reasoning circularly—a logical fallacy. The second limit is:
lim
x
→
0
1
−
cos
x
x
=
0
,
{\displaystyle \lim _{x\rightarrow 0}{\frac {1-\cos x}{x}}=0,}
verified using the identity tan(x /2) = (1 − cos x )/sin x . Having established these two limits, one can use the limit definition of the derivative and the addition theorems to show that (sin x )′ = cos x and (cos x )′ = −sin x . If the sine and cosine functions are defined by their Taylor series , then the derivatives can be found by differentiating the power series term-by-term.
d
d
x
sin
x
=
cos
x
{\displaystyle {d \over dx}\sin x=\cos x}
The rest of the trigonometric functions can be differentiated using the above identities and the rules of differentiation :[ 28] [ 29] [ 30]
d
d
x
sin
x
=
cos
x
,
d
d
x
arcsin
x
=
1
1
−
x
2
d
d
x
cos
x
=
−
sin
x
,
d
d
x
arccos
x
=
−
1
1
−
x
2
d
d
x
tan
x
=
sec
2
x
,
d
d
x
arctan
x
=
1
1
+
x
2
d
d
x
cot
x
=
−
csc
2
x
,
d
d
x
arccot
x
=
−
1
1
+
x
2
d
d
x
sec
x
=
tan
x
sec
x
,
d
d
x
arcsec
x
=
1
|
x
|
x
2
−
1
d
d
x
csc
x
=
−
csc
x
cot
x
,
d
d
x
arccsc
x
=
−
1
|
x
|
x
2
−
1
{\displaystyle {\begin{matrix}{d \over dx}\sin x=&\cos x,&{d \over dx}\arcsin x=&{1 \over {\sqrt {1-x^{2}}}}\\\\{d \over dx}\cos x=&-\sin x,&{d \over dx}\arccos x=&{-1 \over {\sqrt {1-x^{2}}}}\\\\{d \over dx}\tan x=&\sec ^{2}x,&{d \over dx}\arctan x=&{1 \over 1+x^{2}}\\\\{d \over dx}\cot x=&-\csc ^{2}x,&{d \over dx}\operatorname {arccot} x=&{-1 \over 1+x^{2}}\\\\{d \over dx}\sec x=&\tan x\sec x,&{d \over dx}\operatorname {arcsec} x=&{1 \over |x|{\sqrt {x^{2}-1}}}\\\\{d \over dx}\csc x=&-\csc x\cot x,&{d \over dx}\operatorname {arccsc} x=&{-1 \over |x|{\sqrt {x^{2}-1}}}\end{matrix}}}
The integral identities can be found in "list of integrals of trigonometric functions ". Some generic forms are listed below.
∫
d
u
a
2
−
u
2
=
sin
−
1
(
u
a
)
+
C
{\displaystyle \int {\frac {du}{\sqrt {a^{2}-u^{2}}}}=\sin ^{-1}\left({\frac {u}{a}}\right)+C}
∫
d
u
a
2
+
u
2
=
1
a
tan
−
1
(
u
a
)
+
C
{\displaystyle \int {\frac {du}{a^{2}+u^{2}}}={\frac {1}{a}}\tan ^{-1}\left({\frac {u}{a}}\right)+C}
∫
d
u
u
u
2
−
a
2
=
1
a
sec
−
1
|
u
a
|
+
C
{\displaystyle \int {\frac {du}{u{\sqrt {u^{2}-a^{2}}}}}={\frac {1}{a}}\sec ^{-1}\left|{\frac {u}{a}}\right|+C}
The fact that the differentiation of trigonometric functions (sine and cosine) results in linear combinations of the same two functions is of fundamental importance to many fields of mathematics, including differential equations and fourier transformations .
Exponential definitions [ edit ]
Function
Inverse Function[ 31]
sin
θ
=
e
i
θ
−
e
−
i
θ
2
i
{\displaystyle \sin \theta ={\frac {e^{i\theta }-e^{-i\theta }}{2i}}\,}
arcsin
x
=
−
i
ln
(
i
x
+
1
−
x
2
)
{\displaystyle \arcsin x=-i\ln \left(ix+{\sqrt {1-x^{2}}}\right)\,}
cos
θ
=
e
i
θ
+
e
−
i
θ
2
{\displaystyle \cos \theta ={\frac {e^{i\theta }+e^{-i\theta }}{2}}\,}
arccos
x
=
−
i
ln
(
x
+
x
2
−
1
)
{\displaystyle \arccos x=-i\ln \left(x+{\sqrt {x^{2}-1}}\right)\,}
tan
θ
=
e
i
θ
−
e
−
i
θ
i
(
e
i
θ
+
e
−
i
θ
)
{\displaystyle \tan \theta ={\frac {e^{i\theta }-e^{-i\theta }}{i(e^{i\theta }+e^{-i\theta })}}\,}
arctan
x
=
i
ln
(
i
+
x
i
−
x
)
2
{\displaystyle \arctan x={\frac {i\ln \left({\frac {i+x}{i-x}}\right)}{2}}\,}
csc
θ
=
2
i
e
i
θ
−
e
−
i
θ
{\displaystyle \csc \theta ={\frac {2i}{e^{i\theta }-e^{-i\theta }}}\,}
arccsc
x
=
−
i
ln
(
i
x
+
1
−
1
x
2
)
{\displaystyle \operatorname {arccsc} x=-i\ln \left({\tfrac {i}{x}}+{\sqrt {1-{\tfrac {1}{x^{2}}}}}\right)\,}
sec
θ
=
2
e
i
θ
+
e
−
i
θ
{\displaystyle \sec \theta ={\frac {2}{e^{i\theta }+e^{-i\theta }}}\,}
arcsec
x
=
−
i
ln
(
1
x
+
1
−
i
x
2
)
{\displaystyle \operatorname {arcsec} x=-i\ln \left({\tfrac {1}{x}}+{\sqrt {1-{\tfrac {i}{x^{2}}}}}\right)\,}
cot
θ
=
i
(
e
i
θ
+
e
−
i
θ
)
e
i
θ
−
e
−
i
θ
{\displaystyle \cot \theta ={\frac {i(e^{i\theta }+e^{-i\theta })}{e^{i\theta }-e^{-i\theta }}}\,}
arccot
x
=
i
ln
(
i
−
x
i
+
x
)
2
{\displaystyle \operatorname {arccot} x={\frac {i\ln \left({\frac {i-x}{i+x}}\right)}{2}}\,}
cis
θ
=
e
i
θ
{\displaystyle \operatorname {cis} \,\theta =e^{i\theta }\,}
arccis
x
=
ln
x
i
{\displaystyle \operatorname {arccis} \,x={\frac {\ln x}{i}}\,}
The Dirichlet kernel Dn (x ) is the function occurring on both sides of the next identity:
1
+
2
cos
(
x
)
+
2
cos
(
2
x
)
+
2
cos
(
3
x
)
+
⋯
+
2
cos
(
n
x
)
=
sin
[
(
n
+
1
2
)
x
]
sin
(
x
2
)
.
{\displaystyle 1+2\cos(x)+2\cos(2x)+2\cos(3x)+\cdots +2\cos(nx)={\frac {\sin \left[\left(n+{\frac {1}{2}}\right)x\right\rbrack }{\sin \left({\frac {x}{2}}\right)}}.}
The convolution of any integrable function of period 2π with the Dirichlet kernel coincides with the function's n th-degree Fourier approximation. The same holds for any measure or generalized function .
If we set
t
=
tan
(
x
2
)
,
{\displaystyle t=\tan \left({\frac {x}{2}}\right),}
then[ 32]
sin
(
x
)
=
2
t
1
+
t
2
and
cos
(
x
)
=
1
−
t
2
1
+
t
2
and
e
i
x
=
1
+
i
t
1
−
i
t
.
{\displaystyle \sin(x)={\frac {2t}{1+t^{2}}}{\text{ and }}\cos(x)={\frac {1-t^{2}}{1+t^{2}}}{\text{ and }}e^{ix}={\frac {1+it}{1-it}}.}
where e ix is the same as cis(x ).
This substitution of t for tan(x /2), with the consequent replacement of sin(x ) by 2t /(1 + t 2 ) and cos(x ) by (1 − t 2 )/(1 + t 2 ) is useful in calculus for converting rational functions in sin(x ) and cos(x ) to functions of t in order to find their antiderivatives. For more information see tangent half-angle formula .
^ Abramowitz and Stegun, p. 72, 4.3.10
^ a b Weisstein, Eric W. "Trigonometry" . MathWorld .
^ Abramowitz and Stegun, p. 72, 4.3.3
^ Abramowitz and Stegun, p. 73, 4.3.45
^ Abramowitz and Stegun, p. 78, 4.3.147
^ Abramowitz and Stegun, p. 72, 4.3.13–15
^ The Elementary Identities
^ Abramowitz and Stegun, p. 72, 4.3.9
^ Abramowitz and Stegun, p. 72, 4.3.7–8
^ Abramowitz and Stegun, p. 72, 4.3.16
^ a b c Weisstein, Eric W. "Trigonometric Addition Formulas" . MathWorld .
^ Abramowitz and Stegun, p. 72, 4.3.17
^ Abramowitz and Stegun, p. 72, 4.3.18
^ a b Weisstein, Eric W. "Multiple-Angle Formulas" . MathWorld .
^ Abramowitz and Stegun, p. 74, 4.3.48
^ Abramowitz and Stegun, p. 72, 4.3.24–26
^ Weisstein, Eric W. "Double-Angle Formulas" . MathWorld .
^ Abramowitz and Stegun, p. 72, 4.3.27–28
^ Abramowitz and Stegun, p. 72, 4.3.20–22
^ Weisstein, Eric W. "Half-Angle Formulas" . MathWorld .
^ Abramowitz and Stegun, p. 72, 4.3.31–33
^ Abramowitz and Stegun, p. 72, 4.3.34–39
^ Abramowitz and Stegun, p. 74, 4.3.47
^ Abramowitz and Stegun, p. 71, 4.3.2
^ Abramowitz and Stegun, p. 71, 4.3.1
^ Abramowitz and Stegun, p. 75, 4.3.89–90
^ Abramowitz and Stegun, p. 85, 4.5.68–69
^ Abramowitz and Stegun, p. 77, 4.3.105–110
^ Abramowitz and Stegun, p. 82, 4.4.52–57
^ Finney, Ross (2003). Calculus : Graphical, Numerical, Algebraic . Glenview, Illinois: Prentice Hall. pp. 159–161. ISBN 0-13-063131-0 .
^ Abramowitz and Stegun, p. 80, 4.4.26–31
^ Abramowitz and Stegun, p. 72, 4.3.23